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Study Guides > Mathematics for the Liberal Arts Corequisite

Simplifying and Evaluating Expressions Using the Distributive Property

Learning Outcomes

  • Simplify an algebraic expression using the distributive property
  • Evaluate an algebraic expression for a given value using the distributive property

Using the Distributive Property With the Order of Operations

Sometimes we need to use the Distributive Property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.

example

Simplify: [latex]8 - 2\left(x+3\right)[/latex] Solution:
[latex]8--2(x+3)[/latex]
Distribute. [latex]8--2\cdot x--2\cdot 3[/latex]
Multiply. [latex]8--2x--6[/latex]
Combine like terms. [latex]--2x+2[/latex]
 

try it

[ohm_question]146971[/ohm_question] [ohm_question]146972[/ohm_question]
 

example

Simplify: [latex]4\left(x - 8\right)-\left(x+3\right)[/latex]

Answer: Solution:

[latex]4(x--8)--(x+3)[/latex]
Distribute. [latex]4x--32--x--3[/latex]
Combine like terms. [latex]3x--35[/latex]

 

try it

[ohm_question]146973[/ohm_question]
In the following example, we simplify more expressions that require the distributive property. https://youtu.be/STfLvYhDhwk

Evaluate Expressions Using the Distributive Property

Some students need to be convinced that the Distributive Property always works. In the examples below, we will practice evaluating some of the expressions from previous examples; in part 1, we will evaluate the form with parentheses, and in part 2 we will evaluate the form we got after distributing. If we evaluate both expressions correctly, this will show that they are indeed equal.

example

When [latex]y=10[/latex] evaluate: 1. [latex]6\left(5y+1\right)[/latex] 2. [latex]6\cdot 5y+6\cdot 1[/latex]

Answer: Solution:

1.
[latex]6\left(5y+1\right)[/latex]
Substitue [latex]\color{red}{10}[/latex] for y. [latex]6(5\cdot\color{red}{10}+1)[/latex]
Simplify in the parentheses. [latex]6\left(51\right)[/latex]
Multiply. [latex]306[/latex]
2.
[latex]6\cdot 5y+6\cdot 1[/latex]
Substitute [latex]\color {red}{10}[/latex] for y. [latex]6\cdot 5\cdot\color {red}{10}+6\cdot 1[/latex]
Simplify. [latex]300+6[/latex]
Add. [latex]306[/latex]
Notice, the answers are the same. When [latex]y=10[/latex],

[latex]6\left(5y+1\right)=6\cdot 5y+6\cdot 1[/latex]

Try it yourself for a different value of [latex]y[/latex].

 

try it

[ohm_question]146974[/ohm_question]
 

example

When [latex]y=3[/latex], evaluate 1. [latex]-2\left(4y+1\right)[/latex] 2. [latex]-2\cdot 4y+\left(-2\right)\cdot 1[/latex]

Answer: Solution:

1.
[latex]-2\left(4y+1\right)[/latex]
Substitute [latex]\color{red}{3}[/latex] for y. [latex]--2(4\cdot\color{red}{3}+1)[/latex]
Simplify in the parentheses. [latex]-2\left(13\right)[/latex]
Multiply. [latex]-26[/latex]
2.
[latex]-2\cdot 4y+\left(-2\right)\cdot 1[/latex]
Substitute [latex]\color{red}{3}[/latex] for y. [latex]--2\cdot 4\cdot\color{red}{3}+(--2)\cdot 1[/latex]
Multiply. [latex]-24 - 2[/latex]
Subtract. [latex]-26[/latex]
The answers are the same. When [latex]y=3[/latex], [latex]-2\left(4y+1\right)=-8y - 2[/latex]

 

try it

[ohm_question]146975[/ohm_question]
 

example

When [latex]y=35[/latex] evaluate 1. [latex]-\left(y+5\right)[/latex] 2. [latex]-y-5[/latex] to show that [latex]-\left(y+5\right)=-y-5[/latex]

Answer: Solution:

1.
[latex]-\left(y+5\right)[/latex]
Substitute [latex]\color{red}{35}[/latex] for y. [latex]--(\color{red}{35}+5)[/latex]
Add in the parentheses. [latex]-\left(40\right)[/latex]
Simplify. [latex]-40[/latex]
2.
[latex]-y - 5[/latex]
Substitute [latex]\color{red}{35}[/latex] for y. [latex]--\color{red}{35}--5[/latex]
Simplify. [latex]-40[/latex]
The answers are the same when [latex]y=35[/latex], demonstrating that [latex]-\left(y+5\right)=-y-5[/latex]

 

try it

[ohm_question]146986[/ohm_question]
The following video provides another way to show that the distributive property works. https://youtu.be/05jYrJn7W-M

Licenses & Attributions

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  • Question ID 146986, 146975, 146974, 146973, 146972, 146971. Authored by: Lumen Learning. License: CC BY: Attribution.

CC licensed content, Shared previously

  • Evaluate Numerical Expressions Using Order of Operations and Distribution. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Ex 3: Combining Like Terms Requiring Distribution. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.

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