What if the paycheck for your summer job were represented by the algebraic expression \begin{align*}10h + 25\end{align*}, where *h* is the number of hours you work? If you worked 20 hours last week, how could you find the value of your paycheck? After completing this Concept, you'll be able to evaluate algebraic expressions like this one.

### Guidance

When we are given an algebraic expression, one of the most common things we might have to do with it is **evaluate** it for some given value of the variable. The following example illustrates this process.

#### Example A

*Let \begin{align*}x = 12\end{align*}. Find the value of \begin{align*}2x - 7\end{align*}.*

**Solution:**

To find the solution, we substitute 12 for \begin{align*}x\end{align*} in the given expression. Every time we see \begin{align*}x\end{align*}, we replace it with 12.

\begin{align*}2x - 7 &= 2(12) - 7\\ &= 24 - 7\\ &= 17\end{align*}

**Note:** At this stage of the problem, we place the substituted value in parentheses. We do this to make the written-out problem easier to follow, and to avoid mistakes. (If we didn’t use parentheses and also forgot to add a multiplication sign, we would end up turning \begin{align*}2x\end{align*} into 212 instead of 2 times 12!)

#### Example B

*Let \begin{align*}y = -2. \end{align*} Find the value of \begin{align*} \frac {7} {y} - 11 y + 2 \end{align*}.*

**Solution**

\begin{align*}\frac {7} {(-2)} - 11( -2 ) + 2 &= -3 \frac { 1 } { 2 } + 22 + 2\\ &= 24 - 3 \frac { 1 } { 2 }\\ &= 20 \frac { 1 } { 2 }\end{align*}

Many expressions have more than one variable in them. For example, the formula for the perimeter of a rectangle in the introduction has two variables: length \begin{align*}(l)\end{align*} and width \begin{align*}(w)\end{align*}. In these cases, be careful to substitute the appropriate value in the appropriate place.

#### Example C

*The area of a trapezoid is given by the equation \begin{align*} A = \frac{ h } { 2 } (a + b)\end{align*}. Find the area of a trapezoid with bases \begin{align*}a = 10 \ cm\end{align*} and \begin{align*}b = 15 \ cm\end{align*} and height \begin{align*}h = 8 \ cm\end{align*}.*

**Solution:**

To find the solution to this problem, we simply take the values given for the variables \begin{align*}a, \ b,\end{align*} and \begin{align*}h\end{align*}, and plug them in to the expression for \begin{align*}A\end{align*}:

\begin{align*}& A = \frac { h } { 2 }(a + b) \qquad \text{Substitute} \ 10 \ \text{for} \ a, \ 15 \ \text{for} \ b, \ \text{and} \ 8 \ \text{for} \ h.\\ & A = \frac { 8 } { 2 }(10 + 15) \quad \text{Evaluate piece by piece.} \ 10 + 15 = 25; \ \frac { 8 } { 2 } = 4 .\\ & A = 4(25) = 100\end{align*}

*The area of the trapezoid is 100 square centimeters.*

Vocabulary

- When given an algebraic expression, one of the most common things we might have to do with it is
**evaluate**it for some given value of the variable. We substitute the value in for the variable and simplify the expression.

### Guided Practice

*Let \begin{align*}x= 3\end{align*} and \begin{align*}y = -2. \end{align*} Find the value of \begin{align*} 3xy + \frac{6}{y}-2x \end{align*}.*

**Solution**

\begin{align*}3xy + \frac{6}{y}-2x &= 3(3)(-2) + \frac{6}{-2}-2(3)\\ &= -18-3-6)\\ &= -27\end{align*}