Mathematics » Properties of Real Numbers » Distributive Property

# Evaluating Expressions Using the Distributive Property

## Evaluating 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 , we will evaluate the form with parentheses, and in part 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 $$y=10$$ evaluate: $$\phantom{\rule{0.2em}{0ex}}6\left(5y+1\right)$$ $$\phantom{\rule{0.2em}{0ex}}6·5y+6·1.$$

### Solution

 $$6\left(5y+1\right)$$ Simplify in the parentheses. $$6\left(51\right)$$ Multiply. $$306$$
 Simplify. Add.

Notice, the answers are the same. When $$y=10,$$

$$6\left(5y+1\right)=6·5y+6·1.$$

Try it yourself for a different value of $$y.$$

## Example

When $$y=3,$$ evaluate $$\phantom{\rule{0.2em}{0ex}}-2\left(4y+1\right)$$$$\phantom{\rule{0.2em}{0ex}}-2·4y+\left(-2\right)·1.$$

### Solution

 $$\phantom{\rule{0.2em}{0ex}}-2\left(4y+1\right)$$ Simplify in the parentheses. $$-2\left(13\right)$$ Multiply. $$-26$$
 $$\phantom{\rule{0.2em}{0ex}}-2·4y+\left(-2\right)·1$$ Multiply. $$-24-2$$ Subtract. $$-26$$ The answers are the same. When $$y=3,$$ $$-2\left(4y+1\right)=-8y-2$$

## Example

When $$y=35$$ evaluate $$\phantom{\rule{0.2em}{0ex}}\text{−}\left(y+5\right)$$ and $$\phantom{\rule{0.2em}{0ex}}\text{−}\mathit{\text{y}}-5$$ to show that $$-\left(y+5\right)=\text{−}\mathit{\text{y}}-5.$$

### Solution

 $$\phantom{\rule{0.2em}{0ex}}\text{−}\left(y+5\right)$$ Add in the parentheses. $$-\left(40\right)$$ Simplify. $$-40$$
 $$\phantom{\rule{0.2em}{0ex}}-y-5$$ Simplify. $$-40$$ The answers are the same when $$y=35,$$ demonstrating that $$-\left(y+5\right)=\text{−}y-5$$

## Key Concepts

• Distributive Property:
• If $$a,b,c$$ are real numbers then
• $$a\left(b+c\right)=ab+ac$$
• $$\left(b+c\right)a=ba+ca$$
• $$a\left(b\cdot c\right)=ab\cdot ac$$

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