SOLUTION: Smedley has just drawn two rectangles. The length of the first rectangle is 4cm more than the length of the second. The width of the first is 9cm. The width of the second is 6cm. I

Question 507117: Smedley has just drawn two rectangles. The length of the first rectangle is 4cm more than the length of the second. The width of the first is 9cm. The width of the second is 6cm. If the area of the first rectangle is 96 square cm greater than the area of the second, find the length of each.
I tried: I drew rectangles. And my equation was [9(x+4)]+96=6x
9x+36+96=6x
9x+132=6x
132=-3x
Divide by 3.
X=-44
First of all it can't be negative, second... It doesn't work. I checked it and it doesn't work out.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The length of the first rectangle is 4cm more than the length of the second.
The width of the first is 9cm.
The width of the second is 6cm.
If the area of the first rectangle is 96 square cm greater than the area of the second, find the length of each.
:
Let's do it this way.
:
Let a = the length of the 2nd rectangle
then
(a+4) = the length of the 1st rectangle
:
1st rect area - 2nd rect area = 96 sq/cm
9(a+4) - 6a = 96
9a + 36 - 6a = 96
9a - 6a = 96 - 36
3a = 30
a = 30/3
a = 10 cm is the length of the 2nd rectangle
then
10+4 = 14 cm is the length of the 1st rectangle
:
:
Check this
9*14 - 6*10 =
126 - 30 = 96; confirms our solution
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