Question 1164914: Use two equations in two variables to solve the application. See Example 2. (Objective 1)
A rectangle is 3 times as long as it is wide, and its perimeter is 56 centimeters. Find its dimensions.
width
cm
length
cm
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let l = the length and w equal the width.
the perimeter is equal to 2 times (the length and the width).
that would make it 2 * (l + w).
let p = the perimeter.
since p = 56, then your first equation is 56 = 2 * (l + w)
you are given that the rectangle is 3 times as long as it is wide.
this means that l = 3 * w.
that's your second equation.
you need to solve these two equations simultaneously.
that means that the same value of l and w makes both equations true.
start with:
56 = 2 * (l + w)
l = 3 * w
in the first equation, replace l with 3 * w based on its equivalent value from the second equation.
you get:
56 = 2 * (l + w) becomes:
56 = 2 * (3 * w + w)
combine like terms to get:
56 = 2 * 4 * w which is equal to 56 = 8 * w.
solve for w to get:
w = 56 / 8 = 7
since l = 3 * w, then l must be equal to 21.
to confirm, go back to the first equation of 56 = 2 * (l + w)
replace l with 21 and w with 7 to get:
56 = 2 * (21 + 7)
simplify to get:
56 = 56
this confirms the solution is correct.
your solution is that the width of the rectangle is 7 centimeters long and the length of the rectangle is 21 centimeters long.
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