SAT Inequalities Made Easy: The Sign Flip Trick + Hard Questions

 

╔══════════════════════════════════════════════╗
║ 𝐒𝐀𝐓 𝐌𝐀𝐓𝐇 ║
║ 𝐋𝐈𝐍𝐄𝐀𝐑 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒 ║
║ 𝐓𝐡𝐞 𝐒𝐢𝐦𝐩𝐥𝐞 𝐌𝐞𝐭𝐡𝐨𝐝 𝐓𝐡𝐚𝐭 𝐏𝐫𝐞𝐯𝐞𝐧𝐭𝐬 𝐒𝐈𝐋𝐋𝐘 𝐌𝐢𝐬𝐭𝐚𝐤𝐞𝐬 ║
╚══════════════════════════════════════════════╝

𝐖𝐡𝐲 𝐝𝐨 𝐒𝐀𝐓 𝐢𝐧𝐞𝐪𝐮𝐚𝐥𝐢𝐭𝐲 𝐪𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬 𝐜𝐚𝐭𝐜𝐡 𝐬𝐭𝐮𝐝𝐞𝐧𝐭𝐬?

Because they look almost exactly like equations.

You see:

𝟑𝐱 + 𝟒 = 𝟏𝟗

and you know what to do.

Then the SAT changes one symbol:

𝟑𝐱 + 𝟒 ≥ 𝟏𝟗

Now you are no longer looking for just one answer.

You are looking for a whole collection of values.

That is the central idea behind inequalities:

╭──────────────────────────────╮
│ 𝐄𝐐𝐔𝐀𝐓𝐈𝐎𝐍 → 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞 │
│ 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 → 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 │
╰──────────────────────────────╯

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏 — 𝐊𝐍𝐎𝐖 𝐓𝐇𝐄 𝐅𝐎𝐔𝐑 𝐒𝐘𝐌𝐁𝐎𝐋𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

There are four basic inequality signs:

𝐱 < 𝟕
→ x is less than 7

𝐱 > 𝟕
→ x is greater than 7

𝐱 ≤ 𝟕
→ x is less than or equal to 7

𝐱 ≥ 𝟕
→ x is greater than or equal to 7

The tiny horizontal line underneath the symbol is important.

It means:

𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐒 𝐀𝐋𝐋𝐎𝐖𝐄𝐃.

So:

𝐱 < 𝟓

does NOT include 5.

But:

𝐱 ≤ 𝟓

DOES include 5.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐 — 𝐓𝐇𝐄 𝐑𝐔𝐋𝐄 𝐘𝐎𝐔 𝐌𝐔𝐒𝐓 𝐍𝐎𝐓 𝐅𝐎𝐑𝐆𝐄𝐓
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Here is the most important rule in this entire guide:

╔══════════════════════════════════╗
║ 𝐌𝐔𝐋𝐓𝐈𝐏𝐋𝐘 𝐎𝐑 𝐃𝐈𝐕𝐈𝐃𝐄 𝐁𝐘 𝐀 ║
║ 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 𝐍𝐔𝐌𝐁𝐄𝐑? ║
║ ║
║ 𝐅𝐋𝐈𝐏 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍! ║
╚══════════════════════════════════╝

For example:

−𝟐𝐱 > 𝟏𝟎

Divide by −2.

Because −2 is negative:

𝐱 < −𝟓

Notice what happened:

became <

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟑 — 𝐖𝐇𝐄𝐍 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍 𝐃𝐎𝐄𝐒 𝐍𝐎𝐓 𝐅𝐋𝐈𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Do not flip the sign every time you move something.

For example:

𝟓𝐱 − 𝟑 ≤ 𝟏𝟕

Add 3:

𝟓𝐱 ≤ 𝟐𝟎

Divide by +5:

𝐱 ≤ 𝟒

Nothing flips because 5 is positive.

A useful mental test is:

𝐏𝐎𝐒𝐈𝐓𝐈𝐕𝐄 → 𝐒𝐓𝐀𝐘

𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 → 𝐅𝐋𝐈𝐏

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟒 — 𝐓𝐇𝐄 𝐒𝐀𝐓 𝐖𝐀𝐘 𝐓𝐎 𝐒𝐎𝐋𝐕𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Treat an inequality almost like an equation.

Example:

𝟒𝐱 + 𝟕 > 𝟐𝟑

Subtract 7:

𝟒𝐱 > 𝟏𝟔

Divide by 4:

𝐱 > 𝟒

That's it.

But always perform one final check:

𝐃𝐢𝐝 𝐈 𝐝𝐢𝐯𝐢𝐝𝐞 𝐛𝐲 𝐚 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞?

If no, the sign remains unchanged.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟓 — 𝐓𝐇𝐄 𝐓𝐖𝐎-𝐒𝐈𝐃𝐄𝐃 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Sometimes the SAT gives you a sandwich:

𝟐 < 𝐱 + 𝟓 ≤ 𝟏𝟏

Subtract 5 from ALL THREE parts:

𝟐 − 𝟓 < 𝐱 ≤ 𝟏𝟏 − 𝟓

Therefore:

−𝟑 < 𝐱 ≤ 𝟔

The answer contains every number between −3 and 6, except −3 itself.

But 6 IS included.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟔 — 𝐓𝐇𝐄 𝐍𝐔𝐌𝐁𝐄𝐑 𝐋𝐈𝐍𝐄 𝐂𝐎𝐃𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A number-line question can often be solved almost instantly.

𝐎𝐏𝐄𝐍 𝐂𝐈𝐑𝐂𝐋𝐄
→ endpoint NOT included

𝐂𝐋𝐎𝐒𝐄𝐃 𝐂𝐈𝐑𝐂𝐋𝐄
→ endpoint included

So:

𝐱 > 𝟐

means:

○──────→
𝟐

while:

𝐱 ≥ 𝟐

means:

●──────→
𝟐

And direction matters:

←──────○
𝟐

means:

𝐱 < 𝟐

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟕 — 𝐓𝐇𝐄 𝐐𝐔𝐈𝐂𝐊 𝐆𝐑𝐀𝐏𝐇 𝐂𝐇𝐄𝐂𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Whenever you see a number-line graph, ask two questions:

𝐐𝟏. 𝐈𝐬 𝐭𝐡𝐞 𝐞𝐧𝐝𝐩𝐨𝐢𝐧𝐭 𝐨𝐩𝐞𝐧 𝐨𝐫 𝐜𝐥𝐨𝐬𝐞𝐝?

𝐐𝟐. 𝐖𝐡𝐢𝐜𝐡 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐢𝐬 𝐬𝐡𝐚𝐝𝐞𝐝?

That gives you the inequality.

You do not need to guess.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟖 — 𝐓𝐇𝐄 𝐖𝐎𝐑𝐃𝐒 𝐇𝐈𝐃𝐈𝐍𝐆 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

SAT word problems often hide the inequality symbol inside ordinary English.

Memorize these translations:

𝐀𝐓 𝐋𝐄𝐀𝐒𝐓
→ ≥

𝐀𝐓 𝐌𝐎𝐒𝐓
→ ≤

𝐌𝐎𝐑𝐄 𝐓𝐇𝐀𝐍
→ >

𝐋𝐄𝐒𝐒 𝐓𝐇𝐀𝐍
→ <

𝐍𝐎 𝐌𝐎𝐑𝐄 𝐓𝐇𝐀𝐍
→ ≤

𝐍𝐎 𝐋𝐄𝐒𝐒 𝐓𝐇𝐀𝐍
→ ≥

𝐆𝐑𝐄𝐀𝐓𝐄𝐑 𝐓𝐇𝐀𝐍
→ >

𝐅𝐄𝐖𝐄𝐑 𝐓𝐇𝐀𝐍
→ <

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟗 — 𝐓𝐇𝐄 “𝐀𝐓 𝐋𝐄𝐀𝐒𝐓” 𝐓𝐑𝐀𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose a problem says:

“A score of at least 80 is required.”

At least means 80 is acceptable.

Therefore:

𝐱 ≥ 𝟖𝟎

Not:

𝐱 > 𝟖𝟎

This tiny difference can decide the entire answer.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟎 — 𝐓𝐇𝐄 “𝐀𝐓 𝐌𝐎𝐒𝐓” 𝐓𝐑𝐀𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

“At most 25” means 25 is allowed.

Therefore:

𝐱 ≤ 𝟐𝟓

Compare:

“less than 25”

𝐱 < 𝟐𝟓

One word changes the mathematics.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟏 — 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒 𝐈𝐍 𝐓𝐖𝐎 𝐕𝐀𝐑𝐈𝐀𝐁𝐋𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now the SAT can move from a number line to a coordinate plane.

Consider:

𝐲 > 𝟐𝐱 + 𝟏

First draw the boundary:

𝐲 = 𝟐𝐱 + 𝟏

Then determine which side belongs to the solution.

Because the inequality is:

𝐲 > ...

the solution is above the boundary.

Because equality is NOT included, the boundary is dashed.

So remember:

𝐲 > 𝐟(𝐱)
→ above + dashed

𝐲 < 𝐟(𝐱)
→ below + dashed

𝐲 ≥ 𝐟(𝐱)
→ above + solid

𝐲 ≤ 𝐟(𝐱)
→ below + solid

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟐 — 𝐓𝐇𝐄 𝐓𝐄𝐒𝐓-𝐀-𝐏𝐎𝐈𝐍𝐓 𝐌𝐄𝐓𝐇𝐎𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

If you are unsure which side of a boundary is correct, test a point.

Suppose:

𝐲 > 𝐱 + 𝟐

Try the point:

(𝟎,𝟎)

Substitute:

𝟎 > 𝟎 + 𝟐

That becomes:

𝟎 > 𝟐

False.

Therefore, the side containing (0,0) is NOT the solution.

This method is particularly useful when a graph is unfamiliar.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟑 — 𝐒𝐘𝐒𝐓𝐄𝐌𝐒 𝐎𝐅 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose:

𝐱 ≥ 𝟑

and

𝐱 < 𝟖

Both must be true.

