Help me!!!! the dimensions of a rectangle are such that its length is 5 in. more than its width. If the length were doubled and if the width were decreased by 2 in., the area would be increased by 162 in squared. What are the length and width of the rectangle?

Question

Help me!!!!  the dimensions of a rectangle are such that its length is 5 in. more than its width.  If the length were doubled and if the width were decreased by 2 in., the area would be increased by 162 in squared.  What are the length and width of the rectangle?

anonymous · Answer

I got W = 14 in. and L = 19 in. 

I could be wrong...

anonymous · Answer

I haven't gotten anything because this is the equation I came up with (2(x+5))(x-2)=162  and now can't solve it lol

anonymous · Answer

You actually have three equations.  The equation you got would mean that the area is 162 instead of being INCREASED by 162.

Equation 1:
You know that the length is 5 more than the width.
$$L=W+5$$
Equation 2:
Equation for original area:
$$A=LW$$
Equation 3:
The length doubled (2L) times the width decreased by 2 (W-2) would give you 162 more than the original area (A+162)
$$(2L)(W-2)=A+162$$

So here are the 3 equations:
$$L=W+5$$$$A=LW$$$$(2L)(W-2)=A+162$$
Substitute Equation 2 into Equation 3:
$$(2L)(W-2)=LW+162$$Substitute W+5 for L:
$$2(W+5)(W-2)=W(W+5)+162$$Expand both sides:
$$2W^2+6W-20=W^2+5W+162$$
Put this into the form 0=ax^2+bx+c
$$0=W^2+W+182$$
You can factor this:
$$0=(W+14)(W-13)$$
This gives the two solutions W=13 and W=-14.  W=-14 doesn't make sense, because the width must be positive.  Therefore, your width is 13.  Add 5 to get the length (18).