Q52. - Algebra 2

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

Step-by-Step Solution

Answer

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

Step 1 – 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.

Step 2 – 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:

$\begin{array}{c}\text{average annual from 1986 to 2001}=\frac{\left(\begin{array}{l}\text{annual per pupil }\\ \text{spending 2001}\end{array}\right)-\left(\begin{array}{l}\text{annual per pupil }\\ \text{spending 1986}\end{array}\right)}{2001-1986}\\ x=\frac{\left(\begin{array}{l}\text{annual per pupil }\\ \text{spending 2001}\end{array}\right)-\left(\begin{array}{l}\text{annual per pupil }\\ \text{spending 1986}\end{array}\right)}{15}\end{array}$

Step 3 – 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:

$\begin{array}{l}x=\frac{\left(\begin{array}{l}\text{annual per pupil }\\ \text{spending 2001}\end{array}\right)-\left(\begin{array}{l}\text{annual per pupil }\\ \text{spending 1986}\end{array}\right)}{15}\\ x=\frac{7489-3479}{15}\\ x=\frac{4010}{15}\\ x=267.33\end{array}$

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