SOLUTION: So, there's a rectangle named FGHJ. FG(top base) is 4x+11 and GH(right side) is 3x-1. The perimeter is 48. Find x, then find the area of the rectangle. I've tried setting them

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Question 922576: So, there's a rectangle named FGHJ. FG(top base) is 4x+11 and GH(right side) is 3x-1. The perimeter is 48. Find x, then find the area of the rectangle.
I've tried setting them equal to each other, but that didn't work because they both aren't going to be the same thing.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
L = length
W = width


Let's say the horizontal sides are the length, so L+=+4x%2B11. Recall that the parallel sides of a rectangle are congruent.


The vertical sides are the width, so W+=+3x+-+1


The perimeter of the rectangle is found by this formula P+=+2%2A%28L%2BW%29 where P is the perimeter, L is the length, W is the width.


P+=+2%2A%28L%2BW%29


48+=+2%2A%284x%2B11%2B3x+-+1%29 Plug in P+=+48, L+=+4x%2B11, and W+=+3x+-+1. Solve for x.


48+=+2%2A%287x%2B10%29


48+=+14x%2B20


48-20+=+14x


28+=+14x


14x+=+28


x+=+28%2F14


x+=+2


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Let me know if that helps or not. Thanks.

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