Therefore:

𝟑 ≤ 𝐱 < 𝟖

Think of this as finding the common region.

𝐀𝐍𝐃 = 𝐎𝐕𝐄𝐑𝐋𝐀𝐏

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟒 — 𝐖𝐇𝐀𝐓 “𝐎𝐑” 𝐌𝐄𝐀𝐍𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now consider:

𝐱 < −𝟒

OR

𝐱 > 𝟑

These are two separate possibilities.

The solution is:

𝐱 < −𝟒 𝐎𝐑 𝐱 > 𝟑

Do not search for one continuous interval.

Remember:

𝐀𝐍𝐃 → intersection / overlap

𝐎𝐑 → either possibility

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟓 — 𝐀 𝐑𝐄𝐀𝐋 𝐖𝐎𝐑𝐋𝐃 𝐒𝐀𝐓 𝐌𝐎𝐃𝐄𝐋
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Imagine a student has $50.

A ticket costs $12 and each additional item costs $4.

If x represents the number of additional items, the total must not exceed $50.

Write:

𝟏𝟐 + 𝟒𝐱 ≤ 𝟓𝟎

Subtract 12:

𝟒𝐱 ≤ 𝟑𝟖

Divide:

𝐱 ≤ 𝟗.𝟓

But x represents a number of items.

You cannot buy half an item.

Therefore the greatest possible whole-number value is:

𝐱 = 𝟗

This illustrates an important SAT habit:

𝐀𝐋𝐆𝐄𝐁𝐑𝐀 𝐀𝐍𝐒𝐖𝐄𝐑 ≠ 𝐀𝐋𝐖𝐀𝐘𝐒 𝐅𝐈𝐍𝐀𝐋 𝐖𝐎𝐑𝐃

The context matters.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟔 — 𝐀𝐍𝐎𝐓𝐇𝐄𝐑 𝐖𝐎𝐑𝐃 𝐏𝐑𝐎𝐁𝐋𝐄𝐌
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A gym charges $20 to join and $8 per month.

A student can spend no more than $68.

How many months can the student afford?

Let x = number of months.

Write:

𝟐𝟎 + 𝟖𝐱 ≤ 𝟔𝟖

Subtract 20:

𝟖𝐱 ≤ 𝟒𝟖

Divide:

𝐱 ≤ 𝟔

Therefore:

𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐦𝐨𝐧𝐭𝐡𝐬 = 𝟔

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟕 — 𝐓𝐇𝐄 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 𝐓𝐑𝐀𝐏 𝐑𝐄𝐕𝐈𝐒𝐈𝐓𝐄𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Solve:

𝟕 − 𝟑𝐱 ≥ 𝟏𝟔

Subtract 7:

−𝟑𝐱 ≥ 𝟗

Now divide by −3.

𝐒𝐓𝐎𝐏.

This is the danger point.

The sign must reverse:

𝐱 ≤ −𝟑

A useful habit:

Whenever the coefficient of x becomes negative immediately before division, mentally say:

“𝐅𝐋𝐈𝐏.”

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟖 — 𝐖𝐇𝐘 𝐃𝐎𝐄𝐒 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍 𝐅𝐋𝐈𝐏?
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

This is not an arbitrary SAT rule.

Take:

𝟐 < 𝟓

This is true.

Multiply both sides by −1:

−𝟐 > −𝟓

The order on the number line has reversed.

That is why:

< becomes >

and

becomes <

when multiplying or dividing by a negative number.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟗 — 𝐀 𝐅𝐀𝐒𝐓𝐄𝐑 𝐖𝐀𝐘 𝐓𝐎 𝐓𝐇𝐈𝐍𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Instead of memorizing dozens of separate rules, remember this chain:

╔═══════════════════════════════╗
║ 𝟏. 𝐓𝐑𝐀𝐍𝐒𝐋𝐀𝐓𝐄 ║
║ 𝟐. 𝐒𝐈𝐌𝐏𝐋𝐈𝐅𝐘 ║
║ 𝟑. 𝐒𝐎𝐋𝐕𝐄 ║
║ 𝟒. 𝐂𝐇𝐄𝐂𝐊 𝐅𝐎𝐑 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 ║
║ 𝟓. 𝐂𝐇𝐄𝐂𝐊 𝐓𝐇𝐄 𝐂𝐎𝐍𝐓𝐄𝐗𝐓 ║
╚═══════════════════════════════╝

This is much safer than trying to solve everything mentally.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟎 — 𝐓𝐇𝐄 𝐌𝐈𝐍𝐈 𝐒𝐀𝐓 𝐂𝐇𝐀𝐋𝐋𝐄𝐍𝐆𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

𝐐𝟏.

Solve:

𝟔𝐱 − 𝟓 > 𝟏𝟗

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

𝟔𝐱 > 𝟐𝟒

𝐱 > 𝟒

𝐐𝟐.

Solve:

−𝟓𝐱 + 𝟏𝟎 ≤ 𝟑𝟎

Subtract 10:

−𝟓𝐱 ≤ 𝟐𝟎

Divide by −5 and flip:

𝐱 ≥ −𝟒

𝐐𝟑.

Solve:

𝟑 ≤ 𝟐𝐱 + 𝟏 < 𝟏𝟏

Subtract 1:

𝟐 ≤ 𝟐𝐱 < 𝟏𝟎

Divide by 2:

𝟏 ≤ 𝐱 < 𝟓

𝐐𝟒.

A quantity must be no greater than 75.

Which inequality represents the statement?

𝐱 ≤ 𝟕𝟓

𝐐𝟓.

Which value satisfies:

𝐱 > −𝟐?

A. −𝟓
B. −𝟑
C. −𝟐
D. 𝟎

Answer:

𝐃. 𝟎

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟏 — 𝐓𝐇𝐄 𝐅𝐈𝐕𝐄-𝐒𝐄𝐂𝐎𝐍𝐃 𝐅𝐈𝐍𝐀𝐋 𝐂𝐇𝐄𝐂𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Before submitting an inequality answer, run this mental checklist:

✓ Did I translate the words correctly?

✓ Did I distribute brackets correctly?

✓ Did I isolate x?

✓ Did I multiply or divide by a negative?

✓ If yes, did I reverse the sign?

✓ Is the endpoint included?

✓ Does the answer make sense in the real-world situation?

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟐 — 𝐓𝐇𝐄 𝐔𝐋𝐓𝐈𝐌𝐀𝐓𝐄 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐌𝐄𝐌𝐎𝐑𝐘 𝐂𝐀𝐑𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

╭────────────────────────────────╮
│ < → 𝐋𝐄𝐒𝐒 │
│ > → 𝐌𝐎𝐑𝐄 │
│ ≤ → 𝐋𝐄𝐒𝐒 𝐎𝐑 𝐄𝐐𝐔𝐀𝐋 │
│ ≥ → 𝐌𝐎𝐑𝐄 𝐎𝐑 𝐄𝐐𝐔𝐀𝐋 │
│ │
│ 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 → 𝐅𝐋𝐈𝐏 │
│ 𝐏𝐎𝐒𝐈𝐓𝐈𝐕𝐄 → 𝐒𝐓𝐀𝐘 │
│ │
│ 𝐎𝐏𝐄𝐍 → 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐄𝐗𝐂𝐋𝐔𝐃𝐄𝐃 │
│ 𝐂𝐋𝐎𝐒𝐄𝐃 → 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐍𝐂𝐋𝐔𝐃𝐄𝐃│
│ │
│ 𝐀𝐍𝐃 → 𝐎𝐕𝐄𝐑𝐋𝐀𝐏 │
│ 𝐎𝐑 → 𝐄𝐈𝐓𝐇𝐄𝐑 𝐏𝐎𝐒𝐒𝐈𝐁𝐈𝐋𝐈𝐓𝐘 │
╰────────────────────────────────╯

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐅𝐈𝐍𝐀𝐋 𝐒𝐀𝐓 𝐓𝐀𝐊𝐄𝐀𝐖𝐀𝐘
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Linear inequalities are not really about memorizing a large collection of formulas.

They are about controlling one idea:

𝐖𝐇𝐈𝐂𝐇 𝐕𝐀𝐋𝐔𝐄𝐒 𝐀𝐑𝐄 𝐀𝐋𝐋𝐎𝐖𝐄𝐃?

Once you see the question that way, the symbols become easier.

If the SAT says:

“at least”

think:

≥

If it says:

“at most”

think:

≤

If you divide by a negative:

𝐅𝐋𝐈𝐏 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍.

If a graph is involved:

𝐎𝐏𝐄𝐍 = 𝐍𝐎 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘

𝐂𝐋𝐎𝐒𝐄𝐃 = 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐍𝐂𝐋𝐔𝐃𝐄𝐃

And when a word problem produces a mathematical answer, always return to the original situation.

That final step is where many avoidable SAT mistakes disappear.

𝐓𝐡𝐞 𝐛𝐞𝐬𝐭 𝐢𝐧𝐞𝐪𝐮𝐚𝐥𝐢𝐭𝐲 𝐬𝐭𝐫𝐚𝐭𝐞𝐠𝐲 𝐢𝐬:

╔════════════════════════════════════╗
║ 𝐓𝐑𝐀𝐍𝐒𝐋𝐀𝐓𝐄 → 𝐒𝐎𝐋𝐕𝐄 → 𝐅𝐋𝐈𝐏 ║
║ → 𝐂𝐇𝐄𝐂𝐊 → 𝐈𝐍𝐓𝐄𝐑𝐏𝐑𝐄𝐓 ║
╚════════════════════════════════════╝

Master that sequence and a large class of SAT inequality questions becomes much more predictable.

other pages to explore

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 

ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]

SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]

SYSTEM OF EQUATIONS  [ PART  I ]

system of equations part I

SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

SAT system of equations 2

 PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS

to find the parameter, unique solution, no solution infinite number of solutions

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 

Linear Inequality

linear inequalities

PERCENTAGES

PERCENTAGES [introduction]

PERCENTAGES [part 1]

PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]

SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions

geometry notes  [ part 1 ] 

SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster

SAT Word Problems Made Easy

The Big Word-Problem Families:  SAT Math Guide

Big Word-Problem Families

How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier

How SAT word problems hide equations

SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

SAT Word Problems Made Easy

SAT Function Transformations: A Simple Way to Master Shifts, Reflections, Stretches and Compressions

SAT function transformations tips and tricks

SAT Probability Made Easy: Conditional Probability, Tables & Tricky Questions

SAT probability guide

SAT Conditional Probability

Conditional Probability

SAT Conditional Probability Trick: Master “Given That” Questions Fast

 

🎯 SAT Conditional Probability: The Hidden “Given That” Trick

Conditional probability looks complicated on the SAT because the question often hides the mathematics inside a table, survey, experiment, or real-life situation.

