SAT Algebra Linear Equations: Step-by-Step Guide to Solving Linear Equations on the Digital SAT (Part 2)

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SAT Algebra Linear Equations:  Step-by-Step Guide to Solving Linear Equations on the Digital SAT (Part 2)

Solving Equations with Variables on Both Sides, Fractions, Decimals, Ratios, and Proportions

In Part 1, you learned how to solve one-step and two-step linear equations by isolating the variable one operation at a time. Those skills form the foundation of almost every algebra problem on the Digital SAT.

In this chapter, you'll solve equations that look more complicated because variables appear on both sides of the equation. You'll also learn how to work confidently with fractions, decimals, ratios, and proportions—topics that frequently appear in SAT Math questions. Although these problems may seem challenging at first, they all follow the same golden rule of algebra:

Perform the same operation on both sides of the equation while keeping the equation balanced.

Once you understand that principle, every new type of equation becomes much easier to solve.

Solving Equations with Variables on Both Sides

Many SAT questions contain variables on both sides of the equation.

For example,

5x + 8 = 2x + 20

At first glance, students often wonder which variable to solve first. The answer is simple:

Move all the variables to one side and all the numbers to the other side.

Example 1

Solve

5x + 8 = 2x + 20

Step 1: Look at both sides.

The left side contains:

5x + 8

The right side contains:

2x + 20

Both sides contain a variable.

Our first goal is to move all the x terms to the same side.

Step 2: Remove the smaller variable.

Subtract 2x from both sides.

5x + 8 − 2x = 2x + 20 − 2x

Step 3: Simplify.

On the right side,

+2x and −2x cancel.

3x + 8 = 20

Now every variable is on the left side.

Step 4: Remove the constant.

Subtract 8 from both sides.

3x + 8 − 8 = 20 − 8

Step 5: Simplify.

The +8 and −8 cancel.

3x = 12

Step 6: Remove the multiplication.

Divide both sides by 3.

3x ÷ 3 = 12 ÷ 3

Step 7: Simplify.

x = 4

Step 8: Check.

Original equation:

5(4) + 8 = 2(4) + 20

20 + 8 = 8 + 20

28 = 28

The answer is correct.

Example 2

Solve

7x − 5 = 4x + 16

Step 1

Variables appear on both sides.

Subtract 4x from both sides.

7x − 5 − 4x = 4x + 16 − 4x

Step 2

Simplify.

3x − 5 = 16

Step 3

Add 5 to both sides.

3x − 5 + 5 = 16 + 5

Step 4

Simplify.

3x = 21

Step 5

Divide both sides by 3.

3x ÷ 3 = 21 ÷ 3

Step 6

Simplify.

x = 7

Step 7

Check.

7(7) − 5 = 4(7) + 16

49 − 5 = 28 + 16

44 = 44

Correct.

Example 3

Solve

9x + 12 = 6x + 30

Step 1

Subtract 6x from both sides.

9x + 12 − 6x = 6x + 30 − 6x

Step 2

Simplify.

3x + 12 = 30

Step 3

Subtract 12 from both sides.

3x + 12 − 12 = 30 − 12

Step 4

Simplify.

3x = 18

Step 5

Divide both sides by 3.

3x ÷ 3 = 18 ÷ 3

Step 6

Simplify.

x = 6

Step 7

Check.

9(6)+12=6(6)+30

54+12=36+30

66=66

Correct.

Which Variable Should You Move?

Students often ask:

"Should I move the variable on the left or the one on the right?"

Either method works.

However, it is usually easier to move the smaller coefficient.

Example:

9x = 4x + 20

Subtracting 4x produces

5x = 20

which is simpler than subtracting 9x.

Working with Fractions

Fractions appear frequently on the SAT.

Fortunately, the solving process is exactly the same.

Example 4

Solve

x/4 = 9

Step 1

The variable has been divided by 4.

We must undo the division.

Step 2

Multiply both sides by 4.

(x/4) × 4 = 9 × 4

Step 3

The 4 cancels.

x = 36

Step 4

Check.

36 ÷ 4 = 9

Correct.

Example 5

Solve

x/5 + 6 = 14

Step 1

The variable has been divided by 5.

Before removing the division, remove the addition.

