The area of a rectangle is 28 square inches. The length is 8 more than thrice the width. Find the length and the width.

Question

help solve

Answer 1

 You can think of

 

  WL = 28

   L=3w + 8

 

    W( 3W +8) = 28

 

    3 W^2 + 8W + = 28

 

    3W^2 + 8W -28 = 0  (1)

 

    3 W^2 - 6 W + !4 W -28 =0-

     3 W ( W - 2 ) + 14 ( W -2) =0

 

     (3W+ 14) ( w-2) = 0

 

     w = 2     L = 3(2) + 8 = 14 , then LW =28 

   ignore the negative answer of w = -14/3

   

    If  we write all factors of 28 { 2, 4 , 7, 14)

 

          We see that  28 = 2 * 14 ,   14 = 2* 3  + 8

 

     Therefore we can have a rectangle with area of 28 units, with 2, 14 be its dimension , and have the

    

      property of L = 2*3 +8 ( solution by inspection).

                        

  

 

 

Answer 2

Hi Victoria,

 

Area=Length x width= 28 sq in

 

L=8+3w

 

So substitute that back into the equation:

 

Area = L x W

   28 sq in = (8+3w) x w

   28 sq in = 8w +3w²

   28 sq in = w (8 + 3w)

 

Possible factors of 28 are 2 and 14 or 4 and 7, and 1 and 28.  Substituting the possibilities in the equation, we find that the correct set of factors are 2 and (8 + 3*2) or 14. 

Answer 3

w=width

l=3w+8

A=l*w

28=(3w+8)(w)

28=3w^2+8w

3w^2+8w-28=0

(3w+14)(w-2)=0

3w+14=0

3w=-14

w=-14/3, disregard this answer because it is negative

w-2=0

w=2

l=3*2+8

l=14

check: 28=14*2=28

l=14 inches

w=2 inches

14 is 8 more than 3 times 2 !

 

Answer 4

1) A=LW = 28 in

2) L = 3W + 8

 

Substitute L = 28/W into the 2nd equation

 

28/W = 3W + 8

28 = 3W^2 + 8W

3W^2 + 8W - 28 = 0

(3W+ 14)(W - 2) = 0

 

Note that the answer must be positive therefore the second 'zero' is the answer

 

W=2 inches

L=14 inches