But the core idea is surprisingly simple:

When the question says “given that,” your universe becomes smaller.

That one idea can turn a difficult-looking SAT probability question into a short calculation.

---

🔑 1. The Basic Conditional Probability Formula

If the question asks for the probability of A given B, write:

P(A | B) = P(A and B) ÷ P(B)

The vertical bar | means:

“given that”

So:

P(A | B)

means:

“the probability of A, given that B has already happened.”

The most important part is the denominator:

⭐ The denominator is the condition.

If you see:

P(A | B)

start by asking:

“How many outcomes satisfy B?”

That becomes your new total.

---

🧠 2. The SAT Shortcut

Suppose a survey contains 200 students.

• 120 study mathematics
• 80 study physics
• 50 study both mathematics and physics

What is the probability that a randomly selected student studies mathematics given that the student studies physics?

The phrase “given that the student studies physics” changes the problem.

You are no longer choosing from all 200 students.

You are choosing only from the 80 students who study physics.

Among those 80 students, 50 also study mathematics.

Therefore:

P(Math | Physics) = 50 ÷ 80

= 5 ÷ 8

= 0.625

= 62.5%

🚨 SAT trap:

A common mistake is:

50 ÷ 200

That would answer a different question:

What percentage of ALL students study both subjects?

The SAT is testing whether you notice the words:

“given that”

---

📊 3. Two-Way Tables Make Conditional Probability Easier

Many SAT questions present information in a table.

Consider this example:

	Uses Calculator	Does Not Use Calculator	Total
Group A	36	24	60
Group B	28	12	40
Total	64	36	100

Suppose a student is selected from those who use a calculator.

What is the probability that the student belongs to Group A?

The condition is:

Uses Calculator

So the denominator is:

64

The favorable outcomes are Group A students who use a calculator:

36

Therefore:

P(Group A | Uses Calculator)

= 36 ÷ 64

= 9 ÷ 16

= 0.5625

So the answer is:

56.25%

---

⚠️ 4. The Denominator Test

Whenever you see a conditional probability problem, perform this three-second test:

Step ① Find the words after “given that.”

Step ② Find the total number belonging to that condition.

Step ③ Divide the desired intersection by that total.

For example:

P(A | B)

means:

Desired = A and B

Total = B

Therefore:

P(A | B) = (A and B) ÷ B

This is one of the most useful patterns to recognize on the SAT.

---

🔥 5. “And” vs “Given”

These two expressions look similar but mean very different things.

P(A and B)

asks for the probability that both events happen.

P(A | B)

asks for the probability that A happens among the cases where B is already known to happen.

For example, suppose 100 students are surveyed.

• 40 play basketball
• 30 play soccer
• 15 play both

Then:

P(Basketball and Soccer)

= 15 ÷ 100

= 15%

But:

P(Basketball | Soccer)

= 15 ÷ 30

= 50%

Same intersection.

Different denominator.

That is the entire trick.

---

🎯 6. A SAT-Style Example

A school surveys 300 students about whether they participate in music or sports.

The results are:

• 180 participate in sports
• 120 participate in music
• 75 participate in both

A student who participates in music is selected at random.

What is the probability that the student also participates in sports?

The phrase:

“A student who participates in music is selected”

creates the condition.

Therefore, the total possible students are:

120

The students satisfying both conditions are:

75

So:

P(Sports | Music)

= 75 ÷ 120

= 5 ÷ 8

= 0.625

Answer:

62.5%

---

🧩 7. Watch for “Among”

The SAT may avoid the words “given that” and use another phrase.

Watch for:

among

of those who

for students who

from the group that

if the selected student is known to

All of these can signal a restricted sample.

For example:

Among students who own a bicycle, 18 out of 30 ride to school.

The probability that a randomly selected bicycle owner rides to school is:

18 ÷ 30 = 60%

The denominator is 30, not the total number of students in the school.

---

📐 8. Conditional Probability From a Formula

Sometimes the SAT gives probabilities instead of counts.

Suppose:

P(A) = 0.40

P(B) = 0.50

and

P(A and B) = 0.20

Find:

P(A | B)

Use:

P(A | B) = P(A and B) ÷ P(B)

Therefore:

P(A | B) = 0.20 ÷ 0.50

= 0.40

Answer:

40%

Notice that you don't need to create a table.

The same idea works with probabilities, percentages, fractions, or counts.

---

💡 9. The Reverse Conditional Probability Trap

This is an especially important SAT idea.

In general:

P(A | B) ≠ P(B | A)

For example:

Suppose:

• 60 students play soccer
• 40 students play basketball
• 20 play both

Then:

P(Soccer | Basketball)

= 20 ÷ 40

= 50%

But:

P(Basketball | Soccer)

= 20 ÷ 60

= 33⅓%

The numerator is the same.

The denominator changes.

⭐ Remember:

The condition controls the denominator.

---

🧮 10. Turning Percentages Into Counts

SAT questions sometimes give percentages rather than actual numbers.

Suppose 40% of students own a tablet.

Among students who own a tablet, 75% also own a laptop.

What percentage of ALL students own both?

Imagine there are 100 students.

Tablet owners:

40

Of those 40, 75% own a laptop:

0.75 × 40 = 30

Therefore:

30%

of all students own both.

This gives a useful relationship:

P(A and B) = P(B) × P(A | B)

So:

P(A and B) = 0.40 × 0.75

= 0.30

= 30%

---

🚨 11. A Classic SAT Mistake

Suppose:

P(A) = 60%

and

P(B | A) = 25%

A student might incorrectly say:

P(A and B) = 60% + 25%

That is wrong.

The 25% applies only to the group A.

Use:

P(A and B) = P(A) × P(B | A)

Therefore:

= 0.60 × 0.25

= 0.15

So:

15%

of the entire population belongs to both groups.

---

📈 12. Conditional Probability and Tables

A table can often be converted directly into a probability.

Suppose:

	Passed	Did Not Pass	Total
Studied	72	18	90
Did Not Study	24	36	60
Total	96	54	150

Question:

What is the probability that a student studied, given that the student passed?

The condition is:

Passed

So use the Passed column.

Total who passed:

96

Passed and studied:

72

Therefore:

P(Studied | Passed)

= 72 ÷ 96

= 3 ÷ 4

= 75%

---

🧠 13. The “Shrink the Universe” Method

Here is a powerful way to think about every conditional probability problem.

Imagine that the entire group is a large circle.

When the question says:

“given that B”

you throw away everything outside B.

Now your entire universe is:

B

Then ask:

How much of B is also A?

That gives:

A ∩ B ÷ B

or:

P(A | B) = P(A ∩ B) ÷ P(B)

This mental picture is often easier than memorizing a formula.

---

🎯 14. SAT Challenge Question

A survey of 400 students found:

• 240 students use a particular study app.
• 160 students use a particular online course.
• 100 students use both.

If a student who uses the online course is selected at random, what is the probability that the student also uses the study app?

Step 1: Identify the condition.

The student uses the online course.

So the denominator is:

160

Step 2: Find the intersection.

Both:

100

Step 3: Divide.

100 ÷ 160

= 5 ÷ 8

= 0.625

Answer:

62.5%

---

🔍 15. The Fastest Way to Solve These Questions

When you see a conditional probability question, don't immediately calculate.

First write:

Condition = ______

Then:

Total in condition = ______

Then:

Favorable within condition = ______

Finally:

Probability = favorable ÷ condition

For:

P(A | B)

write:

Condition → B

Total → B

Favorable → A ∩ B

Answer → (A ∩ B) ÷ B

This prevents one of the most common denominator errors.

---

🏆 16. What to Remember on Test Day

You do not need a complicated strategy.

Remember these five rules:

① “Given that” means the sample space changes.

② The condition becomes the denominator.

③ “And” usually points toward the intersection.

④ P(A | B) and P(B | A) are usually different.

⑤ For a two-way table, locate the condition's row or column first.

---

⚡ The 10-Second SAT Method

When the clock is running:

GIVEN → DENOMINATOR

AND → INTERSECTION

CONDITIONAL → SHRINK THE GROUP

PROBABILITY → FAVORABLE ÷ TOTAL

So if you see:

P(A | B)

think immediately:

“Within B, how many are also A?”

That single question can solve many conditional probability problems quickly.

---

📌 Final Takeaway

Conditional probability is not really about complicated probability.

It is about choosing the correct universe.

The SAT may disguise the condition inside:

• a survey
• a two-way table
• percentages
• a real-world situation
• a probability statement
• or a sentence containing “among”

But the underlying structure remains:

P(A | B) = P(A and B) ÷ P(B)

And the most important memory trick is:

⭐ GIVEN = DENOMINATOR ⭐

Once that becomes automatic, many SAT probability questions become much shorter.

explore more pages

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 

ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]

SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]

SYSTEM OF EQUATIONS  [ PART  I ]

system of equations part I

SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

SAT system of equations 2

 PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS

to find the parameter, unique solution, no solution infinite number of solutions

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 

PERCENTAGES

PERCENTAGES [introduction]

PERCENTAGES [part 1]

PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]

SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions

geometry notes  [ part 1 ] 

SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster

SAT Word Problems Made Easy

The Big Word-Problem Families:  SAT Math Guide

Big Word-Problem Families

How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier

How SAT word problems hide equations

SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

SAT Word Problems Made Easy

SAT Function Transformations: A Simple Way to Master Shifts, Reflections, Stretches and Compressions

SAT function transformations tips and tricks

SAT Probability Made Easy: Conditional Probability, Tables & Tricky Questions

SAT probability guide

SAT Conditional Probability

Conditional Probability

SAT Word Problems Made Easy: The Tricks for Speed, Work, Age & Profit

 

SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

🎯 Why SAT Word Problems Feel Harder Than They Are

A long SAT word problem can look intimidating because it contains a story, several numbers, units, percentages, and extra information.

