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.^2. what are the length and width of the rectangle?

If the length is l and the width is w, we have w+5=l.
Next, 2*l (l doubled)*(w-2) =162+x due to that the area formula is l*w (and given that x= the original area = l*w) Substituting w+5 for l, we have
2*(w+5)*(w-2)= 2*(w^2-2w+5w-10)=2*(w^2+3w-10)=2w^2+6w-20=162+x. Subtracting 162 from both sides, we get 2w^2+6w-182=x = l*w=(w+5)(w) by substituting w+5 for l. We then expand to get 
2w^2+6w-182 = w^2+5w, and subtract w^2+5w by both sides to get
w^2+w-182=0. Googling the factors of 182, we see that -13 and 14 add up to 1 (since it is the coefficient of w, meaning that 1*w=w. For 2w, the coefficient would be 2) while multiplying to -182. Factoring that, we get
w^2-13w+14w-182=0 and w(w-13)+14(w-13)=(w+14)(w-13), therefore making the zeros of the equation -14 and 13 (e.g. -14+14=0, and 0* anything =0), making the possible widths -14 and 13. Since it can't be -14, it would have to be 13.  13+5=18=the length, and making sure it works, we get
2*18*(13-2)=36*11=162+(18*13)=396=396

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