SOLUTION: The length of the rectangle is twice its width. If its length is decrease by 3m and its width is increase by 2m, the area resulting rectangle is 30 square mater. Find the dimension

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Question 885576: The length of the rectangle is twice its width. If its length is decrease by 3m and its width is increase by 2m, the area resulting rectangle is 30 square mater. Find the dimensions of the rectangle
Found 2 solutions by lwsshak3, algebrapro18:
Answer by lwsshak3(11628) (Show Source):
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The length of the rectangle is twice its width. If its length is decrease by 3m and its width is increase by 2m, the area resulting rectangle is 30 square mater. Find the dimensions of the rectangle
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let x=width of rectangle
2x=length of rectangle
..
(2x-3)(x+2)=30
2x^2+x-6=30
2x^2+x-36=0
(2x+9)(x-4)=0
2x+9=0
x=-9/2(reject, x>0)
or
x-4=0
x=4
2x=8
dimensions of the rectangle: 4m by 8m

Answer by algebrapro18(249) (Show Source):
You can put this solution on YOUR website!
Well to do this we will need to set up and solve a non-linear system of equations. Lets let L be the length of the rectangle and W be the width of the rectangle. First look at the fact that the length is 2 times the width. An equation for this would be:

L=2W

So that's going to be our first equation. Now to find the second equation we need the fact that area of a rectangle equals its lenght times its width and the fact that if its length is decrease by 3m and its width is increase by 2m, the area resulting rectangle is 30 square meters. As an equation this would be:

(L-3)(W+2) = 30

So we have our two equations and we know that L=2W so we can plug that into the second equation and solve for W.

(L-3)(W+2) = 30 plug in 2w for L
(2W-3)(W+2) = 30 F.O.I.L. the left hand side
2W^2+2W-3W-6 = 30 Combine like terms
2W^2-W-6 = 30 Subtract 30 from both sides to get the equation into a solvable form.
2W^2-W-36 = 0 Factor the left hand side
(2W+9)(W-4) = 0

Now normally we would set each factor equal to zero and solve but lets look at what happens when 2W+9 = 0. The first step of solving that would be to subtract 9 which would make W negative. We can't have a negative width so we ignore that factor and just work with the other factor. Solving W-4=0 we get that w = 4. So we know that the width is 4 meters.

Now we can plug back into the first equation that told us that L=2W. Plugging 4 in for 2 we get that L = 2(4) = 8. So we know that the length of the Rectangle is 8 and the width is 4.