Subtract 6 from both sides.

x/5 + 6 − 6 = 14 − 6

Step 2

Simplify.

x/5 = 8

Step 3

Undo the division.

Multiply both sides by 5.

(x/5) × 5 = 8 × 5

Step 4

Simplify.

x = 40

Step 5

Check.

40/5 + 6

8 + 6

14

Correct.

Example 6

Solve

2 + x/3 = 11

Step 1

Subtract 2 from both sides.

2 + x/3 − 2 = 11 − 2

Step 2

Simplify.

x/3 = 9

Step 3

Multiply both sides by 3.

(x/3) × 3 = 9 × 3

Step 4

Simplify.

x = 27

Step 5

Check.

2 + 27/3

2 + 9

11

Correct.

Clearing Fractions

Sometimes every term contains a fraction.

Instead of solving with fractions, remove them first.

Example 7

Solve

x/2 + x/3 = 10

Step 1

Find the Least Common Denominator (LCD).

The denominators are:

2 and 3

The LCD is:

6

Step 2

Multiply every term by 6.

6(x/2)+6(x/3)=6(10)

Step 3

Simplify.

3x+2x=60

Step 4

Combine like terms.

5x=60

Step 5

Divide both sides by 5.

5x÷5=60÷5

Step 6

Simplify.

x=12

Step 7

Check.

12/2 +12/3

6+4

10

Correct.

Working with Decimals

SAT questions sometimes contain decimals instead of fractions.

Many students become nervous when they see decimals, but decimals follow exactly the same algebra rules.

Example 8

Solve

0.5x = 9

Step 1

The variable has been multiplied by 0.5.

Undo the multiplication.

Divide both sides by 0.5.

0.5x ÷0.5 =9÷0.5

Step 2

Simplify.

x=18

Step 3

Check.

0.5 ×18

9

Correct.

Example 9

Solve

2.4x +1.2 =13.2

Step 1

Subtract 1.2 from both sides.

2.4x +1.2−1.2 =13.2−1.2

Step 2

Simplify.

2.4x=12

Step 3

Divide both sides by 2.4.

2.4x÷2.4 =12÷2.4

Step 4

Simplify.

x=5

Step 5

Check.

2.4(5)+1.2

12+1.2

13.2

Correct.

Ratios and Proportions

Many SAT word problems involve ratios.

A proportion is simply two equal fractions.

Example:

x/8 = 6/12

Example 10

Solve

x/8 =6/12

Step 1

Notice that

6/12 simplifies to

1/2

However, we can solve directly.

Step 2

Cross multiply.

12 × x =8 ×6

12x=48

Step 3

Divide both sides by 12.

12x÷12=48÷12

Step 4

Simplify.

x=4

Step 5

Check.

4/8=6/12

1/2=1/2

Correct.

SAT Strategy

Whenever fractions appear,

ask yourself,

"Can I remove the fractions first?"

Whenever decimals appear,

ask yourself,

"Would converting them to fractions make this easier?"

Many difficult SAT algebra questions become much simpler after removing fractions or decimals.

Practice Questions

Solve each equation.

  1. 6x + 9 = 3x + 24

  2. 8x − 7 = 5x + 20

  3. x/6 = 8

  4. x/4 + 7 = 15

  5. 2 + x/5 = 10

  6. 0.25x = 12

  7. 3.5x + 7 = 28

  8. x/3 + x/6 = 15

  9. x/10 = 7/14

  10. 4x + 18 = 2x + 34

Answers

  1. x = 5

  2. x = 9

  3. x = 48

  4. x = 32

  5. x = 40

  6. x = 48

  7. x = 6

  8. x = 30

  9. x = 5

  10. x = 8


As equations become more complex, the underlying algebra never changes. Whether variables appear on both sides, fractions need to be cleared, decimals are involved, or ratios must be solved using proportions, the objective is always to isolate the variable while keeping the equation balanced. By practicing these methods carefully and checking each solution, you'll develop the accuracy and confidence needed for more challenging SAT algebra, Digital SAT math, linear equation solving, and SAT word problem questions. In Part 3, you'll apply these equation-solving skills to linear functions, slope, intercepts, graphs, and systems of linear equations—the concepts that connect algebra with coordinate geometry and mathematical modeling on the Digital SAT.

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