Friday, July 10, 2026

Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

Three pipes A, B and C can together fill a tank in 8 hours. After working at it together for 2 hours, B is closed and A and C fill the remaining part in 9 hours. Determine the time in which pipe B alone can fill the tank.

A + B + C together fill the tank in 8 hours

So, rate of (A + B + C) = [1/8] tank per hour

Work done by A + B + C in 2 hours = 2 × 1/8 =[ 1/4 ]tank

Remaining work = 1 − 1/4 = [3/4 ]tank

Remaining [3/4] tank is filled by A + C in 9 hours

So, rate of (A + C) = 3/4 × 1/9 = 1/12 tank per hour

Rate of B = Rate of (A + B + C) − Rate of (A + C)

= 1/8 − 1/12

= 3/24 − 2/24 =[ 1/24 [tank per hour

Time taken by B alone = 24 hours

see this video for more explanation

pipes problem, cbse 12th applied maths old board exam question paper 2025 2026

Solving Quadratic Equations by Completing the Square The Digital SAT Math Guide to Quadratic Equations (Part 2)

The Digital SAT Math Guide to Quadratic Equations (Part 2)

Solving Quadratic Equations by Completing the Square

In the previous chapter, you learned how to solve quadratic equations by factoring. Factoring is often the quickest method, but many quadratic equations on the Digital SAT cannot be factored easily. Some have large coefficients, some produce fractional values, and others have no integer factors at all.

For these equations, completing the square provides a systematic method that always works. Unlike factoring, you do not have to guess factor pairs or recognize patterns. Instead, you follow the same sequence of algebraic steps every time.

Before learning the procedure, remember one important rule.

The coefficient of x² should be 1 before you begin completing the square.

A quadratic equation whose coefficient of x² is 1 is called a monic quadratic equation.

If the equation is not monic, divide every term on both sides of the equation by the coefficient of x². This makes the remaining steps much easier and reduces mistakes.

Universal Method for Completing the Square

Always follow these steps.

Write the equation in standard form.
If the coefficient of x² is not 1, divide every term on both sides by that coefficient.
Move the constant term to the opposite side.
Take half of the coefficient of x.
Square that number.
Add the squared value to both sides.
Rewrite the left side as the square of a binomial.
Take the square root of both sides.
Remember both the positive and negative square roots.
Solve the resulting linear equations.
Check every solution in the original equation.

Example 1

Solve

2x² + 12x + 4 = 0

Step 1

The coefficient of x² is 2.

Divide every term on both sides by 2.

2x² ÷ 2 + 12x ÷ 2 + 4 ÷ 2 = 0 ÷ 2

Simplify.

x² + 6x + 2 = 0

The equation is now monic.

Step 2

Subtract 2 from both sides.

x² + 6x + 2 − 2 = 0 − 2

Simplify.

x² + 6x = −2

Step 3

Take half of 6.

6 ÷ 2 = 3

Square it.

3² = 9

Step 4

Add 9 to both sides.

x² + 6x + 9 = −2 + 9

Simplify.

x² + 6x + 9 = 7

Step 5

Rewrite the left side.

(x + 3)² = 7

Step 6

Take square roots.

√((x + 3)²) = ±√7

Simplify.

x + 3 = ±√7

Step 7

Subtract 3 from both sides.

Positive solution:

x = −3 + √7

Negative solution:

x = −3 − √7

Example 2

Now solve a question that produces fractions immediately after making the quadratic monic.

4x² + 10x − 3 = 0

Step 1

The coefficient of x² is 4.

Divide every term by 4.

4x² ÷ 4 + 10x ÷ 4 − 3 ÷ 4 = 0 ÷ 4

Simplify.

x² + ⁵⁄₂x − ³⁄₄ = 0

Notice that fractions are perfectly acceptable. Do not convert them to decimals because exact fractions make later calculations more accurate.

Step 2

Move the constant term.

Add ³⁄₄ to both sides.

x² + ⁵⁄₂x = ³⁄₄

Step 3

Take half of the coefficient of x.

The coefficient is ⁵⁄₂.

Half of ⁵⁄₂ is

⁵⁄₂ ÷ 2 = ⁵⁄₄

Now square the result.

(⁵⁄₄)² = ²⁵⁄₁₆

Step 4

Add ²⁵⁄₁₆ to both sides.

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁄₄ + ²⁵⁄₁₆

Convert ³⁄₄ to sixteenths.

³⁄₄ = ¹²⁄₁₆

Now add.

