Question

A teacher spends $150 on school supplies for the classroom, folders are $0.25 each and notebooks are $2.50 each This linear relationship can be represented by 0.25x + 2.50y = 150, where I represents the number of folders and y represents the number of notebooks. .
Identify the z-intercept Explain what this means in the context of the problem.

Identify the y-intercept. Explain what this means in the context of the problem.

Identify one additional combination of school supplies that the teacher could have purchased with their budget.

Explain what this means in the context of the problem.​

Answer 1

Answer:

a. $x = 600$

b. $y = 60$

c. x = 150 and y = 45

Step-by-step explanation:

Given

$0.25x + 2.50y = 150$

Solving (a): The x intercept;

To do this; set y to 0

$0.25x + 2.50y = 150$

$0.25x + 2.50 * 0 = 150$

$0.25x + 0 = 150$

$0.25x = 150$

Solve for x

$x = 150/0.25$

$x = 600$

This implies the amount spent on folders when no notebook was purchased

Solving (b): The y intercept

To do this; set x to 0

$0.25x + 2.50y = 150$

$0.25 * 0 + 2.50y = 150$

$0 + 2.50y = 150$

$2.50y = 150$

Solve for y

$y = 150/2.5$

$y = 60$

This implies the amount spent on notebooks when no folder was purchased

Solving (c): Other values of x and y

This will be solved using trial by error method;

Set x = 150

Solve for y

$0.25x + 2.50y = 150$

$0.25 * 150 + 2.50y = 150$

$37.5+ 2.50y = 150$

Solve for 2.50y

$2.50y = 150 - 37.5$

$2.50y = 112.5$

Solve for y

$y = 112.5/2.50$

$y = 45$

Hence;

x = 150 and y = 45

This implies that with the budget of $150, the teacher can afford to purchase 150 folders and 45 notebooks