Introduction

In this type of problems, we use linear equations of the form Ax = B.

B is given as a quantity which is a fractional part A of another unknown quantity x.

So, A and B are given and x is to be found here.

Rules to be followed while solving linear equations of the form Ax = B

• The quantity A is usually a fraction and B is a whole number or decimal.
• There is multiplying/dividing on both sides of the equation to isolate the unknown variable x.
• Cross multiplying is done, if required, and the equation is solved to get the value of x.

Some solved examples given below will help in understanding the lesson.

Example 1

A store sold 35 video game consoles on a single day which is one-ninth of their monthly sales. What is then the monthly sales of the store?

Solution

Step 1:

Let the number of game consoles sold in a month = x; Then given 35=1×9

Step 2:

Cross multiplying

x = 35 × 9 = 315

So, total number of game consoles sold in a month = 315

Example 2

Jeanie traveled 15 kilometers of her journey, which is one-eighth of total journey. How long is her total journey then?

Solution

Step 1:

Let the total journey in kilometers be = x; Then given 15=1×8

Step 2:

Cross multiplying

x = 15 × 8 = 120

So, total journey in kilometers = 120 km

Introduction

We come across problems about solutions of equations with parentheses.

In such cases, the parentheses are simplified by using the distributive property of multiplication over addition and subtraction. After simplification, the equations are solved as discussed in previous lesson by following the given rules in such cases.

Let us recall the distributive property of multiplication over addition and subtraction.

For any three numbers a, b, and c

1. ab+c = ab + ac

2. abc = ab ac

The example given below will make it easy to understand how to solve equations with parentheses.

Example 1

Solve for w

7w3 = 28

Solution

Step 1:

Given 7w3 = 28

Using the distributive property of multiplication

7w 7×3 = 28; 7w 21 = 28

Step 2:

The variable to be solved for is w.

Adding 21 to both sides

7w 21 + 21 = 28 + 21 = 49; 7w = 49

Step 3:

Dividing both sides by 7

7w7=497

w = 7 is the solution

Step 4:

Checking the solution

Plugging in w = 7 in the original equation

7w 21 = 28

7 7 21 = 28

49 21 = 28

28 = 28

So, the solution is verified to be correct.

Example 2

Solve for w

4z8 = 20

Solution

Step 1:

Given 4z8 = 20

Dividing both sides of the equation by 4

4(z8)4=204

z 8 = 5

Step 2:

The variable to be solved for is z.

Adding 8 to both sides

z 8 + 8 = 5 + 8 = 13

So, z = 13 is the solution

Step 3:

Checking the solution

Plugging in z = 13 in the original equation

4z8 = 20

4138 = 20

45 = 20

20 = 20

So, the solution is verified to be correct.

Introduction

When we solve an equation, we are solving to find the number that is missing. This missing number is usually represented by a letter. We find the value of that letter or variable to solve the equation.

Rules for Solving 2-Step Equations:

• Identify the variable.We look for the letter in the problem. The variable letter can be any letter, not just x and y2x + 3 = 7, x is the variable; 5w 9 = 17, w is the variableTo solve the equation, we need to isolate the variable or get the variable by itself.
• Add/Subtract whole numbers so theyre all on one side.For example, in the equation 4x 7 = 21, we add 7 to both sides to get the whole numbers all on ones side.4x 7 + 7 = 21 + 7; \: So 4x = 28
• Multiply /Divide to get the variable by itself.For example, 4x = 28; Here we divide both sides of the equation by 44×4=284;x=7
• We check our workWe plug the value of the variable got as solution in the equation to check our work as follows.Given equation is 4x 7 = 21; we plug in the solutionx = 74×7 7 = 2128 7 = 2121 = 21So, the solution is verified to be correct.

Example 1

Solve the following two step equation:

7g + 3 = 24

Solution

Step 1:

We first identify the variable in the given equation

7g + 3 = 24

The only letter in the equation is g and it is the variable.

Step 2:

We add/subtract whole numbers to the equation so all are one side.

Here we subtract 3 from both sides of the equation.

7g + 3 3 = 24 3;

7g = 21

Step 3:

We multiply/divide on both sides of the equation to get the variable by itself

We divide both sides of the equation by 7

7g7=217

g = 3

So, the solution of the equation is g = 3

Step 4:

We check our work by plugging the numbers into the equation.

Here, we plug g = 3 in the equation, 7g + 3 = 24

7 3 + 3 = 24

21 + 3 = 24

So the solution is verified to be correct.

