Question

two brothers went shopping at a back - to - school sale where all shirts and shorts were the same prices. the younger brother spent $175 on 7 new shirts and 7 pairs of shorts. the older brother purchased 6 new shirts and 7 pairs of shorts and paid a total of $165. the system of equations is as follows: x = new shirt and y = new shorts: 7x + 7y = $175 6x + 7y = $165 how much did each shirt and each pair of shorts cost? shirts $7; shorts $7 shirts $7; shorts $6 shirts $15; shorts $10 shirts $10; shorts $15

Explanation:

Step1: Subtract the two equations

We have the system of equations:
$$7x + 7y = 175$$
$$6x + 7y = 165$$
Subtract the second equation from the first equation to eliminate \(y\):
$$(7x + 7y)-(6x + 7y)=175 - 165$$
Simplify the left - hand side: \(7x+7y - 6x - 7y=x\), and the right - hand side: \(175 - 165 = 10\). So we get \(x = 10\).

Step2: Substitute \(x = 10\) into one of the original equations

Substitute \(x = 10\) into the first equation \(7x+7y=175\).
We have \(7\times10+7y = 175\).
First, calculate \(7\times10 = 70\), so the equation becomes \(70+7y=175\).
Subtract 70 from both sides: \(7y=175 - 70\).
Calculate \(175 - 70 = 105\), so \(7y = 105\).
Divide both sides by 7: \(y=\frac{105}{7}=15\).

Answer:

shirts \$10; shorts \$15