Lesson Objectives

- Learn how to solve equations using the Addition Property of Equality
- Learn how to solve equations using the Multiplication Property of Equality
- Learn how to solve any linear equation in one variable using the four-step Method

## How to Solve a Linear Equation

Let's begin by reviewing how to solve an equation using addition or subtraction.

7 + (-7) = 0

-10 + 10 = 0

343 + (-343) = 0

19 + 0 = 19

0 + (-189) = -189

x + 0 = x

0 + z = z

x + 3 = 5

To isolate the variable x, we think about what is being done to x. In this case, we have 3 that is being added to x. If we want to isolate x, we need to get rid of the (+ 3) part on the left side of the equation. We do this by adding (-3) to both sides of the equation since:

3 + (-3) = 0

We can also think about this as subtracting 3 away from each side of the equation:

x + 3 - 3 = 5 - 3

On the left side of the equation:

x + 3 - 3 » x + 0 » x

On the right side of the equation:

5 - 3 » 2

Our equation becomes:

x = 2 ✔

We can always check the solution to an equation by plugging in our solution for the variable in the original equation. If the answer is correct, the left and right sides will simplify to the same value:

x + 3 = 5

2 + 3 = 5

5 = 5

Example 1: Solve each equation

x - 9 = 20

To isolate our variable x, we want to get rid of the (- 9) on the left side of the equation. We can do this by adding 9 to both sides of the equation:

x - 9 + 9 = 20 + 9

x + 0 = 29

x = 29

Check:

x - 9 = 20

29 - 9 = 20

20 = 20 ✔

Example 2: Solve each equation

x + 15 = -14

To isolate our variable x, we want to get rid of the (+ 15) on the left side of the equation. We can do this by adding (-15) to both sides of the equation. We can also think about this as subtracting 15 away from each side of the equation:

x + 15 - 15 = -14 - 15

x = -29

Check:

x + 15 = -14

-29 + 15 = -14

-14 = -14 ✔

Next, let's review how to solve an equation using multiplication or division.

92 • 1 = 92

1 • -55 = -55

1 • x = 1x = x

z • 1 = 1z = z

Example 3: Solve each equation

6x = -42

To isolate our variable x, we want to get rid of the (• 6) on the left side of the equation (6x is the same as 6 • x or x • 6). We can do this by multiplying both sides of the equation by (1/6). We can also think about this as dividing both sides of the equation by 6: $$6x=-42$$ $$\frac{6x}{6}=\frac{-42}{6}$$ $$\frac{\cancel{6}}{\cancel{6}}x=\frac{-7 \cancel{42}}{\cancel{6}}$$ $$x=-7$$ Check:

6x = -42

6(-7) = -42

-42 = -42 ✔

Example 4: Solve each equation $$\frac{4}{5}x=13$$ To isolate our variable x, we want to get rid of the (• 4/5) on the left side of the equation (4/5 x is the same as 4/5 • x or x • 4/5). We can do this by multiplying both sides of the equation by 5/4: $$\frac{5}{4}\cdot \frac{4}{5}x=13 \cdot \frac{5}{4}$$ $$\frac{1\cancel{5}}{1\cancel{4}}\cdot \frac{1\cancel{4}}{1\cancel{5}}x=\frac{65}{4}$$ $$x=\frac{65}{4}$$ Check: $$\frac{4}{5}x=13$$ $$\frac{4}{5}\cdot \frac{65}{4}=13$$ $$\frac{1\cancel{4}}{1\cancel{5}}\cdot \frac{13\cancel{65}}{1\cancel{4}}=13$$ 13 = 13 ✔

ax + b = c

If we follow our steps above:

Step 1) Simplify each side - Each side is simplified

Step 2) Use the addition property of equality to move the variable terms to one side of the equation and the constants to the other:

ax + b = c

ax + b - b = c - b

ax = c - b

Step 3) Use the multiplication property of equality to get the variable by itself: $$\frac{a}{a}x=\frac{c - b}{a}$$ $$\frac{1\cancel{a}}{1\cancel{a}}x=\frac{c - b}{a}$$ $$x=\frac{c - b}{a}$$ Step 4) Check: $$a \cdot \frac{c - b}{a}+ b=c$$ $$\cancel{a}\cdot \frac{c - b}{\cancel{a}}+ b=c$$ $$c - b + b=c$$ $$c=c \hspace{.5em}✔$$ Therefore, if we have a linear equation in one variable in the format of:

ax + b = c

The solution will be:

$$x=\frac{c - b}{a}$$ Let's look at some examples.

