SOLUTION: Please help me on this problem, thanks! The perimeter of the rectangle below is 78 feet. Find it's Dimension. One of the side on the rectangle is 2x-4 and the other side is 2x+3

Algebra ->  Rectangles -> SOLUTION: Please help me on this problem, thanks! The perimeter of the rectangle below is 78 feet. Find it's Dimension. One of the side on the rectangle is 2x-4 and the other side is 2x+3       Log On


   

Question 891537: Please help me on this problem, thanks!
The perimeter of the rectangle below is 78 feet. Find it's Dimension. One of the side on the rectangle is 2x-4 and the other side is 2x+3 and the other two are blank
Found 2 solutions by Theo, josgarithmetic:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
P = perimeter
L = length
W = width

P = 2L + 2W

P = 78

L = 2x-4
W = 2x+3

P = 2L + 2W becomes:

78 = 2(2x-4) + 2(2x+3)

simplify to get:

78 = 4x - 8 + 4x + 6

simplify to get 78 = 8x - 2

add 2 to both sides of the equation to get:

80 = 8x

divide both sides of the equation by 8 to get:

x = 10

when x = 10, your original equation of:

78 = 2(2x-4) + 2(2x+3) becomes:

78 = 2(2*10-4) + 2(2*10+3) which becomes:

78 = 2 * 16 + 2 * 23 which becomes:

78 = 32 + 46 which becomes:

78 = 78

this confirms the solution os x = 10 is correct.





Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Perimeter 2w+2L=p.
Using one expression for each dimension, 2%282x-4%29%2B2%282x%2B3%29=78.
2x-4+2x+3=39
4x-1=39
4x=40
x=10