¹²⁄₁₆ + ²⁵⁄₁₆ = ³⁷⁄₁₆

The equation becomes

x² + ⁵⁄₂x + ²⁵⁄₁₆ = ³⁷⁄₁₆

Step 5

Rewrite the left side.

(x + ⁵⁄₄)² = ³⁷⁄₁₆

Step 6

Take square roots.

√((x + ⁵⁄₄)²) = ±√(³⁷⁄₁₆)

Simplify.

x + ⁵⁄₄ = ±√37⁄4

Step 7

Subtract ⁵⁄₄ from both sides.

x = −⁵⁄₄ ± √37⁄4

These are the exact solutions.

Notice that completing the square works just as well with fractions as it does with whole numbers. On the Digital SAT, leaving answers in exact fractional or radical form is often the correct approach unless the question specifically asks for a decimal approximation.

Thursday, July 9, 2026

The Digital SAT Math Guide to Quadratic Equations (Part 1)

Master Quadratic Equations for the Digital SAT with Step-by-Step Explanations

Quadratic equations are one of the most important algebra topics on the Digital SAT. They appear in many forms, from straightforward equation-solving questions to graph interpretation, mathematical modeling, and real-world word problems. A strong understanding of quadratics also makes it much easier to learn functions, parabolas, coordinate geometry, and polynomial expressions.

Unlike linear equations, which produce straight lines when graphed, quadratic equations create curved graphs called parabolas. Learning how these equations behave will help you answer a wide variety of SAT Math questions quickly and accurately.

This guide is written for students who want to build a solid understanding of quadratics from the ground up. Every solution is explained one step at a time, with no skipped steps or unexplained shortcuts. By the time you finish this chapter, you'll understand what quadratic equations are, how to recognize them, and how to solve many of them by factoring.

Learning Goals

In this chapter, you will learn how to:

Recognize a quadratic equation.
Understand why quadratic equations are different from linear equations.
Identify the standard form of a quadratic equation.
Understand quadratic expressions and quadratic functions.
Solve simple quadratic equations by factoring.
Apply the Zero Product Property.
Check your answers correctly.
Avoid common mistakes made on the Digital SAT.

These concepts form the foundation for more advanced methods such as completing the square and using the quadratic formula, which will be covered in later chapters.

What Is a Quadratic Equation?

A quadratic equation is an equation in which the highest exponent of the variable is 2.

Examples include:

x² = 49

x² + 5x + 6 = 0

2x² − 7x + 3 = 0

4x² = 100

Notice that each equation contains x².

That squared variable is what makes the equation quadratic.

Compare these two equations.

Linear equation:

2x + 7 = 13

Highest exponent = 1

Quadratic equation:

x² + 2x − 15 = 0

Highest exponent = 2

The difference may seem small, but it changes how the equation behaves. A linear equation usually has one solution, while a quadratic equation can have two solutions, one solution, or no real solutions.

The Standard Form of a Quadratic Equation

Most quadratic equations on the SAT are written in standard form:

ax² + bx + c = 0

Each letter has a meaning.

a is the coefficient of x².

b is the coefficient of x.

c is the constant term.

For example,

3x² + 8x − 11 = 0

Here,

a = 3

b = 8

c = −11

Learning to identify these three values is important because later methods, especially the quadratic formula, use them directly.

Understanding the Parts of a Quadratic Equation

Consider

2x² + 9x − 18 = 0

This equation has three terms.

First term:

2x²

This is called the quadratic term because it contains x².

Second term:

This is called the linear term because it contains x.

Third term:

−18

This is the constant term because it contains no variable.

Recognizing these parts helps you identify which solving method to use.

What Does It Mean to Solve a Quadratic Equation?

Solving a quadratic equation means finding every value of the variable that makes the equation true.

For example,

x² = 25

Which numbers produce 25 when squared?

5² = 25

(−5)² = 25

Therefore,

x = 5

and

x = −5

Unlike linear equations, quadratic equations often have more than one correct answer.

Why Are There Two Answers?

Many students are surprised to discover that one equation can have two solutions.

The reason is simple.

Squaring removes the negative sign.

Positive example:

5 × 5 = 25

Negative example:

−5 × −5 = 25

Both calculations produce the same answer.

Whenever you solve an equation involving x², always ask yourself whether both a positive and a negative solution are possible.

Introduction to Factoring

Factoring x² + bx + c type when a = 1

Factoring is one of the fastest methods for solving many quadratic equations on the Digital SAT.

Factoring means rewriting an expression as the product of two smaller expressions.

