Solving Equations Using the Distributive Property

πŸ†Practice solving quadratic equations using factoring

Solving an equation using the distributive property is related to the need to open the parentheses as the first step to then be able to simplify similar members. When an equation contains one or more pairs of parentheses, we must start by opening them all and then proceed to the next phase.Β 

Below, we provide you with some examples where this method is applied.

2(X+3)=8 2\left(X+3\right)=8

In this equation, we can clearly see some parentheses. To start, we must open them (that is, apply the distributive property) and then we can proceed with the following phases of the exercise.

2X+6=8 2X+6=8

2X=2 2X=2

X=1 X=1

The result of the equation is 1 1 .

Solving equations using the distributive property


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Solve for x:

\( 2(4-x)=8 \)

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Another example

5(X+2)=3(X+4) 5\left(X+2\right)=3\left(X+4\right)

In this equation, we clearly see that there are two pairs of parentheses, one on each side. To begin, we must open them (that is, apply the distributive property) and then we can proceed with the following phases of the exercise.

5X+10=3X+12 5X+10=3X+12

2X=2 2X=2

X=1 X=1

The result of the equation is 1 1 .


If this article interested you, you might also be interested in the following articles

  • First-degree equations with one unknown
  • What is the unknown of a mathematical equation?
  • Equivalent equations
  • Transposition of terms
  • Solving equations by adding or subtracting the same number from both sides
  • Solving equations by multiplying or dividing both sides by the same number
  • Solving equations by simplifying like terms
  • Solution of an equation
  • Exponential equations

On the Tutorela website, you will find a wide variety of articles about mathematics


Examples and exercises with solutions for solving equations using the distributive property

examples.example_title

Solve for x:

7(βˆ’2x+5)=77 7(-2x+5)=77

examples.explanation_title

To open parentheses we will use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

(7Γ—βˆ’2x)+(7Γ—5)=77 (7\times-2x)+(7\times5)=77

We multiply accordingly

βˆ’14x+35=77 -14x+35=77

We will move the 35 to the right section and change the sign accordingly:

βˆ’14x=77βˆ’35 -14x=77-35

We solve the subtraction exercise on the right side and we will obtain:

βˆ’14x=42 -14x=42

We divide both sections by -14

βˆ’14xβˆ’14=42βˆ’14 \frac{-14x}{-14}=\frac{42}{-14}

x=βˆ’3 x=-3

examples.solution_title

-3

examples.example_title

Solve x:

5(x+3)=0 5(x+3)=0

examples.explanation_title

We open the parentheses according to the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

5Γ—x+5Γ—3=0 5\times x+5\times3=0

5x+15=0 5x+15=0

We will move the 15 to the right section and keep the corresponding sign:

5x=βˆ’15 5x=-15

Divide both sections by 5

5x5=βˆ’155 \frac{5x}{5}=\frac{-15}{5}

x=βˆ’3 x=-3

examples.solution_title

βˆ’3 -3

examples.example_title

Solve for x:

βˆ’3(12x+4)=12 -3(\frac{1}{2}x+4)=\frac{1}{2}

examples.explanation_title

We open the parentheses on the left side by the distributive property and use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

βˆ’32xβˆ’12=12 -\frac{3}{2}x-12=\frac{1}{2}

We multiply all terms by 2 to get rid of the fractions:

βˆ’3xβˆ’12Γ—2=1 -3x-12\times2=1

βˆ’3xβˆ’24=1 -3x-24=1

We will move the minus 24 to the right section and keep the corresponding sign:

βˆ’3x=24+1 -3x=24+1

βˆ’3x=25 -3x=25

Divide both sections by minus 3:

βˆ’3xβˆ’3=25βˆ’3 \frac{-3x}{-3}=\frac{25}{-3}

x=βˆ’253 x=-\frac{25}{3}

examples.solution_title

βˆ’253 -\frac{25}{3}

examples.example_title

Solve for x:

βˆ’9(2βˆ’x)=(x+4)β‹…3 -9(2-x)=(x+4)\cdot3

examples.explanation_title

We open the parentheses in both sections by the distributive property and use the formula:

a(x+b)=ax+ab a(x+b)=ax+ab

βˆ’18+9x=3x+12 -18+9x=3x+12

We move 3X to the left section, and 18 to the right section and maintain the corresponding signs:

9xβˆ’3x=12+18 9x-3x=12+18

We add the terms:

6x=30 6x=30

We divide both sections by 6:

6x6=306 \frac{6x}{6}=\frac{30}{6}

x=5 x=5

examples.solution_title

5

examples.example_title

Solve for x:

βˆ’8(2x+4)=6(xβˆ’4)+3 -8(2x+4)=6(x-4)+3

examples.explanation_title

To open parentheses we will use the formula:

a(x+b)=ax+ab a\left(x+b\right)=ax+ab

(βˆ’8Γ—2x)+(βˆ’8Γ—4)=(6Γ—x)+(6Γ—βˆ’4)+3 (-8\times2x)+(-8\times4)=(6\times x)+(6\times-4)+3

We multiply accordingly:

βˆ’16xβˆ’32=6xβˆ’24+3 -16x-32=6x-24+3

Calculate the elements on the right section:

βˆ’16xβˆ’32=6xβˆ’21 -16x-32=6x-21

In the left section we enter the elements with the X and in the left section those without the X, remember to change the plus and minus signs as appropriate when transferring:

βˆ’32+21=6x+16x -32+21=6x+16x

Calculate the elements accordingly

βˆ’11=22x -11=22x

We divide the two sections by 22

βˆ’1122=22x22 -\frac{11}{22}=\frac{22x}{22}

βˆ’12=x -\frac{1}{2}=x

examples.solution_title

βˆ’12 -\frac{1}{2}

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