But underneath the wording, there is usually one simple mathematical relationship.

The real SAT skill is not just calculation.

It is translation.

You need to turn:

Words → Quantities → Relationship → Equation → Answer

Once you learn to recognize the hidden structure, many word problems become much quicker.

---

🔹 1. The Golden Rule: Find the Relationship Before Calculating

Suppose the SAT says:

A cyclist travels 84 miles in 4 hours. What is the cyclist's average speed?

Do not immediately start calculating.

Identify:

Distance = 84 miles

Time = 4 hours

Unknown = speed

The relationship is:

Speed = Distance ÷ Time

Therefore:

r = d/t

r = 84/4

r = 21 miles per hour

✅ Answer: 21 miles per hour

The important step was not the division.

It was recognizing the relationship.

---

🔹 2. The SAT Distance Formula You Should Know Instantly

For motion problems:

d = rt

where:

d = distance

r = rate

t = time

From this:

r = d/t

and:

t = d/r

Think of the three quantities as connected:

Distance = Rate × Time

If the SAT gives you any two, you can find the third.

---

🔹 3. Units Can Quietly Destroy a Correct Solution

Suppose a car travels at:

60 miles per hour

for:

30 minutes

A common mistake is:

60 × 30

That would be wrong because the rate is measured in hours, while the time is given in minutes.

Convert:

30 minutes = 30/60 hour

= 1/2 hour

Now:

d = rt

d = 60 × 1/2

d = 30 miles

🧠 SAT habit

Before calculating, ask:

“Are my units speaking the same language?”

If the rate is in miles per hour, the time should be in hours.

---

🔹 4. Useful Time Conversions

Memorize these:

15 min = 1/4 hr

20 min = 1/3 hr

30 min = 1/2 hr

40 min = 2/3 hr

45 min = 3/4 hr

60 min = 1 hr

90 min = 3/2 hr

Fractions can often make SAT calculations easier than decimals.

---

🔹 5. When a Trip Has More Than One Speed

Suppose a driver travels:

60 miles at 30 mph

and then:

90 miles at 45 mph

To find the total travel time, handle each section separately.

First section:

t₁ = 60/30

t₁ = 2 hours

Second section:

t₂ = 90/45

t₂ = 2 hours

Therefore:

Total time = 2 + 2

= 4 hours

⭐ Key idea

When a journey has different speeds, break it into sections.

Do not try to force the entire trip into one speed equation.

---

🔥 6. The Average-Speed Trap

This is one of the most useful SAT traps to understand.

Suppose a car travels:

100 miles at 50 mph

and then:

100 miles at 100 mph

A tempting answer is:

(50 + 100)/2 = 75 mph

❌ Not correct.

Average speed means:

Total Distance ÷ Total Time

First journey:

100/50 = 2 hours

Second journey:

100/100 = 1 hour

Total distance:

200 miles

Total time:

3 hours

Therefore:

Average speed = 200/3

≈ 66.67 mph

🚨 Remember

Average speed ≠ average of speeds

Instead:

Average speed = Total Distance ÷ Total Time

---

🔹 7. Catch-Up Problems: Think “Gap”

Suppose Runner A is:

20 meters ahead

Runner A runs at:

5 m/s

Runner B runs at:

7 m/s

How long does B take to catch A?

The important quantity is not either speed by itself.

It is the speed at which B closes the gap.

Relative speed = 7 − 5

= 2 m/s

Initial gap:

20 m

Therefore:

Time = Gap ÷ Relative Speed

t = 20/2

t = 10 seconds

🧠 Shortcut

For objects moving in the same direction:

Relative speed = faster speed − slower speed

---

🔹 8. Objects Moving Toward Each Other

Suppose two cyclists are:

120 miles apart

One travels at:

30 mph

The other travels at:

50 mph

They move toward each other.

Their separation decreases at:

30 + 50 = 80 mph

Therefore:

t = 120/80

= 1.5 hours

Remember

Same direction:

Subtract speeds

Opposite directions:

Add speeds

---

🔹 9. Work Problems Are Really Rate Problems

Work questions sometimes look completely different from distance questions.

They aren't.

They use the same basic idea:

Amount completed = Rate × Time

If a worker completes an entire job in 6 hours, the worker completes:

1/6 of the job per hour

So:

Work rate = 1/6

If another worker completes the same job in 3 hours:

Work rate = 1/3

Together:

1/6 + 1/3

= 1/6 + 2/6

= 1/2

Together they complete:

1/2 of the job per hour

Therefore:

Time = 1 ÷ 1/2

= 2 hours

⭐ The key idea

In work problems:

Add rates, not times.

---

🔹 10. A Work-Rate Template Worth Memorizing

If someone completes a job in T hours:

Rate = 1/T

For multiple workers:

Combined Rate = Rate₁ + Rate₂ + Rate₃ + ...

Then:

Time = Total Work ÷ Combined Rate

If the entire job is represented by 1:

Time = 1 ÷ Combined Rate

---

🔹 11. Pipes and Tanks Use the Same Trick

A pipe fills a tank in:

4 hours

Its rate is:

1/4 tank per hour

Another pipe fills it in:

6 hours

Its rate is:

1/6 tank per hour

Together:

1/4 + 1/6

= 3/12 + 2/12

= 5/12

So they fill:

5/12 of the tank per hour

Therefore:

Time = 1 ÷ 5/12

= 12/5 hours

= 2.4 hours

---

🔥 12. What If There Is a Drain?

A drain removes water, so its rate is subtracted.

Suppose:

Pipe A = 1/4 tank/hour

Pipe B = 1/6 tank/hour

Drain = 1/12 tank/hour

Net rate:

1/4 + 1/6 − 1/12

Convert to twelfths:

3/12 + 2/12 − 1/12

= 4/12

= 1/3

Therefore the tank fills at:

1/3 tank per hour

and takes:

3 hours

---

🔹 13. Age Problems: Define the Present Age

Age questions become much easier when you choose one person's age as x.

Suppose Maya is:

4 years older than Leo

Let Leo's age be:

x

Then Maya's age is:

x + 4

If their ages add to 30:

x + (x + 4) = 30

2x + 4 = 30

2x = 26

x = 13

Therefore:

Leo = 13

Maya = 17

---

🔹 14. Future-Age Problems

Suppose a father is currently three times as old as his son.

Let the son's age be:

x

Father's age:

3x

Five years later:

Son:

x + 5

Father:

3x + 5

If the father will then be twice the son's age:

3x + 5 = 2(x + 5)

Expand:

3x + 5 = 2x + 10

Therefore:

x = 5

So the son is currently:

5 years old

and the father is:

15 years old

---

💡 15. The Age Difference Never Changes

This is one of the most useful observations in age problems.

If two people have an age difference of:

12 years

today, their difference will still be:

12 years

in:

5 years

10 years

20 years

The numbers change.

The difference does not.

This can sometimes eliminate the need for a long equation.

---

🔹 16. Mixture Problems: Find the Amount of Pure Substance

Mixture questions often involve concentration.

The fundamental relationship is:

Amount of pure substance = Total amount × Concentration

Suppose you have:

50 liters

of a:

20% salt solution

Then:

0.20 × 50 = 10 liters

of the mixture is salt.

The remaining:

50 − 10 = 40 liters

is the other component.

---

🔥 17. Mixing Two Different Concentrations

Suppose:

20 liters of a 30% solution

are mixed with:

x liters of a 50% solution

to create a:

40% solution

First solution contributes:

0.30(20) = 6

Second solution contributes:

0.50x

Total amount:

20 + x

Final amount of pure substance:

0.40(20 + x)

Therefore:

6 + 0.50x = 0.40(20 + x)

Expand:

6 + 0.50x = 8 + 0.40x

0.10x = 2

x = 20

✅ Answer:

20 liters

---

🔹 18. The Universal Mixture Equation

When appropriate, use:

Amount₁ × Concentration₁ + Amount₂ × Concentration₂

= Total Amount × Final Concentration

This single pattern can solve many mixture questions.

And always convert:

25% → 0.25

40% → 0.40

7% → 0.07

---

🔹 19. Profit: Revenue Minus Cost

Profit questions are usually straightforward once you identify the two quantities.

Profit = Revenue − Cost

Suppose a store buys an item for:

$40

and sells it for:

$55

Then:

Profit = 55 − 40

= $15

---

🔥 20. Profit Percentage

Profit percentage is calculated relative to the cost.

Formula:

Profit % = Profit/Cost × 100

Using the previous example:

Profit % = 15/40 × 100

= 37.5%

🚨 SAT warning

Do not divide the profit by the selling price unless the question specifically asks for a percentage based on the selling price.

For ordinary profit percentage:

Cost is the reference value.

---

🔹 21. Markups and Discounts

Suppose an item costs:

$80

The store marks it up by:

25%

Markup:

0.25 × 80 = 20

New price:

80 + 20 = $100

Now suppose the store gives a:

20% discount

Discount:

0.20 × 100 = 20

Final price:

100 − 20 = $80

This example illustrates an important SAT idea:

Percentages are applied to the current value.