Example:

x² + 5x + 6

can be written as

(x + 2)(x + 3)

When multiplied together,

(x + 2)(x + 3)

= x² + 3x + 2x + 6

= x² + 5x + 6

The original expression and its factored form are mathematically identical.

The Zero Product Property

Factoring works because of an important algebra rule.

A × B = 0

then

A = 0

B = 0

or both.

This rule is called the Zero Product Property.

Example:

(x + 4)(x − 7) = 0

Either

x + 4 = 0

x − 7 = 0

Solve each equation separately.

First equation:

x + 4 = 0

Subtract 4 from both sides.

x + 4 − 4 = 0 − 4

Simplify.

x = −4

Second equation:

x − 7 = 0

Add 7 to both sides.

x − 7 + 7 = 0 + 7

Simplify.

x = 7

Therefore,

the two solutions are

x = −4

and

x = 7

Example 1

Solve

x² + 7x + 12 = 0

Step 1

Write the equation.

x² + 7x + 12 = 0

Step 2

Find two numbers whose product is 12 and whose sum is 7.

Possible factor pairs of 12 are

1 and 12

2 and 6

3 and 4

Only

3 and 4

add to 7.

Step 3

Write the factors.

(x + 3)(x + 4) = 0

Step 4

Apply the Zero Product Property.

Either

x + 3 = 0

x + 4 = 0

Step 5

Solve the first equation.

Subtract 3 from both sides.

x + 3 − 3 = 0 − 3

Simplify.

x = −3

Step 6

Solve the second equation.

Subtract 4 from both sides.

x + 4 − 4 = 0 − 4

Simplify.

x = −4

Final Answer

x = −3

x = −4

Example 2

Solve

x² − 9x + 20 = 0

Step 1

Find two numbers whose product is 20.

1 and 20

2 and 10

4 and 5

Step 2

Which pair adds to −9?

Since the product is positive and the sum is negative,

both numbers must be negative.

−4 and −5

Step 3

Write the factors.

(x − 4)(x − 5) = 0

Step 4

Set each factor equal to zero.

x − 4 = 0

x − 5 = 0

Step 5

Solve.

Add 4 to both sides.

x = 4

Add 5 to both sides.

x = 5

Check

4² − 9(4) + 20

16 − 36 + 20

Correct.

Now check 5.

25 − 45 + 20

Correct.

Both answers satisfy the equation.

Example 3

Solve

x² + x − 12 = 0

Step 1

Find two numbers whose product is −12.

Possible pairs include

1 and −12

2 and −6

3 and −4

Step 2

Find the pair whose sum equals 1.

4 and −3

Step 3

Write the factors.

(x + 4)(x − 3) = 0

Step 4

Set each factor equal to zero.

x + 4 = 0

x − 3 = 0

Step 5

Solve.

Subtract 4 from both sides.

x = −4

Add 3 to both sides.

x = 3

Final Answer

x = −4

x = 3

A Quick Factoring Strategy

Whenever you see

x² + bx + c

ask yourself two questions.

Question 1

Which two numbers multiply to give c?

Question 2

Do those same numbers add to give b?

If the answer is yes,

you have found the correct factors.

With practice, this process becomes much faster.

Forgetting to Check

Substitute every solution back into the original equation.

If the equation balances,

your solution is correct.

Practice Questions

Solve by factoring.

x² + 5x + 6 = 0
x² − 8x + 15 = 0
x² + 9x + 20 = 0
x² − 7x + 10 = 0
x² + 2x − 15 = 0
x² − x − 12 = 0
x² + 11x + 24 = 0
x² − 10x + 24 = 0

Answers

x = −2, −3
x = 3, 5
x = −4, −5
x = 2, 5
x = 3, −5
x = 4, −3
x = −3, −8
x = 4, 6

A quadratic equation is an equation whose highest exponent is two. Before attempting to solve it, identify whether it is already in standard form and determine the values of a, b, and c. When the equation can be factored, rewriting it as the product of two binomials often provides the quickest solution. The Zero Product Property then allows each factor to be solved separately, producing all possible solutions. As you continue practicing, you'll begin to recognize common factor patterns quickly, an essential skill for success on the Digital SAT Math section.

Wednesday, July 8, 2026

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

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

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:

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

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

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.

6x + 9 = 3x + 24
8x − 7 = 5x + 20
x/6 = 8
x/4 + 7 = 15
2 + x/5 = 10
0.25x = 12
3.5x + 7 = 28
x/3 + x/6 = 15
x/10 = 7/14
4x + 18 = 2x + 34

Answers

x = 5
x = 9
x = 48
x = 32
x = 40
x = 48
x = 6
x = 30
x = 5
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.

