a rectangle has an area of 52 inches squared. its length is 3 inches less than four times its width. find the length and width

Question

please help me with this problem 

Answer 1

Area of a rectangle is length times width:  A = LW

 

LW = 52

 

L = 4W - 3

 

Substitute 4W-3 in place of L in 1st equation

(4W-3)(W) = 52

4W² - 3W = 52

4W² - 3W - 52 = 0

 

This can be done with factor by grouping

4(-52) = -208

Factors of -208 that add to -3 are (-16)(13)

Replace -3W with -16W+13W

 

4W² - 16W + 13W - 52 = 0

4W(W-4) + 13(W-4) = 0

(4W+13)(W-4) = 0

 

Either 4W+13 = 0 and W = -13/4

OR W-4 = 0 and W = 4

 

Discarding the negative, the width is 4 inches

 

L = 4W-3 = 4(4)-3 = 16-3 = 13 inches

 

The rectangle is 13 by 4 inches