SOLUTION: a rectangle has a width of x+9 and its width is x+7. Find the dimensions of the rectangle if its area is 143 square meters.

Algebra ->  Equations -> SOLUTION: a rectangle has a width of x+9 and its width is x+7. Find the dimensions of the rectangle if its area is 143 square meters.      Log On


   

Question 1036138: a rectangle has a width of x+9 and its width is x+7. Find the dimensions of the rectangle if its area is 143 square meters.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is A+=+L%2AW where A is the area, L is the length and W is the width.


In this case, A = 143, L = x+7 and W = x+9.


Let's plug these expressions in. Then let's solve for x.


A+=+L%2AW


A+=+%28x%2B7%29%2A%28x%2B9%29


143+=+%28x%2B7%29%2A%28x%2B9%29


143+=+x%2A%28x%2B9%29%2B7%2A%28x%2B9%29


143+=+x%5E2%2B9x%2B7x%2B63


143+=+x%5E2%2B16x%2B63


143-143+=+x%5E2%2B16x%2B63-143


x%5E2%2B16x-80=0


%28x-4%29%28x%2B20%29=0


x-4=0 or x%2B20=0


x=4 or x=-20


-------------------------


The possible solutions are x=4 or x=-20


-------------------------


If x=4, then


L = x+7 = 4+7 = 11


W = x+9 = 4+9 = 13


So if x = 4, then the rectangle is 13 meters by 11 meters. Notice how 11*13 = 143


-------------------------


If x+=+-20, then


L = x+7 = -20+7 = -13


W = x+9 = -20+9 = -11


but having negative side lengths doesn't make much sense. So we ignore the solution x = -20


-------------------------


The only practical solution is x = 4


That leads to the dimensions of the rectangle being 13 meters by 11 meters