SAT Algebra Linear Equations Step-by-Step Guide to Solving f Linear Equations on the Digital SAT (Part 1)

SAT Algebra Linear Equations: Step-by-Step Guide to Solving Linear Equations on the Digital SAT (Part 1)

Build a Strong Algebra Foundation for a High SAT Math Score

Success on the Digital SAT Math section begins with mastering algebra. Among all the algebra topics tested, linear equations are the most fundamental because they appear directly in equation-solving questions and indirectly in linear functions, graph interpretation, systems of equations, coordinate geometry, mathematical modeling, and many real-world word problems.

Many students believe algebra is about memorizing formulas. In reality, algebra is about logical thinking. Every equation tells a mathematical story, and every solution follows a sequence of logical steps. Once you understand those steps, even difficult-looking SAT questions become manageable.

This guide explains every concept carefully, assuming no prior knowledge beyond basic arithmetic. Each example is solved one step at a time, with an explanation for every operation performed. Instead of simply showing the answer, you'll learn why each step works. By the end of this chapter, you'll be able to solve one-step and two-step linear equations confidently, avoid common mistakes, and build the foundation needed for more advanced SAT algebra topics.

What Is a Linear Equation?

A linear equation is an equation in which every variable has an exponent of one. When the equation is represented on a graph, it forms a straight line rather than a curve.

Some examples of linear equations are:

x + 5 = 12
3x − 7 = 20
4y = 36
2a + 9 = 19

Although these equations look different, they all follow exactly the same mathematical principles.

Every linear equation contains an unknown value called a variable. Your goal is to determine the value of the variable that makes the equation true.

Understanding Variables

A variable is simply a symbol that represents an unknown number.

Instead of writing

□ + 8 = 15

mathematicians write

x + 8 = 15

The letter x can represent any number.

If x equals 7, then

7 + 8 = 15

Both sides are equal, so the equation is true.

The variable does not always have to be x.

It may also be

The letter changes, but the method of solving the equation never changes.

What Does It Mean to Solve an Equation?

To solve an equation means to find the value of the variable that makes both sides exactly equal.

Imagine an old-fashioned balance scale.

If both sides contain the same weight, the scale remains perfectly balanced.

If you remove weight from one side only, the balance tips.

To keep the balance level, whatever you do to one side must also be done to the other side.

This simple idea is the foundation of all algebra.

The Golden Rule of Algebra

Whatever operation you perform on one side of an equation must also be performed on the other side.

This rule never changes.

If you add 6 to one side, add 6 to the other side.

If you subtract 10 from one side, subtract 10 from the other side.

If you multiply one side by 4, multiply the other side by 4.

If you divide one side by 7, divide the other side by 7.

Following this rule ensures that both sides remain equal throughout the solution.

Understanding Inverse Operations

An inverse operation is an operation that reverses another operation.

Operation	Inverse Operation
Addition	Subtraction
Subtraction	Addition
Multiplication	Division
Division	Multiplication

For example,

if 8 has been added,

subtract 8.

If a number has been multiplied by 5,

divide by 5.

Inverse operations allow us to remove numbers one by one until only the variable remains.

Solving One-Step Linear Equations

One-step equations require only one operation to isolate the variable.

Although these questions are among the easiest on the SAT, learning them thoroughly makes later topics much easier.

Example 1

Solve

x + 9 = 18

Step 1: Identify the operation attached to the variable.

The variable x has 9 added to it.

Since our goal is to leave x by itself, we must remove the +9.

Step 2: Choose the inverse operation.

The opposite of adding 9 is subtracting 9.

Therefore, subtract 9 from both sides of the equation.

x + 9 − 9 = 18 − 9

Step 3: Simplify both sides.

On the left side,

+9 and −9 cancel each other.

x = 9

On the right side,

18 − 9 = 9

Therefore,

x = 9

Step 4: Check the answer.

Substitute 9 into the original equation.

9 + 9 = 18

18 = 18

Both sides are equal.

The solution is correct.

Example 2

Solve

x − 14 = 23

Step 1

The variable has 14 subtracted from it.

We must remove the −14.

Step 2

The opposite of subtracting 14 is adding 14.

Add 14 to both sides.

x − 14 + 14 = 23 + 14

Step 3

The −14 and +14 cancel.

x = 37

Step 4

Check.

37 − 14 = 23

23 = 23

The answer is correct.

