Ellsworth J. answered 11/07/23
LEARN FROM THE ONE WHO ACTUALLY TAUGHT THE CLASS!
Definitely looks like a simultaneous-equations problem (two equations, two unknowns).
You are given two conditions involving the dimensions (length L and width W) of the rectangle:
1. The total perimeter is 256 feet, and
2. The garden is 8 feet longer than it is wide.
Translate these into equations, solve for L and W, and you’re there!
1) Total perimeter is 256 feet
Write an equation equating the total distance around the rectangle to 256 feet:
Perimeter = L + W + L + W = 256
or
2L + 2W = 256
We can simplify a bit by dividing through by 2:
L + W = 128
2) Garden is 8 feet longer than it is wide
Easy.
L = W + 8
We can solve this extremely system in multiple ways.
The most obvious and straightforward method is substitution, where we substitute one variable in terms of the other, and solve.
In this case, the substitution is handed to us, ready-made!
(You often have to work to get one of the equations into substitution-ready form… that’s probably what’s coming next… shhhh…. :) )
Substitute in the second equation into the first, and solve:
L + W = 128
(W + 8) + W = 128
2W + 8 = 128
2W = 120
W = 60
Now that we have the length, use the other equation to find the width:
L = W + 8
= 60 + 8
= 68
So the length of the garden is 68 feet, and the width is 60 feet.
Let’s check our answer in the original equation (ALWAYS STRONGLY SUGGESTED):
2L + 2W = 256
2(68) + 2(60) = 256
2 56 = 256
It works!