Question

Define a variable and write an equation that can be used to determine on average how much the annual per-pupil spending has increased from 1986 to 2001. Solve the problem

Short answer

The value of the average annual per-pupil spending has increased from 1986 to 2001 is $ 267.33

Step-by-step solution

– Describe a variable.

To solve the word problems, it is necessary to define a variable and form an equation.

For this problem, choose a variable that gives the average annual per-pupil spending has increased from 1986 to 2001.

– Write an equation that describes the given situation.

From the graph, the annual per-pupil spending in 1986 is $\$3,479$and the annual per-pupil spending in 2001 is $\$7,489.$

Write the equation as follows:

$\begin{array}{c}\text{averageannualfrom1986to2001}=\frac{\left(\begin{array}{l}\text{annualperpupil}\\ \text{spending2001}\end{array}\right)-\left(\begin{array}{l}\text{annualperpupil}\\ \text{spending1986}\end{array}\right)}{2001-1986}\\ x=\frac{\left(\begin{array}{l}\text{annualperpupil}\\ \text{spending2001}\end{array}\right)-\left(\begin{array}{l}\text{annualperpupil}\\ \text{spending1986}\end{array}\right)}{15}\end{array}$

– Solve the equation

Substitute $7,489$for the annual per-pupil spending 2001 and $3,479$for the annual per-pupil spending 1986 in the equation:

localid="1648725829493" $\begin{array}{l}x=\frac{\left(\begin{array}{l}\text{annualperpupil}\\ \text{spending2001}\end{array}\right)-\left(\begin{array}{l}\text{annualperpupil}\\ \text{spending1986}\end{array}\right)}{15}\\ x=\frac{7489-3479}{15}\\ x=\frac{4010}{15}\\ x=267.33\end{array}$

So, the average annual per-pupil spending has increased from 1986 to 2001 is $ 267.33