14.0k views
2 votes
Solve #118. Algebra ASAP

Solve #118. Algebra ASAP-example-1

1 Answer

4 votes

To solve this, we need to set up two different equations, then use them to find what we are looking for.

First, we need to change the words in the problem into variables and numbers. "The length of a rectangle is 3mm less than four times the width." This translates into: L=4W-3

Now, we are given the area and are told to solve for the dimensions (meaning the height and width. We need to use the area equation for a rectangle to finish the problem. A=L x W

Substitute the equation we created for the length into the area equation and we can solve for the width.

A= L x W --> 1387=(4W-3) x W --> 1387=4W^2-3W --> 0=4W^2-3W-1387

0=4(W-19)(W+18.25)

So our width either equals 19 or -18.25, and since a distance can't really be negative, our width is 19mm. Now we can use this to solve for length.

L=4W-3 --> L=4(19)-3 --> L=76-3 --> L=73

Our length is 73 mm.

Check: We can check this answer by multiplying the two numbers to see if they provide the area given in the problem. A=73x19=1387mm^2

User Bbholzbb
by
7.6k points