---

🔥 22. Successive Percentage Changes

Suppose a price increases by:

20%

and then decreases by:

20%

Start with:

100

After the increase:

100 × 1.20 = 120

After the decrease:

120 × 0.80 = 96

Final value:

96

So the overall change is:

−4%

⭐ The multiplier method

Increase by 20%:

× 1.20

Decrease by 20%:

× 0.80

Combined:

1.20 × 0.80 = 0.96

Therefore:

96% of the original value remains.

---

🔹 23. Commission Problems

Suppose a salesperson earns:

6% commission

on sales.

If the salesperson sells:

$4,000

then:

Commission = 0.06 × 4000

= $240

If the salesperson also has a fixed salary:

Total earnings = Salary + Commission

The SAT may hide this simple structure inside a longer story.

---

🔹 24. Tax Problems

An item costs:

$500

and the tax rate is:

8%

Tax:

0.08 × 500 = 40

Total:

500 + 40 = $540

Or use the multiplier:

500 × 1.08 = 540

🧠 Quick rule

Tax added:

Original × (1 + tax rate)

Discount applied:

Original × (1 − discount rate)

---

🔥 25. Combined Word Problems

The SAT may combine several ideas in one question.

Example:

A cyclist travels the first:

30 miles at 15 mph

and the remaining:

45 miles at 30 mph

What is the average speed?

First section:

30/15 = 2 hours

Second section:

45/30 = 1.5 hours

Total distance:

30 + 45 = 75 miles

Total time:

2 + 1.5 = 3.5 hours

Average speed:

75/3.5 ≈ 21.43 mph

Notice what happened.

The problem looked complicated.

But it was simply:

Distance ÷ Rate → Time

followed by:

Total Distance ÷ Total Time

---

🔹 26. Decode SAT Wording

Certain words provide mathematical clues.

“Per”

Usually indicates a rate.

240 miles per 4 hours

means:

240/4

“Each”

Often indicates multiplication or a unit rate.

“Of”

Usually indicates multiplication.

30% of 80

means:

0.30 × 80

“At least”

means:

≥

“At most”

means:

≤

“More than”

means:

>

“Less than”

means:

<

---

🔹 27. “Difference” Questions

The difference between two quantities is generally represented by:

|A − B|

For example:

A = 19

B = 12

Difference:

|19 − 12| = 7

The absolute value makes the result nonnegative.

---

🔥 28. “How Much Greater?” vs. “How Much Greater Percent?”

These questions are not the same.

Suppose:

A = 30

B = 20

“How much greater is A than B?”

30 − 20 = 10

But:

“What percent greater is A than B?”

Use B as the reference:

(30 − 20)/20 × 100

= 50%

🚨 Always ask:

“Percent relative to what?”

---

🔹 29. The Five-Step SAT Translation System

When a word problem looks enormous, use this system.

STEP 1 — Identify the quantities

What numbers and units are given?

STEP 2 — Define the unknown

Write:

x = ...

STEP 3 — Identify the relationship

Is it:

distance?

rate?

work?

percentage?

mixture?

age?

profit?

STEP 4 — Build the equation

Translate the words into mathematics.

STEP 5 — Check the result

Ask:

Does this answer make sense?

This final question is surprisingly powerful.

---

🔥 30. Six SAT Word-Problem Traps to Avoid

❌ Trap 1: Averaging speeds directly

Do not automatically calculate:

(r₁ + r₂)/2

Use:

Total Distance ÷ Total Time

when appropriate.

❌ Trap 2: Ignoring units

Minutes, hours, seconds, miles, kilometers, and meters must be handled consistently.

❌ Trap 3: Adding work times

Workers' rates are added.

❌ Trap 4: Treating percentage changes as ordinary addition

A 20% increase followed by a 20% decrease does not equal 0%.

❌ Trap 5: Using the wrong reference value

Profit percentage normally uses:

Cost

❌ Trap 6: Solving for the wrong thing

You may find x correctly and still choose the wrong answer if the question asks for another quantity.

---

🔹 31. Practice Challenge #1

A train travels:

180 miles

at:

60 miles per hour

How long does the trip take?

A) 2 hours
B) 3 hours
C) 4 hours
D) 6 hours

Use:

t = d/r

t = 180/60

t = 3

✅ Answer: B

---

🔹 32. Practice Challenge #2

One worker can complete a task in:

10 hours

Another can complete it in:

15 hours

How long will they take together?

A) 5 hours
B) 6 hours
C) 7.5 hours
D) 25 hours

Rates:

1/10

and:

1/15

Combined:

1/10 + 1/15

= 3/30 + 2/30

= 5/30

= 1/6

Therefore:

Time = 6 hours

✅ Answer: B

---

🔹 33. Practice Challenge #3

A mother is:

24 years older

than her daughter.

In 4 years, the mother will be twice the daughter's age.

How old is the daughter now?

Let:

Daughter = x

Mother:

x + 24

Four years later:

Daughter:

x + 4

Mother:

x + 28

Equation:

x + 28 = 2(x + 4)

x + 28 = 2x + 8

x = 20

✅ Answer: 20 years old

---

🔹 34. Practice Challenge #4

A 20% solution is mixed with a 50% solution to produce:

30 liters of a 40% solution

How many liters of the 50% solution are needed?

Let:

x = liters of 50% solution

Then:

30 − x = liters of 20% solution

Equation:

0.20(30 − x) + 0.50x = 0.40(30)

Expand:

6 − 0.20x + 0.50x = 12

0.30x = 6

x = 20

✅ Answer: 20 liters

---

🔹 35. Practice Challenge #5

A store buys an item for:

$60

and sells it for:

$75

What is the profit percentage?

Profit:

75 − 60 = 15

Profit percentage:

15/60 × 100

= 25%

✅ Answer: 25%

---

🔹 36. Practice Challenge #6

A car travels:

120 miles at 40 mph

and then:

180 miles at 60 mph

What is its average speed?

First section:

120/40 = 3 hours

Second section:

180/60 = 3 hours

Total distance:

300 miles

Total time:

6 hours

Average speed:

300/6 = 50 mph

✅ Answer: 50 miles per hour

---

🧠 37. The SAT Word-Problem Formula Bank

🚗 Motion

d = rt

r = d/t

t = d/r

⚡ Relative motion

Same direction:

difference of rates

Opposite directions:

sum of rates

🛠 Work

Rate = 1/time

Combined rate = sum of individual rates

🧪 Mixtures

Amount × Concentration

💰 Profit

Profit = Revenue − Cost

📈 Profit percentage

Profit/Cost × 100

📊 Percentage increase

Increase/Original × 100

📉 Percentage decrease

Decrease/Original × 100

👨‍👩‍👧 Ages

Future age = Current age + Years Passed

🔢 Inequalities

At least → ≥

At most → ≤

More than → >

Less than → <

---

🚀 38. The 10-Second SAT Word-Problem Scan

When you see a long word problem, mentally ask:

① What quantities are given?

② What is unknown?

③ What are the units?

④ What relationship connects the quantities?

⑤ What exactly does the question ask me to find?

Then solve.

Do not let the story control your thinking.

You control the story by translating it into mathematics.

---

🏆 Final SAT Takeaway

The SAT can wrap a simple equation inside a paragraph designed to make you hesitate.

A distance question may hide:

d = rt

A work question may hide:

rate = 1/time

A mixture question may hide:

amount × concentration

A profit question may hide:

revenue − cost

An age question may hide:

x + years

A percentage question may hide:

original × multiplier

The secret is not memorizing dozens of special tricks.

It is learning to recognize the mathematical structure hiding inside the language.

Remember this sequence:

READ → IDENTIFY → TRANSLATE → EQUATE → SOLVE → CHECK

When you can consistently turn an SAT word problem into a clean equation, the paragraph stops being the problem.

The equation is the problem.

And equations are much easier to solve.

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ALGEBRA

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LINEAR EQUATIONS [PART 1]

SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]

SYSTEM OF EQUATIONS  [ PART  I ]

system of equations part I

SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

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QUADRATIC EQUATIONS [PART II] 

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PERCENTAGES

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SAT Quadratic Equations Part 2: Hard Questions, Tricks & Shortcuts

 SAT Quadratic Equations Part 2: Hard Questions, Parabolas, Parameters, Word Problems & Shortcuts

Quadratic equations become much more interesting on the harder SAT questions.

You may be asked to find a root, interpret a parabola, determine a maximum value, identify a parameter, connect an equation to a graph, or solve a real-world problem.

The key is recognizing that these are not separate topics.

They are different ways of using the same quadratic relationship.

A quadratic equation generally has the form:

ax² + bx + c = 0

where:

a ≠ 0

The graph of a quadratic function has the shape of a parabola.

This guide focuses on the harder patterns you should recognize quickly.

---

1. The Three Forms of a Quadratic

A quadratic can appear in three especially useful forms.

Standard form

f(x) = ax² + bx + c

Useful for identifying:

• a
• b
• c
• y-intercept

Factored form

f(x) = a(x − r₁)(x − r₂)

Useful for finding:

• roots
• zeros
• x-intercepts

Vertex form

f(x) = a(x − h)² + k

Useful for finding:

• vertex
• maximum or minimum
• axis of symmetry

Learning to move between these forms is one of the most useful SAT quadratic skills.

---

2. Roots and Zeros

Suppose:

f(x) = (x − 3)(x + 5)

The zeros occur when:

f(x) = 0

Therefore:

(x − 3)(x + 5) = 0

So:

x − 3 = 0

or:

x + 5 = 0

Therefore:

x = 3 or x = −5

These values are called:

• roots
• zeros
• solutions

They all refer to the x-values where the quadratic equals zero.

---

3. X-Intercepts

The x-intercepts of a parabola occur where:

y = 0

Therefore, finding the x-intercepts means solving:

f(x) = 0

For:

y = (x − 2)(x + 6)

the x-intercepts are:

(2, 0)

and:

(−6, 0)

The roots are:

2 and −6

---

4. The Vertex

The vertex is the highest or lowest point of a parabola.

