### Multiple Variable Expressions

When given an algebraic expression, one of the most common things to do with it is **evaluate** it for some given value of the variable.

Take a look at this example to see how this works:

Let x = 12. Find the value of 2x - 7.

To find the solution, substitute 12 in place of \begin{align*}x\end{align*} in the given expression.

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

**Note:** In the first step of the problem, keep the substituted value in parentheses. This makes the written-out problem easier to follow, and helps 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!")

#### Evaluating an Expression

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

\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, \begin{align*}P=2l+2w\end{align*}, has two variables: length \begin{align*}(l)\end{align*} and width \begin{align*}(w).\end{align*} Be careful to substitute the appropriate value in the appropriate place.

#### Evaluating an Expression with Multiple Variables

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*}.

To find the solution to this problem, substitute the given values for variables \begin{align*}a, \ b,\end{align*} and \begin{align*}h\end{align*} in place of the appropriate letters in the equation.

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

### Example

#### Example 1

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*}.

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

### Review

Evaluate questions 1 through 8, using \begin{align*}a = -3, \ b = 2, \ c = 5,\end{align*} and \begin{align*}d = -4.\end{align*}

- \begin{align*}2a + 3b\end{align*}
- \begin{align*}4c + d\end{align*}
- \begin{align*}5ac - 2b\end{align*}
- \begin{align*} \frac { 2a } { c - d }\end{align*}
- \begin{align*} \frac { 3b } { d }\end{align*}
- \begin{align*} \frac { a - 4b } { 3c + 2d }\end{align*}
- \begin{align*} \frac { 1 } { a + b }\end{align*}
- \begin{align*} \frac { ab } {cd }\end{align*}

For questions 9 through 11, the weekly cost \begin{align*}C\end{align*} of manufacturing \begin{align*}x\end{align*} remote controls is given by the formula \begin{align*}C = 2000 + 3x\end{align*}, where the cost is given in dollars.

- What is the cost of producing 1000 remote controls?
- What is the cost of producing 2000 remote controls?
- What is the cost of producing 2500 remote controls?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 1.2.