Example 5: Solve each equation

-x + 5 = -6x - 15

Step 1) Simplify each side - Each side is simplified

Step 2) Use the addition property of equality to move the variable terms to one side of the equation and the constants to the other:

Let's add (6x) to each side of the equation:

-x + 6x + 5 = -6x + 6x - 15

5x + 5 = -15

Let's subtract 5 away from each side of the equation:

5x + 5 - 5 = -15 - 5

5x = -20

Step 3) Use the multiplication property of equality to get the variable by itself:

Divide each side of the equation by 5: $$\frac{5}{5}x=\frac{-20}{5}$$ $$\frac{1\cancel{5}}{\cancel{5}}x=\frac{-4\cancel{20}}{\cancel{5}}$$ $$x=-4$$ Step 4) Check:

-x + 5 = -6x - 15

-(-4) + 5 = -6(-4) - 15

4 + 5 = 24 - 15

9 = 9 ✔

Example 6: Solve each equation

-2(x + 2) - 8(x + 3) = -6x

Step 1) Simplify each side:

Let's clear the parentheses to start:

-2x - 4 - 8x - 24 = -6x

Now we can combine like terms on each side:

-10x - 28 = -6x

Step 2) Use the addition property of equality to move the variable terms to one side of the equation and the constants to the other:

Let's add (6x) to each side of the equation:

-10x + 6x - 28 = -6x + 6x

-4x - 28 = 0

Let's add 28 to each side of the equation:

-4x - 28 + 28 = 0 + 28

-4x = 28

Step 3) Use the multiplication property of equality to get the variable by itself:

Divide each side of the equation by (-4): $$\frac{-4}{-4}x=\frac{28}{-4}$$ $$\frac{1\cancel{-4}}{\cancel{-4}}x=\frac{-7\cancel{28}}{\cancel{-4}}$$ $$x=-7$$ Step 4) Check:

-2(x + 2) - 8(x + 3) = -6x

-2(-7 + 2) - 8(-7 + 3) = -6(-7)

-2(-5) + (-8)(-4) = 42

10 + 32 = 42

42 = 42 ✔

### Additive Inverse Property

The additive inverse property will be used to solve equations. This property tells us that a number plus its opposite results in 0. Remember, we can obtain the opposite or additive inverse of a number by changing the sign of the number.7 + (-7) = 0

-10 + 10 = 0

343 + (-343) = 0

### Zero is the Additive Identity

When we add zero to any number, the number remains unchanged. For this reason, zero is the "Additive Identity". This may seem really obvious, but when we work with variables it can cause some confusion:19 + 0 = 19

0 + (-189) = -189

x + 0 = x

0 + z = z

### Addition Property of Equality

The addition property of equality tells us we may add the same value to both sides of an equation without changing the solution set (a set whose elements are solutions to the given equation). Since subtraction can be rewritten as a related addition statement, this property covers subtraction as well. When we solve an equation, our goal is to isolate the variable. Let's look at an example.x + 3 = 5

To isolate the variable x, we think about what is being done to x. In this case, we have 3 that is being added to x. If we want to isolate x, we need to get rid of the (+ 3) part on the left side of the equation. We do this by adding (-3) to both sides of the equation since:

3 + (-3) = 0

We can also think about this as subtracting 3 away from each side of the equation:

x + 3 - 3 = 5 - 3

On the left side of the equation:

x + 3 - 3 » x + 0 » x

On the right side of the equation:

5 - 3 » 2

Our equation becomes:

x = 2 ✔

We can always check the solution to an equation by plugging in our solution for the variable in the original equation. If the answer is correct, the left and right sides will simplify to the same value:

x + 3 = 5

2 + 3 = 5

5 = 5

Example 1: Solve each equation

x - 9 = 20

To isolate our variable x, we want to get rid of the (- 9) on the left side of the equation. We can do this by adding 9 to both sides of the equation:

x - 9 + 9 = 20 + 9

x + 0 = 29

x = 29

Check:

x - 9 = 20

29 - 9 = 20

20 = 20 ✔

Example 2: Solve each equation

x + 15 = -14

To isolate our variable x, we want to get rid of the (+ 15) on the left side of the equation. We can do this by adding (-15) to both sides of the equation. We can also think about this as subtracting 15 away from each side of the equation:

x + 15 - 15 = -14 - 15

x = -29

Check:

x + 15 = -14

-29 + 15 = -14

-14 = -14 ✔

Next, let's review how to solve an equation using multiplication or division.