If:

f(x) = a(x − h)² + k

the vertex is:

(h, k)

For example:

f(x) = (x − 4)² + 7

the vertex is:

(4, 7)

Because the coefficient of the squared term is positive, the parabola opens upward.

Therefore, the vertex represents a:

minimum

---

5. When the Parabola Opens Downward

Consider:

f(x) = −(x − 3)² + 10

The vertex is:

(3, 10)

The leading coefficient is negative.

Therefore, the parabola opens downward.

The vertex is therefore the:

maximum

value.

Quick rule

a > 0 → minimum

a < 0 → maximum

---

6. Finding the Vertex From Standard Form

Suppose:

f(x) = x² − 8x + 11

The x-coordinate of the vertex is:

x = −b/(2a)

Here:

a = 1

b = −8

Therefore:

x = −(−8)/(2)

x = 4

Now substitute x = 4:

f(4) = 16 − 32 + 11

f(4) = −5

Therefore, the vertex is:

(4, −5)

---

7. The Axis of Symmetry

The axis of symmetry passes through the vertex.

For:

f(x) = ax² + bx + c

the axis is:

x = −b/(2a)

For:

f(x) = 2x² − 12x + 7

we have:

a = 2

b = −12

Therefore:

x = 12/4

x = 3

Axis of symmetry:

x = 3

---

8. A Useful Symmetry Trick

Suppose the roots of a quadratic are:

2 and 10.

The axis of symmetry lies exactly halfway between them.

Therefore:

(2 + 10)/2 = 6

Axis of symmetry:

x = 6

This can be faster than using:

−b/(2a)

when the roots are already known.

---

9. The Vertex From Two Roots

Suppose:

f(x) = (x − 2)(x − 10)

The roots are:

2 and 10.

Therefore, the axis of symmetry is:

x = 6

Now substitute:

f(6) = (6 − 2)(6 − 10)

= 4(−4)

= −16

Therefore:

Vertex = (6, −16)

---

10. Completing the Square

Completing the square is useful for converting standard form into vertex form.

Consider:

x² + 6x + 5

Take half of 6:

3

Square it:

9

Rewrite:

x² + 6x + 9 − 9 + 5

Group:

(x + 3)² − 4

Therefore:

x² + 6x + 5 = (x + 3)² − 4

The vertex is:

(−3, −4)

---

11. Completing the Square With a Leading Coefficient

Consider:

2x² + 12x + 7

First factor 2 from the quadratic terms:

2(x² + 6x) + 7

Complete the square:

2[(x + 3)² − 9] + 7

Expand:

2(x + 3)² − 18 + 7

Therefore:

2(x + 3)² − 11

The vertex is:

(−3, −11)

---

12. Quadratic Formula

When factoring is difficult, use the quadratic formula.

For:

ax² + bx + c = 0

the solutions are:

x = [−b ± √(b² − 4ac)]/(2a)

For example:

x² − 5x + 6 = 0

Here:

a = 1

b = −5

c = 6

Therefore:

x = [5 ± √(25 − 24)]/2

x = [5 ± 1]/2

So:

x = 3

or:

x = 2

---

13. The Discriminant

The expression:

b² − 4ac

is called the discriminant.

It tells you how many real solutions the quadratic has.

If:

b² − 4ac > 0

There are:

two distinct real solutions

If:

b² − 4ac = 0

There is:

one real solution

If:

b² − 4ac < 0

There are:

no real solutions

This connects algebra directly to the graph.

---

14. Discriminant and the Graph

If a quadratic has two real roots, its parabola crosses the x-axis twice.

If it has one real root, the parabola touches the x-axis once.

If it has no real roots, the parabola does not cross the x-axis.

Therefore:

Two roots

→ Two x-intercepts

One root

→ One x-intercept

No real roots

→ No x-intercepts

This is a powerful connection between algebra and graphs.

---

15. Parameter Questions

Hard SAT questions may include a variable parameter.

For example:

x² − 6x + k = 0

For what value of k does the equation have exactly one real solution?

Exactly one real solution means:

discriminant = 0

Therefore:

b² − 4ac = 0

Here:

a = 1

b = −6

c = k

So:

36 − 4k = 0

4k = 36

k = 9

Answer:

9

---

16. Parameter Question: Two Real Solutions

Consider:

x² − 4x + k = 0

For two distinct real solutions:

b² − 4ac > 0

Therefore:

16 − 4k > 0

−4k > −16

k < 4

Answer:

k < 4

This is an example where you are not looking for one number.

You are looking for a range of values.

---

17. Parameter Question: No Real Solutions

Consider:

x² + 2x + k = 0

For no real solutions:

b² − 4ac < 0

Therefore:

4 − 4k < 0

−4k < −4

k > 1

Answer:

k > 1

---

18. Quadratic Equations From Graphs

Suppose a graph crosses the x-axis at:

x = −2

and:

x = 5

Then the quadratic can be written as:

f(x) = a(x + 2)(x − 5)

If the graph also passes through a known point, you can determine a.

Suppose the graph passes through:

(0, 20)

Then:

20 = a(2)(−5)

20 = −10a

a = −2

Therefore:

f(x) = −2(x + 2)(x − 5)

---

19. Finding a Quadratic From Three Points

A quadratic can be written:

y = ax² + bx + c

If three points are given, substitute each point to create three equations.

For example, suppose:

(0, 2)

(1, 6)

(2, 12)

From (0, 2):

c = 2

From (1, 6):

a + b + 2 = 6

a + b = 4

From (2, 12):

4a + 2b + 2 = 12

4a + 2b = 10

Solving gives:

a = 1

b = 3

c = 2

Therefore:

y = x² + 3x + 2

---

20. Quadratic Word Problems

Many SAT quadratic word problems involve area.

Suppose a rectangle has length:

x + 5

and width:

x

Its area is:

x(x + 5)

Therefore:

A = x² + 5x

If the area is 36:

x² + 5x = 36

or:

x² + 5x − 36 = 0

Factor:

(x + 9)(x − 4) = 0

Therefore:

x = −9

or:

x = 4

Since a length cannot be negative:

x = 4

---

21. Why Negative Roots Can Be Rejected

Algebraically, x = −9 is a solution to the equation.

But if x represents a physical length, negative 9 is impossible.

Therefore:

x = 4

This is a common SAT skill.

The equation may produce multiple mathematical solutions, but the context may allow only one.

---

22. Area Problems

Suppose a rectangle has dimensions:

x + 2

and:

x + 5

Its area is 48.

Then:

(x + 2)(x + 5) = 48

Expand:

x² + 7x + 10 = 48

x² + 7x − 38 = 0

At this point, you can use factoring if possible or the quadratic formula.

The important step is translating the geometry into an equation.

---

23. Projectile Problems

Quadratics frequently model height.

A typical model might be:

h(t) = −16t² + 64t + 5

where:

t is time and:

h(t) is height.

The negative coefficient tells you the parabola opens downward.

Therefore, the object eventually reaches a maximum height.

---

24. Finding Maximum Height

For:

h(t) = −16t² + 64t + 5

the time at maximum height is:

t = −b/(2a)

Therefore:

t = −64/[2(−16)]

t = 2

Now calculate:

h(2) = −16(4) + 64(2) + 5

= −64 + 128 + 5

= 69

Maximum height:

69

---

25. Finding When an Object Hits the Ground

If height is zero, set:

h(t) = 0

For:

h(t) = −16t² + 64t + 5

solve:

−16t² + 64t + 5 = 0

The resulting solutions represent times when the height is zero.

A negative time would normally be rejected because the physical situation begins at:

t = 0

---

26. Maximum and Minimum Questions

Whenever the SAT asks:

• greatest value

• maximum

• highest point

• minimum

• lowest value

and the equation is quadratic, immediately think:

vertex

For:

f(x) = ax² + bx + c

find:

x = −b/(2a)

Then substitute to find the corresponding y-value.

---

27. Vertex Form Makes Maximum Questions Easy

Suppose:

f(x) = −2(x − 4)² + 13

The vertex is:

(4, 13)

Because the parabola opens downward, the maximum value is:

13

No calculation is required.

---

28. Comparing Two Quadratics

Suppose:

f(x) = x²

and:

g(x) = x² − 6x + 8

To find where they are equal:

x² = x² − 6x + 8

Subtract x²:

0 = −6x + 8

6x = 8

x = 4/3

This is another example of finding an intersection between two functions.

---

29. Quadratic vs. Linear

Suppose:

f(x) = x² − 4x + 3

and:

g(x) = x + 3

To find where they intersect:

x² − 4x + 3 = x + 3

x² − 5x = 0

x(x − 5) = 0

Therefore:

x = 0

or:

x = 5

There are two intersection points.

---

30. A Common SAT Graph Question

A parabola crosses the x-axis at:

−3 and 7.

Without doing any algebra, what is the axis of symmetry?

Take the midpoint:

(−3 + 7)/2

= 4/2

= 2

Answer:

x = 2

This is much faster than deriving the entire equation.

---

31. Finding the Constant From a Root

Suppose:

f(x) = x² + kx − 12

and x = 3 is a root.

Then:

f(3) = 0

Substitute:

9 + 3k − 12 = 0

3k − 3 = 0

k = 1

Answer:

1

Whenever the question says a particular value is a solution or root, substitute it directly.

---

32. Finding a Parameter From a Point

Suppose:

f(x) = x² + kx + 4

and the graph passes through:

(2, 10)

Therefore:

f(2) = 10

4 + 2k + 4 = 10

2k + 8 = 10

2k = 2

k = 1

Answer:

1

The same principle works for many function questions.

---

33. Root Relationships

If:

ax² + bx + c = 0

has roots:

r₁ and r₂

then:

r₁ + r₂ = −b/a

and:

r₁r₂ = c/a

These relationships can save time.

---

34. Example Using Root Sum

Suppose the roots of:

2x² − 10x + 7 = 0

are r₁ and r₂.

What is:

r₁ + r₂?

Use:

r₁ + r₂ = −b/a

Therefore:

10/2 = 5

Answer:

5

You do not need to solve the quadratic.