Example 3

Solve

6x = 48

Step 1

The variable has been multiplied by 6.

Our goal is to remove the multiplication.

Step 2

The opposite of multiplying by 6 is dividing by 6.

Divide both sides by 6.

6x ÷ 6 = 48 ÷ 6

Step 3

On the left side,

the 6 in the numerator and denominator cancel.

x = 8

On the right side,

48 ÷ 6 = 8

Therefore,

x = 8

Step 4

Check.

6 × 8 = 48

48 = 48

The answer is correct.

Example 4

Solve

x ÷ 5 = 12

Step 1

The variable has been divided by 5.

We need to undo the division.

Step 2

The opposite of division is multiplication.

Multiply both sides by 5.

(x ÷ 5) × 5 = 12 × 5

Step 3

The 5 in the numerator and denominator cancel.

x = 60

Step 4

Check.

60 ÷ 5 = 12

12 = 12

The solution is correct.

Solving Two-Step Linear Equations

Two-step equations require removing one operation before removing another.

Always work from the outside toward the variable.

Never try to remove the multiplication before removing the addition or subtraction.

Example 5

Solve

3x + 7 = 25

Step 1

The variable x has first been multiplied by 3.

After that,

7 has been added.

Since addition happened last,

we remove the addition first.

Step 2

Subtract 7 from both sides.

3x + 7 − 7 = 25 − 7

Step 3

Simplify.

+7 and −7 cancel.

3x = 18

Step 4

The variable is still multiplied by 3.

Undo the multiplication by dividing both sides by 3.

3x ÷ 3 = 18 ÷ 3

Step 5

Simplify.

The 3 cancels.

x = 6

Step 6

Check.

3(6) + 7 = 25

18 + 7 = 25

25 = 25

The solution is correct.

Example 6

Solve

5x − 20 = 35

Step 1

The variable has 20 subtracted.

Remove the subtraction first.

Add 20 to both sides.

5x − 20 + 20 = 35 + 20

Step 2

Simplify.

−20 and +20 cancel.

5x = 55

Step 3

The variable is multiplied by 5.

Divide both sides by 5.

5x ÷ 5 = 55 ÷ 5

Step 4

Simplify.

x = 11

Step 5

Check.

5 × 11 − 20 = 35

55 − 20 = 35

35 = 35

The answer is correct.

Example 7

Solve

8x + 16 = 64

Step 1

Subtract 16 from both sides.

8x +16 −16 =64 −16

Step 2

Simplify.

8x =48

Step 3

Divide both sides by 8.

8x ÷8 =48 ÷8

Step 4

Simplify.

x =6

Step 5

Check.

8(6)+16=64

48+16=64

64=64

Correct.

If an equation says

4x + 12

the multiplication happened first,

then the addition.

When solving,

remove the addition first,

then the multiplication.

Thinking this way makes multi-step equations much easier.

Working with Negative Numbers

Negative numbers scare many students, but the solving process never changes.

Treat them exactly like positive numbers while paying close attention to the signs.

Example

Solve

−4x = 28

Step 1

The variable is multiplied by −4.

Undo the multiplication by dividing both sides by −4.

−4x ÷ −4 = 28 ÷ −4

Step 2

Simplify.

The −4 cancels.

x = −7

Step 3

Check.

−4(−7)=28

28=28

Correct.

Tuesday, July 7, 2026

Show that f: R → R defined as f(x) = x / sqrt(1 + x^2) is one-one but not onto.

Show that f: R → R defined as f(x) = x / √(1 + x²) is one-one but not onto.

f: R → R, f(x) = x / √(1 + x²)

To check One-One / Injective

Let x₁, x₂ ∈ R

such that f(x₁) = f(x₂)

x₁ / √(1 + x₁²) = x₂ / √(1 + x₂²) --------[1]

Squaring both sides:

x₁² / (1 + x₁²) = x₂² / (1 + x₂²)

x₁²(1 + x₂²) = x₂²(1 + x₁²)

x₁² + x₁²x₂² = x₂² + x₁²x₂²

x₁² = x₂²

x₁ = ± x₂

[1] is possible only if x₁, x₂ have the same sign

x₁ = -x₂, is rejected

we have to conclude that x₁ =x₂,

Therefore f is one-one.