### Multiplicative Inverse Property

The multiplicative inverse property will also be used to solve equations. This property tells us when we multiply a non-zero number by its reciprocal, the result is always 1. $$\require{cancel}12 \cdot \frac{1}{12}=\frac{\cancel{12}\cdot 1}{\cancel{12}}=1$$ $$\frac{5}{17}\cdot \frac{17}{5}=\frac{1\cancel{5}\cdot 1\cancel{17}}{1\cancel{17}\cdot 1\cancel{5}}=1$$### One is the Multiplicative Identity

Similar to having zero as the additive identity, we have 1 as the multiplicative identity. When we multiply a number by one, the number remains unchanged. Again, this may seem really obvious, but when we work with variables it may cause some confusion:92 • 1 = 92

1 • -55 = -55

1 • x = 1x = x

z • 1 = 1z = z

### Multiplication Property of Equality

The multiplication property of equality tells us we may multiply both sides of the equation by the same non-zero number without changing the solution set. Since division can be rewritten as a related multiplication statement, this property covers division as well. Again, when we solve an equation, our goal is to isolate the variable. Let's look at an example. $$\frac{1}{3}x=-12$$ To isolate the variable x, we think about what is being done to x. In this case, we are multiplying x by 1/3. If we want to isolate x, we need to get rid of the (• 1/3) part on the left side of the equation. We do this by multiplying both sides of the equation by 3 since: $$\frac{1}{3}\cdot 3=\frac{1 \cdot \cancel{3}}{\cancel{3}}=1$$ Let's multiply each side of the equation by 3: $$3 \cdot \frac{1}{3}x=-12 \cdot 3$$ $$\frac{1 \cdot \cancel{3}}{\cancel{3}}x=-36$$ $$x=-36$$ Check: $$\frac{1}{3}x=-12$$ $$\frac{1}{3}\cdot -36=-12$$ $$\frac{1 \cdot -12\cancel{36}}{\cancel{3}}=-12$$ -12 = -12 ✔Example 3: Solve each equation

6x = -42

To isolate our variable x, we want to get rid of the (• 6) on the left side of the equation (6x is the same as 6 • x or x • 6). We can do this by multiplying both sides of the equation by (1/6). We can also think about this as dividing both sides of the equation by 6: $$6x=-42$$ $$\frac{6x}{6}=\frac{-42}{6}$$ $$\frac{\cancel{6}}{\cancel{6}}x=\frac{-7 \cancel{42}}{\cancel{6}}$$ $$x=-7$$ Check:

6x = -42

6(-7) = -42

-42 = -42 ✔

Example 4: Solve each equation $$\frac{4}{5}x=13$$ To isolate our variable x, we want to get rid of the (• 4/5) on the left side of the equation (4/5 x is the same as 4/5 • x or x • 4/5). We can do this by multiplying both sides of the equation by 5/4: $$\frac{5}{4}\cdot \frac{4}{5}x=13 \cdot \frac{5}{4}$$ $$\frac{1\cancel{5}}{1\cancel{4}}\cdot \frac{1\cancel{4}}{1\cancel{5}}x=\frac{65}{4}$$ $$x=\frac{65}{4}$$ Check: $$\frac{4}{5}x=13$$ $$\frac{4}{5}\cdot \frac{65}{4}=13$$ $$\frac{1\cancel{4}}{1\cancel{5}}\cdot \frac{13\cancel{65}}{1\cancel{4}}=13$$ 13 = 13 ✔

### Solving Linear Equations in One Variable Using More Than One Property of Equality

Now that we have reviewed how to solve equations using addition or subtraction and how to solve equations with multiplication or division, we should be able to solve any linear equation in one variable using the four-step process.#### Four-Step Method for Solving a Linear Equation in One Variable