---

35. Example Using Root Product

For:

3x² + 8x − 10 = 0

the product of the roots is:

c/a

Therefore:

−10/3

Answer:

−10/3

Again, there is no need to use the quadratic formula.

---

36. A Powerful SAT Shortcut

If the question asks for:

sum of roots

→ use:

−b/a

If it asks for:

product of roots

→ use:

c/a

Do not solve the equation unless the actual roots are required.

---

37. Transformations of Quadratics

Suppose:

f(x) = x²

Then:

f(x) + 5

moves the graph:

5 units upward

while:

f(x) − 3

moves it:

3 units downward

Similarly:

f(x − 4)

moves the graph:

4 units right

and:

f(x + 2)

moves it:

2 units left

The signs inside the parentheses can be especially tricky.

---

38. The Horizontal Shift Trap

Compare:

f(x − 3)

and:

f(x + 3)

The first shifts right by 3.

The second shifts left by 3.

Remember:

Inside the parentheses, the direction appears reversed.

---

39. Effect of the Leading Coefficient

In:

f(x) = ax²

the value of |a| affects the width.

If:

|a| > 1

the parabola is narrower.

If:

0 < |a| < 1

the parabola is wider.

If:

a < 0

the parabola opens downward.

---

40. A Difficult Parameter Problem

Consider:

f(x) = x² − 8x + k

For the graph to touch the x-axis exactly once, what must k equal?

Touching the x-axis once means:

one real root

Therefore:

discriminant = 0

b² − 4ac = 0

64 − 4k = 0

k = 16

Answer:

16

---

41. A Harder Interpretation

Suppose:

f(x) = x² − 8x + k

For the graph to have no x-intercepts:

64 − 4k < 0

Therefore:

k > 16

So:

k > 16

This means the parabola lies entirely above the x-axis because its leading coefficient is positive.

---

42. Another Hard Question

Suppose:

f(x) = −x² + 6x + k

For the maximum value of f(x) to equal 10, find k.

The vertex occurs at:

x = −6/[2(−1)]

= 3

Now substitute:

10 = −(3²) + 6(3) + k

10 = −9 + 18 + k

10 = 9 + k

k = 1

Answer:

1

---

43. The SAT “Which Equation?” Question

Sometimes you are shown a graph and several equations.

Look for:

Direction

Parabola opens upward:

a > 0

Parabola opens downward:

a < 0

Vertex

Compare the vertex coordinates.

X-intercepts

Check the roots.

Y-intercept

Set:

x = 0

The correct equation must match all the available information.

---

44. Use the Cheapest Information First

If the graph clearly shows:

• x-intercepts
• vertex
• y-intercept

do not calculate everything.

Use the easiest visible feature to eliminate incorrect answers.

For example, if the parabola opens downward, immediately eliminate every equation with a positive x² coefficient.

This is a powerful multiple-choice strategy.

---

45. SAT Quadratic Checklist

Before test day, make sure you can:

✓ Factor quadratics.

✓ Use the quadratic formula.

✓ Find roots.

✓ Find x-intercepts.

✓ Find the vertex.

✓ Find the axis of symmetry.

✓ Complete the square.

✓ Use the discriminant.

✓ Determine the number of real solutions.

✓ Solve parameter questions.

✓ Interpret quadratic graphs.

✓ Find maximum and minimum values.

✓ Solve quadratic word problems.

✓ Reject impossible contextual solutions.

✓ Use root sum and root product.

✓ Recognize transformations.

✓ Compare quadratic functions.

✓ Find a parameter using a known root or point.

---

Final SAT Quadratic Strategy

When you see a quadratic, do not automatically reach for the quadratic formula.

First ask:

What is the question actually asking?

If it asks for the roots:

→ factor or use the quadratic formula.

If it asks for the x-intercepts:

→ find the roots.

If it asks for the maximum or minimum:

→ find the vertex.

If it asks for the axis of symmetry:

→ use −b/(2a), or the midpoint of the roots.

If it asks how many real solutions exist:

→ use the discriminant.

If it gives a parameter and asks for exactly one solution:

→ set the discriminant equal to zero.

If it gives a known root:

→ substitute it directly.

If it asks for the sum of roots:

→ use −b/a.

If it asks for the product of roots:

→ use c/a.

If it gives a graph:

→ use the vertex, intercepts, direction, and symmetry.

The biggest SAT quadratic skill is therefore not memorizing more formulas.

It is recognizing which piece of the quadratic contains the answer.

Once you can identify that quickly, many of the hardest-looking quadratic questions become short problems.

SAT Systems of Equations: Tricks & Hard Questions You Need to Know

 

SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

If you already understand the basic methods for solving systems of equations, the next step is learning how the SAT hides systems inside unfamiliar-looking problems.

The harder questions are often not difficult because the algebra is advanced. They are difficult because you must recognize what the equations represent, decide what information matters, and sometimes determine the answer without completely solving the system.

This guide focuses on those harder SAT systems of equations questions.

---

1. The SAT Can Hide a System in Plain Sight

A question does not always give you something obvious like:

x + y = 20

2x + 3y = 50

Instead, you might see:

• a table
• a graph
• a word problem
• a relationship involving a constant
• two different rates
• ticket prices
• mixtures
• consecutive quantities
• equations with unknown coefficients

Your first job is therefore not to calculate.

Your first job is to ask:

“What are the two relationships in this problem?”

Once you identify them, the system often becomes much easier.

---

2. The Hidden-System Word Problem

Consider this example:

A store sells notebooks for $4 each and pens for $2 each. A customer buys 18 items and spends $54. How many notebooks does the customer buy?

Let:

n = number of notebooks

p = number of pens

The total number of items is:

n + p = 18

The total cost is:

4n + 2p = 54

Now you have a system.

From:

n + p = 18

we get:

p = 18 − n

Substitute:

4n + 2(18 − n) = 54

4n + 36 − 2n = 54

2n = 18

n = 9

Answer:

9 notebooks

The important part was not the algebra.

It was recognizing the two independent pieces of information.

---

3. The “Total” Pattern

SAT word problems frequently contain a total.

Watch for:

total number

total cost

total distance

total amount

total revenue

total points

total weight

A total often gives you one equation.

For example:

There are 75 students in two groups.

Let:

x = students in Group A

y = students in Group B

Then:

x + y = 75

If Group A has 9 more students than Group B:

x = y + 9

Now you have a system.

---

4. The “More Than” Trap

Suppose:

A is 12 more than B.

The correct equation is:

A = B + 12

Not:

B = A + 12

Similarly:

A is 12 less than B

means:

A = B − 12

These small wording differences can completely change the answer.

Quick translation rule

“X is more than Y”

→ X = Y + amount

“X is less than Y”

→ X = Y − amount

---

5. Consecutive Numbers and Systems

Systems can also appear in questions involving consecutive quantities.

Suppose two numbers have a sum of 41, and the larger number is 7 more than the smaller.

Let:

x = smaller number

y = larger number

Then:

x + y = 41

y = x + 7

Substitute:

x + x + 7 = 41

2x = 34

x = 17

y = 24

Answer:

17 and 24

The SAT may replace ordinary numbers with quantities such as ages, scores, lengths, or amounts.

The underlying structure remains the same.

---

6. Systems Involving Ages

Age problems can look complicated because the wording is long.

Suppose two siblings have a combined age of 31 years. One sibling is 5 years older than the other.

Let:

x = younger sibling's age

y = older sibling's age

Then:

x + y = 31

y = x + 5

Substitute:

x + x + 5 = 31

2x = 26

x = 13

Therefore:

y = 18

Answer:

13 and 18

---

7. Rate Problems

Remember the fundamental relationship:

Distance = Rate × Time

or:

d = rt

Suppose two people travel for the same amount of time but at different speeds.

You may need to write equations such as:

d₁ = r₁t

d₂ = r₂t

If the problem gives a relationship between their distances, a system may result.

The important question is:

Which quantities are equal, and which are different?

---

8. Mixture Problems

Mixture questions can also create systems.

Suppose a solution contains two liquids with different concentrations.

Let:

x = amount of first liquid

y = amount of second liquid

The total amount might give:

x + y = 20

The amount of pure substance might give:

0.10x + 0.40y = 5

Now you have a system.

The decimal numbers may look intimidating, but the structure is still:

quantity equation + concentration equation

---

9. Systems From Graphs

A graph can sometimes give you the answer faster than algebra.

Suppose two lines intersect at:

(4, 6)

If the question asks:

“What is the solution to the system?”

the answer is simply:

(4, 6)

You do not need to calculate anything else.

The intersection represents the values of x and y that satisfy both equations.

---

10. What If the Graph Shows Parallel Lines?

If two lines never meet, the system has:

No solution

This means there is no ordered pair that satisfies both equations.

Look for:

• same slope
• different y-intercepts
• no intersection

These are all signals for no solution.

---

11. What If the Graph Shows the Same Line?

If both equations produce the exact same line, there are:

Infinitely many solutions

Every point on the line satisfies both equations.

This is different from having one intersection point.

---

12. The Three Graph Patterns You Must Know

Pattern 1: Crossing lines

→ One solution

Pattern 2: Parallel lines

→ No solution

Pattern 3: Same line

→ Infinitely many solutions

If you can identify these three patterns instantly, you can answer many graphical systems questions quickly.

---

13. A Powerful Slope Trick

Suppose you are given:

y = 5x + 2

and:

y = 5x − 9

Both slopes are:

5

The intercepts are different.

Therefore:

No solution

You do not need to solve for x.

Now suppose:

y = 5x + 2

and:

2y = 10x + 4

Divide the second equation by 2:

y = 5x + 2

The equations are identical.

Therefore:

Infinitely many solutions

---

14. Parameter Questions

Parameter questions are among the most useful systems questions to practice.

Consider:

2x + 4y = 12

x + ky = 6

For infinitely many solutions, the second equation must be exactly half of the first.