To check Onto / Surjective

Let y = x / √(1 + x²)

squaring

y² = x² / (1 + x²)

y²(1 + x²) = x²

y² + y²x² = x²

y² = x² - y²x² = x²(1 - y²)

x² = y² / (1 - y²)

For x to be real, RHS ≥ 0

Since y² ≥ 0, we need 1 - y² > 0

⇒ y² < 1

⇒ -1 < y < 1

Range of f = (-1, 1) ≠ R

Therefore f is not onto.

see this video for more explanation

cbse 12th maths old board exam question paper 2025 2026 one to one injective function onto function

SAT Math Formula Cheat Sheet for Quick Reference

SAT Math Formulas

The Digital SAT includes a reference sheet with some geometry formulas, but it does not include everything you'll need. Knowing the most common formulas before test day helps you solve problems more quickly and reduces the chance of making simple mistakes.

Exponent Rules

For any nonzero number a:

a⁰ = 1

a¹ = a

aᵐ × aⁿ = aᵐ⁺ⁿ

aᵐ ÷ aⁿ = aᵐ⁻ⁿ

(aᵐ)ⁿ = aᵐⁿ

(ab)ⁿ = aⁿbⁿ

(a/b)ⁿ = aⁿ/bⁿ

a⁻ⁿ = 1/aⁿ

Remember

Multiply → add exponents.
Divide → subtract exponents.
A negative exponent means take the reciprocal.

Radicals

√a × √b = √(ab)

√a ÷ √b = √(a/b)

Examples

√49 = 7

√81 = 9

∛125 = 5

Perfect squares worth memorizing:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

121, 144, 169, 196, 225, 256, 289, 324, 361, 400

Linear Equations

Slope

m = (y₂ − y₁)/(x₂ − x₁)

Slope-intercept form

y = mx + b

Point-slope form

y − y₁ = m(x − x₁)

Standard form

Ax + By = C

Parallel lines have the same slope.

Perpendicular lines have negative reciprocal slopes.

Coordinate Geometry

Distance Formula

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Midpoint Formula

((x₁ + x₂)/2, (y₁ + y₂)/2)

Quadratic Equations

Standard form

ax² + bx + c = 0

Quadratic Formula

x = (−b ± √(b² − 4ac))/2a

Discriminant

b² − 4ac

Positive → two real solutions

Zero → one real solution

Negative → no real solutions

Vertex

x = −b/(2a)

Factoring Identities

Difference of Squares

a² − b² = (a + b)(a − b)

Perfect Square Trinomials

a² + 2ab + b² = (a + b)²

a² − 2ab + b² = (a − b)²

Circle Formulas

Circumference

C = 2πr

Area

A = πr²

Arc Length

(θ/360) × 2πr

Sector Area

(θ/360) × πr²

Diameter = 2r

Rectangles and Squares

Rectangle

Area = length × width

Perimeter = 2(length + width)

Square

Area = side²

Perimeter = 4 × side

Diagonal = side√2

Triangles

Area

½ × base × height

Pythagorean Theorem

a² + b² = c²

The angles inside every triangle add up to 180°.

An exterior angle equals the sum of the two opposite interior angles.

Special Right Triangles

45°–45°–90°

1 : 1 : √2

30°–60°–90°

1 : √3 : 2

These ratios are tested regularly.

Trigonometry

SOH CAH TOA

sin θ = opposite/hypotenuse

cos θ = adjacent/hypotenuse

tan θ = opposite/adjacent

tan θ = sin θ/cos θ

Polygons

Interior Angle Sum

(n − 2) × 180°

Each Interior Angle of a Regular Polygon of n vertics

[(n − 2) × 180°]/n

Volume

Cube

s³

Rectangular Prism

lwh

Cylinder

πr²h

Cone

⅓πr²h

Sphere

⁴⁄₃πr³

Surface Area

Cube

6s²

Cylinder

2πr² + 2πrh

Sphere

4πr²

Mean

Sum of all values ÷ Number of values

Weighted Mean

Σ(value × weight) ÷ Σ(weights)

Probability

P(Event)

Number of Favorable Outcomes ÷ Total number of Outcomes

Complement Rule

P(complement event ) =1 − P(Event)

A probability is always between 0 and 1.

Percents

Increase

Original × (1 + rate)

Decrease

Original × (1 − rate)

Percent Change

(New − Original)/Original × 100%

Simple Interest

I = Prt

P = Principal

r = Interest Rate

t = Time

Exponential Growth and Decay

Growth

A = P(1 + r)ᵗ

Decay

A = P(1 − r)ᵗ

Functions

Example

f(x) = 2x + 3

f(5) = 13

Replace x with the given value.

Useful Constants

π ≈ 3.14

√2 ≈ 1.414

√3 ≈ 1.732