- Simplify each side separately
- Clear parentheses or grouping symbols and combine any like terms on each side

- Isolate the variable terms on one side of the equation
- Use the addition property of equality to move all variable terms to one side and all numbers (constants) to the other

- Isolate the variable
- Use the multiplication property of equality to get the variable by itself

- Check
- Replace the variable with the solution in the original equation; the left and right sides should be the same value

ax + b = c

If we follow our steps above:

Step 1) Simplify each side - Each side is simplified

Step 2) Use the addition property of equality to move the variable terms to one side of the equation and the constants to the other:

ax + b = c

ax + b - b = c - b

ax = c - b

Step 3) Use the multiplication property of equality to get the variable by itself: $$\frac{a}{a}x=\frac{c - b}{a}$$ $$\frac{1\cancel{a}}{1\cancel{a}}x=\frac{c - b}{a}$$ $$x=\frac{c - b}{a}$$ Step 4) Check: $$a \cdot \frac{c - b}{a}+ b=c$$ $$\cancel{a}\cdot \frac{c - b}{\cancel{a}}+ b=c$$ $$c - b + b=c$$ $$c=c \hspace{.5em}✔$$ Therefore, if we have a linear equation in one variable in the format of:

ax + b = c

The solution will be:

$$x=\frac{c - b}{a}$$ Let's look at some examples.

Example 5: Solve each equation

-x + 5 = -6x - 15

Step 1) Simplify each side - Each side is simplified

Step 2) Use the addition property of equality to move the variable terms to one side of the equation and the constants to the other:

Let's add (6x) to each side of the equation:

-x + 6x + 5 = -6x + 6x - 15

5x + 5 = -15

Let's subtract 5 away from each side of the equation:

5x + 5 - 5 = -15 - 5

5x = -20

Step 3) Use the multiplication property of equality to get the variable by itself:

Divide each side of the equation by 5: $$\frac{5}{5}x=\frac{-20}{5}$$ $$\frac{1\cancel{5}}{\cancel{5}}x=\frac{-4\cancel{20}}{\cancel{5}}$$ $$x=-4$$ Step 4) Check:

-x + 5 = -6x - 15

-(-4) + 5 = -6(-4) - 15

4 + 5 = 24 - 15

9 = 9 ✔

Example 6: Solve each equation

-2(x + 2) - 8(x + 3) = -6x

Step 1) Simplify each side:

Let's clear the parentheses to start:

-2x - 4 - 8x - 24 = -6x

Now we can combine like terms on each side:

-10x - 28 = -6x

Step 2) Use the addition property of equality to move the variable terms to one side of the equation and the constants to the other:

Let's add (6x) to each side of the equation:

-10x + 6x - 28 = -6x + 6x

-4x - 28 = 0

Let's add 28 to each side of the equation:

-4x - 28 + 28 = 0 + 28

-4x = 28

Step 3) Use the multiplication property of equality to get the variable by itself:

Divide each side of the equation by (-4): $$\frac{-4}{-4}x=\frac{28}{-4}$$ $$\frac{1\cancel{-4}}{\cancel{-4}}x=\frac{-7\cancel{28}}{\cancel{-4}}$$ $$x=-7$$ Step 4) Check:

-2(x + 2) - 8(x + 3) = -6x

-2(-7 + 2) - 8(-7 + 3) = -6(-7)

-2(-5) + (-8)(-4) = 42

10 + 32 = 42

42 = 42 ✔

#### Skills Check:

Example #1

Solve each equation. $$10 - 4x=5(7x + 2)$$

Please choose the best answer.

A

$$x=4$$

B

$$x=0$$

C

$$x=-2$$

D

$$x=8$$

E

$$x=13$$

Example #2

Solve each equation. $$-(7x + 6)=-3x - 2$$

Please choose the best answer.

A

$$x=4$$

B

$$x=7$$

C

$$x=-10$$

D

$$x=2$$

E

$$x=-1$$

Example #3

Solve each equation. $$-6(2x - 3)=-8x + 30$$

Please choose the best answer.

A

$$x=-3$$

B

$$x=13$$

C

$$x=5$$

D

$$x=0$$

E

$$x=-7$$

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test