Question

the perimeter of a rectangle is 32 meters and the length is 4 meters longer than the width. find the dimensions of the rectangle. let (x) be the length and (y) be the width. the corresponding modeling system is (\begin{cases}2x + 2y=32\\x - y = 4end{cases}). use the method of substitution to solve this system.

Explanation:

Step1: Isolate x from the second - equation

From $x - y=4$, we get $x=y + 4$.

Step2: Substitute x into the first - equation

Substitute $x=y + 4$ into $2x+2y = 32$. Then $2(y + 4)+2y=32$. Expand the left - hand side: $2y+8 + 2y=32$. Combine like terms: $4y+8 = 32$. Subtract 8 from both sides: $4y=32 - 8=24$. Divide both sides by 4: $y = 6$.

Step3: Find the value of x

Substitute $y = 6$ into $x=y + 4$. Then $x=6 + 4=10$.

Answer:

Length of the rectangle = 10 meters
Width of the rectangle = 6 meters