1.6: Evaluating Expressions

If a family pays \(\$ 30\) per phone line, and \(\$ 20\) per \(1 G B\) data, per month to a phone service provider, we can write a mathematical expression to represent the cost that this family pays per month.

Let's use \(x\) to represent the number of phone lines that the family has, and \(y\) to represent the number of \(G B\) of data the family uses. Then \(30 x+20 y\) is a mathematical expression that represents the cost of the family phone services per month.

\(30 x+20 y\) is an algebraic expression. A mathematical expression that consists of variables, numbers and algebraic operations is called an algebraic expression.

Each algebraic expression can contain several terms. For example, the expression above contains two terms: \(30 x\) and \(20 y .\) The numerical factor of each term is called a coefficient. The coefficients of the terms above are \(30\) and \(20,\) respectively. When considering a variable term, we see that it is composed of a numerical coefficient and a variable part. In the term \(30 x\) the numerical coefficient is 30 and the variable part is \(x\).

The value of an algebraic expression can vary. For example, the value of the expression above can vary depending on the number of phone lines and the number of \(\mathrm{GB}\) of data the family uses.

If \(x=2\) (the family uses 2 lines), and \(y=3\) (the family uses \(3 G B\) of data), then the cost of the family phone services is \(30 \cdot 2+20 \cdot 3=\$ 120\) for this month.

If \(x=4,\) and \(y=2,\) then the cost of the family phone devices is \(30 \cdot 4+20 \cdot 2=\) \(\$ 160\) for this month.

Finding the value of the expressions when the variables are substituted by given values is called evaluating an algebraic expression.

Example 4.1

The algebraic expression \(5 x^{3} y-2 y^{2}-z+4,\) which we write using only addition as \(5 x^{3} y+\left(-2 y^{2}\right)+(-z)+4,\) contains four terms: \(5 x^{3} y,-2 y^{2},-z\) and \(+4\). The first three terms are variable terms and the 4 is the constant term. Notice that the coefficient of \(-z\) is \(-1,\) and we usually write \(-z\) instead of \(-1 z .\) In the same way, if the coefficient is \(1,\) we usually omit it.

This process of finding the value of an algebraic expression for particular values of its variables is called evaluating an expression.

Evaluating an expression

1. Replace each variable by the given numerical value.
2. Simplify the resulting expression. Be careful to follow the order of operations.

Helpful tip: When evaluating a variable expression containing a fraction bar, don’t forget to work out the numerator and denominator separately (being careful to follow the order of operations as you do so); finally, divide the numerator by the denominator.

Example 4.2

Evaluate if \(a=1, b=2, c=4,\) and \(d=-1:\)

a) \(\begin{array}{l}
8 b \\
=8(2) \\
=16
\end{array}\)

b) \(\begin{array}{l}
a+c \\
=1+4 \\
=5
\end{array}\)

c) \(\begin{array}{l}
a-d \\
=1-(-1) \\
=1+1 \\
=2
\end{array}\)

d) \(\begin{array}{l}
5 a b \\
=5(1)(2) \\
=10
\end{array}\)

e) \(\begin{array}{l}
\frac{d+a}{b} \\
=\frac{(-1)+1}{2} \\
=\frac{0}{2} \\
=0
\end{array}\)

f) \(\begin{array}{l}
\frac{7 b+c}{a-d} \\
=\frac{7(2)+4}{1-(-1)} \\
=\frac{14+4}{1+1} \\
=\frac{18}{2} \\
=9
\end{array}\)

Example 4.3

Evaluate if \(a=-3, b=5, c=-2,\) and \(d=7:\)

a) \(\begin{array}{l}
4 c-2 b \\
=4(-2)-2(5) \\
=-8-10 \\
=(-8)+(-10) \\
=-18
\end{array}\)

b) \(\begin{array}{l}
b^{2}+b \\
=5^{2}+5 \\
=25+5 \\
=30
\end{array}\)

c) \(\begin{array}{l}
3 c^{2} \\
=3(-2)^{2} \\
=3(4) \\
=12
\end{array}\)

d) \(\begin{array}{l}
(c+a)\left(c^{2}-a c+a^{2}\right) \\
=((-2)+(-3))\left((-2)^{2}-(-3)(-2)+(-3)^{2}\right) \\
=(-5)(4-(-3)(-2)+9) \\
=(-5)(4-6+9) \\
=(-5)(7) \\
=-35
\end{array}\)

e) \(\begin{array}{l}
4 b+5 d-\frac{c}{a} \\
=4(5)+5(7)-\frac{(-2)}{(-3)} \\
=4(5)+5(7)-\frac{2}{3} \\
=20+35-\frac{2}{3} \\
=\frac{60}{3}+\frac{105}{3}-\frac{2}{3} \\
=\frac{60+105-2}{3} \\
=\frac{163}{3} \\
=54 \frac{1}{3}
\end{array}\)

f) \(-d^{2}=-(7)^{2}=-49\)

Application

Here are some applications that come from geometry.

Example 4.4

a) Find the perimeter of a rectangle with length \(33 \mathrm{cm}\) and width \(17 \mathrm{cm}\).

\(\begin{align*}
P &=2 l+2 w \\
&=2(33 \mathrm{cm})+2(17 \mathrm{cm}) \\
&=66 \mathrm{cm}+34 \mathrm{cm} \\
&=100 \mathrm{cm}
\end{align*}\)

b) Find the area of a rectangle that is \(12 \mathrm{cm}\) long and \(16 \mathrm{cm}\) wide.

\(\begin{align*}
A &=l \cdot w \\
&=12 \mathrm{cm} \cdot 16 \mathrm{cm} \\
&=192 \mathrm{cm}^{2}
\end{align*}\)

c) Find the area of a triangle with height of \(20 \mathrm{in}\) and a base of \(30 \mathrm{in}\).

\(\begin{align*}
A &=\frac{1}{2} b \cdot h \\
&=\frac{1}{2} 20 \mathrm{in} \cdot 30 \mathrm{in} \\
&=300 \mathrm{in}^{2}
\end{align*}\)

d) Find the area of a circle with radius \(7 \mathrm{cm},\) round your answer to the nearest tenth.

\(\begin{align*}
A &=\pi r^{2} \\
&=\pi(7 \mathrm{cm})^{2} \\
& \approx 3.14 \cdot 49 \mathrm{cm}^{2} \\
&=153.86 \mathrm{cm}^{2} \\
& \approx 153.9 \mathrm{cm}^{2}
\end{align*}\)

Exit Problem

Evaluate if \(a=-2:-3 a^{2}-4 a-16\)