Divide the first equation by 2:

x + 2y = 6

Therefore:

k = 2

Answer:

2

The key phrase to watch for is:

“has infinitely many solutions.”

That tells you the two equations must represent the same line.

---

15. Parameter Questions With No Solution

Consider:

2x + 6y = 12

x + ky = 5

Divide the first equation by 2:

x + 3y = 6

For the system to have no solution, the second equation needs the same left-side relationship but a different constant.

Therefore:

k = 3

The equations become:

x + 3y = 6

x + 3y = 5

They cannot both be true.

Therefore:

No solution

---

16. Why the Constant Matters

Compare:

x + 2y = 7

x + 2y = 7

These describe the same line.

→ Infinitely many solutions.

Now compare:

x + 2y = 7

x + 2y = 3

The left sides are identical, but the constants differ.

→ No solution.

This is one of the fastest patterns to recognize on the SAT.

---

17. A Question That Looks Harder Than It Is

Suppose:

4x + 8y = 20

2x + 4y = 10

You might start solving.

Don't.

Notice that the first equation is exactly twice the second.

Therefore, the equations represent the same line.

Answer:

Infinitely many solutions

Recognizing proportional equations can save valuable time.

---

18. Systems With Fractions

Fractions do not change the underlying method.

Consider:

x/2 + y = 7

x/2 − y = 1

Add the equations:

x = 8

Then:

8/2 + y = 7

4 + y = 7

y = 3

Answer:

(8, 3)

Tip

If fractions are making elimination difficult, multiply every equation by the least common denominator.

---

19. Systems With Decimals

Suppose:

0.5x + y = 8

x − y = 4

Multiply the first equation by 2:

x + 2y = 16

Now combine with:

x − y = 4

Subtract:

3y = 12

y = 4

Then:

x − 4 = 4

x = 8

Answer:

(8, 4)

Converting decimals to simpler forms can make the system much easier.

---

20. Systems With Negative Coefficients

Consider:

3x − 2y = 7

−3x + 5y = 8

Add:

3y = 15

y = 5

Then:

3x − 10 = 7

3x = 17

x = 17/3

Answer:

(17/3, 5)

Do not let negative coefficients make the problem appear more advanced than it is.

Look for cancellation.

---

21. When You Do Not Need Both Variables

Sometimes the question asks for an expression such as:

x + y

or:

2x − y

You may not need to calculate x and y separately.

For example:

x + y = 12

2x − y = 8

Suppose the question asks for:

3x

Add the equations:

3x = 20

Therefore:

3x = 20

You can answer immediately.

SAT lesson:

Solve for what the question asks, not necessarily for every variable.

---

22. This Can Save a Lot of Time

Suppose:

3x + 2y = 17

5x − 2y = 23

Question:

What is the value of 8x?

Add the equations:

8x = 40

Therefore:

8x = 40

You do not need to calculate y.

This is exactly the kind of shortcut worth recognizing.

---

23. Eliminate the Variable You Do Not Need

If the question asks for x, eliminate y.

If the question asks for y, eliminate x.

If the question asks for x + y, look for a combination that produces x + y.

Your goal is not:

“Solve everything.”

Your goal is:

“Find the requested quantity as efficiently as possible.”

---

24. Systems and Expressions

Consider:

x + y = 15

2x − y = 9

Question:

What is x?

Add:

3x = 24

x = 8

You do not need y.

This approach becomes particularly useful when the SAT gives answer choices that involve only one variable.

---

25. When a System Is Embedded in a Function

You might see:

f(x) = 2x + 3

g(x) = 11 − x

Question:

For what value of x is f(x) = g(x)?

Set them equal:

2x + 3 = 11 − x

3x = 8

x = 8/3

The function notation does not fundamentally change the problem.

You are still finding where two relationships are equal.

---

26. Function Intersection

If:

f(x) = 4x − 1

g(x) = 2x + 7

The intersection occurs when:

f(x) = g(x)

Therefore:

4x − 1 = 2x + 7

2x = 8

x = 4

Then:

y = 15

So the intersection is:

(4, 15)

This is another way the SAT can test systems thinking without explicitly calling it a system.

---

27. A Difficult Word Problem Pattern

A company sells two types of memberships.

Basic membership costs $20.

Premium membership costs $35.

The company sells 120 memberships and receives $3,300.

Let:

b = basic memberships

p = premium memberships

Then:

b + p = 120

20b + 35p = 3300

From the first equation:

b = 120 − p

Substitute:

20(120 − p) + 35p = 3300

2400 − 20p + 35p = 3300

15p = 900

p = 60

Therefore:

b = 60

Answer:

60 basic and 60 premium memberships

---

28. The Hidden “Average” System

Suppose a class has 20 students.

The average score of all students is 78.

The average score of one group of 8 students is 85.

The remaining 12 students have an average score of x.

The total score of all students is:

20 × 78 = 1560

The first group's total score is:

8 × 85 = 680

Therefore:

680 + 12x = 1560

12x = 880

x = 220/3

This example demonstrates an important principle:

Convert averages into totals.

Average × number of items = total.

That often exposes the underlying equation.

---

29. SAT Systems and Percentages

Percent questions can also produce equations.

Suppose there are x adults and y students.

If 40% of the adults and 25% of the students participate, and the total number participating is known, you can create equations using:

0.40x + 0.25y = total participating

Combined with:

x + y = total population

Again, the same system-solving techniques apply.

---

30. Use Estimation Before Exact Calculation

Suppose your equations suggest:

x ≈ 20

and:

y ≈ 5

but your calculation produces:

x = −200

That should immediately make you suspicious.

Use the context.

If x represents the number of students, a negative answer is impossible.

If x represents a length, a negative value usually does not make physical sense.

The SAT often provides enough context to reject an incorrect interpretation.

---

31. Check Units in Word Problems

If x represents:

• dollars → answer should be in dollars

• miles → answer should be in miles

• tickets → answer should be a number of tickets

• hours → answer should be a time

A mathematically correct number can still be the wrong answer if it represents the wrong quantity.

---

32. Hard SAT Strategy: Read the Question First

Before doing the algebra, read what the question actually asks.

Suppose you are given a system and asked:

“What is the value of x + y?”

Do not automatically solve for x and y separately.

Look for a way to obtain x + y directly.

This habit can turn a multi-step calculation into one or two steps.

---

33. Another Direct-Expression Example

Given:

2x + 3y = 18

4x − 3y = 12

What is the value of 6x?

Add:

6x = 30

Therefore:

6x = 30

No need to solve for y.

---

34. SAT Mistake: Dividing Only Part of an Equation

Suppose:

2x + 4y = 10

Dividing by 2 gives:

x + 2y = 5

Every term must be divided.

Do not write:

x + 4y = 5

or:

2x + 2y = 5

The operation applies to the entire equation.

---

35. SAT Mistake: Changing the Equation Incorrectly

If you multiply:

x + 2y = 5

by 3, you get:

3x + 6y = 15

not:

3x + 2y = 15

Whatever operation you perform must affect every term.

---

36. SAT Mistake: Forgetting That Ordered Pairs Have an Order

If the solution is:

x = 3

y = 8

the ordered pair is:

(3, 8)

not:

(8, 3)

The first coordinate is always x.

The second coordinate is always y.

---

37. Hard Practice Question

Consider:

3x + 2y = 16

6x + 4y = 32

How many solutions does the system have?

Multiply the first equation by 2:

6x + 4y = 32

The equations are identical.

Answer:

Infinitely many solutions

Do not waste time solving for x and y.

---

38. Hard Practice Question

Consider:

4x + 8y = 20

2x + 4y = 7

Multiply the second equation by 2:

4x + 8y = 14

Now compare:

4x + 8y = 20

4x + 8y = 14

The left sides are identical but the constants differ.

Answer:

No solution

---

39. Hard Practice Question

Solve:

5x + 2y = 24

3x − 2y = 8

Add:

8x = 32

x = 4

Substitute:

5(4) + 2y = 24

20 + 2y = 24

y = 2

Answer:

(4, 2)

---

40. The Ultimate SAT Systems Strategy

When you see a system, use this sequence:

① Identify the structure

Are you looking at equations, a graph, a table, or a word problem?

② Identify the variables

What does x represent?

What does y represent?

③ Identify the target

Does the question ask for x, y, x + y, a point, or the number of solutions?

④ Choose the shortest method

Substitution?

Elimination?

Graph?

Coefficient comparison?

⑤ Check the answer

Does it satisfy the original conditions?

This five-step process is much more powerful than memorizing isolated tricks.

---

Final SAT Systems Mastery Checklist

Before test day, make sure you can:

✓ Solve systems using substitution.

✓ Solve systems using elimination.

✓ Recognize when graphing gives the answer.

✓ Identify one solution.

✓ Identify no solution.

✓ Identify infinitely many solutions.

✓ Translate word problems into systems.

✓ Solve ticket and cost problems.

✓ Solve age and consecutive-number problems.

✓ Handle rates and mixtures.

✓ Work with fractions and decimals.

✓ Recognize proportional equations.

✓ Solve parameter questions.

✓ Find an expression without solving every variable.

✓ Interpret intersections.

✓ Check answers using the original equations.

✓ Recognize when a negative answer contradicts the context.

---

The One Idea to Remember

The hardest SAT systems questions often become easy once you stop asking:

“Which formula should I use?”

Instead ask:

“What two relationships must be true at the same time?”

Those two relationships are your system.

Once you see them, the rest is usually just algebra.

Find the relationships.
Choose the shortest path.
Answer exactly what the question asks.

Explore the following

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 

ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]

SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]

SYSTEM OF EQUATIONS  [ PART  I ]

system of equations part I

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 

PERCENTAGES

PERCENTAGES [introduction]

PERCENTAGES [part 1]

PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]

SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions

geometry notes  [ part 1 ] 

SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster

SAT Word Problems Made Easy

The Big Word-Problem Families:  SAT Math Guide

Big Word-Problem Families

How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier

How SAT word problems hide equations