Introduction

The multiplicative property of equality states that we can multiply ordivide both sides of an equation by the same nonzero fractional number oralgebraicexpression without changing the solution.

If a, b and c are any three fractional numbers

If a = b, and c 0, then

1. a × c = b × c

2. a ÷ c = b ÷ c

Example 1

Solve for w

14=2w3

Solution

Step 1:

In this equation, w is multiplied by 23

We can undo this by multiplying both sides of equation by reciprocal 32.

Step 2:

Then, we simplify

14×32=2w3×32

21=1w

Step 3:

w=21

The solution is w=21

Example 2

Solve for w

5w=209

Solution

Step 1:

In this equation, w is multiplied by 5

We can undo this by dividing both sides of equation by 5.

Step 2:

Then, we simplify

5w5=209÷5

Step 3:

1w=209×15

w=49

The solution is w=49

Definition

The multiplicative property of equality states that we can multiply ordivide both sides of an equation by the same nonzero number oralgebraicexpression without changing the solution.

If a, b and c are any three numbers

If a = b, and c 0, then

1. a × c = b × c

2. a ÷ c = b ÷ c

Example 1

Solve for w

12 = 7w

Solution

In this equation, w is multiplied by 7

We can undo this by dividing both sides of equation by 7. Then, we simplify

127=7w7

127=1w

w=127

The solution is w=127, which is a fractional answer

Example 2

Solve for w

5w = 9

Solution

In this equation, w is multiplied by 5

We can undo this by dividing both sides of equation by 5. Then, we simplify

5w5=95

1w=95

w=95

The solution is w=95, which is a fractional answer

Definition

The multiplicative property of equality states that we can multiply ordivide both sides of an equation by the same nonzero number oralgebraicexpression without changing the solution.

If a, b and c are any three numbers

If a = b, and c 0, then

1. a × c = b × c

2. a ÷ c = b ÷ c

Example 1

Solve for x

2x = 3.58

Solution

Step 1:

To solve for x, we must isolate x. On left side of equation, we have 2x; to isolate x, we must divide by 2.

Step 2:

From the multiplicative property of equality with decimals we must divide both sides of an equation by the same number. So, we divide the both sides by 2 to get

2xx=3.582

Step 3:

Simplifying

3.582=1.79

So, the solution is x = 1.79

Example 2

Solve for x

x3=4.27

Solution

Step 1:

To solve for x, we must isolate x. On left side of equation, we have x3; to isolate x, we must multiply by 3.

Step 2:

From the multiplicative property of equality with decimals we must multiply both sides of an equation by the same number. So, we multiply both sides by 3 to get

x3×3=4.27×3

Step 3:

Simplifying

4.27 × 3 = 1281

So, the solution is x = 12.81

Definition

The additive property of equality states that we can add orsubtract the same number oralgebraicexpression to both sides of an equation without changing the solution.

If a, b and c are any three numbers

if a = b, then

1. a + c = b + c

2. a c = b c

Example 1

Solve for x

x + 2.7 = 8.9

Solution

Step 1:

To solve for x, we must isolate x. On left side of equation, we have x + 2.7; to isolate x, we must subtract 2.7.

Step 2:

From the additive property of equality with decimals we must subtract from both sides of an equation the same number. So, we subtract 2.7 from both sides as follows

x + 2.7 2.7 = 8.9 2.7

Step 3:

Simplifying

x = 8.9 2.7 = 6.2

So, x = 6.2

Example 2

Solve for x

x 1.3 = 11.7

Solution

Step 1:

To solve for x, we must isolate x. On left side of equation, we have x 1.3; to isolate x, we must add 1.3

Step 2:

From the additive property of equality with decimals we must add to both sides of an equation the same number. So, we add 1.3 to both sides as follows

x 1.3 + 1.3 = 11.7 + 1.3

Step 3:

Simplifying

x = 11.7 + 1.3 = 13.0

So, x = 13

This tutorial provides comprehensive coverage of Equations and Applications based on Common Core CCSS and State Standards and its prerequisites. Students can navigate learning paths based on their level of readiness. Institutional users may customize the scope and sequence to meet curricular needs. This simple tutorial uses appropriate examples to help you understand Equations and Applications in a general and quick way.

Audience

This tutorial has been prepared for beginners to help them understand the basics of Equations and Applications. After completing this tutorial, you will find yourself at a moderate level of expertise in Equations and Applications, from where you can advance further.

Prerequisites

Before proceeding with this tutorial, you need a basic knowledge of elementary math concepts such as number sense, addition, subtraction, multiplication, division, whole numbers, fractions, decimals, algebraic concepts, equations, and so on.