Question 179488
Let L=original length, W=original width, and A=original area of the rectangle



"a rectangle is 12 cm longer than it is wide." translates to {{{L=W+12}}}


and 


"if its length and width are both decreased by 2 cm its area is decreased by 108 cm" means that the new area is {{{A-108=(L-2)(W-2)}}}



Note: the original area of the rectangle is {{{A=LW}}}



{{{A-108=(L-2)(W-2)}}} Start with the second equation.



{{{LW-108=(L-2)(W-2)}}} Plug in {{{A=LW}}}



{{{(W+12)W-108=(W+12-2)(W-2)}}} Plug in {{{L=W+12}}}



{{{W(W+12)-108=(W+12-2)(W-2)}}} Rearrange the terms.



{{{W(W+12)-108=(W+10)(W-2)}}} Combine like terms.



{{{W(W+12)-108=W^2+8W-20}}} FOIL



{{{W^2+12W-108=W^2+8W-20}}} Distribute



{{{4W=88}}} Subtract {{{W^2}}} from both sides. Subtract {{{8W}}} from both sides. Add 108 to both sides.



{{{W=88/4}}} Divide both sides by 4 to isolate W.



{{{W=22}}} Divide. So the original width of the rectangle is 22 cm




{{{L=W+12}}} Go back to the first equation



{{{L=22+12}}} Plug in {{{W=22}}} 



{{{L=34}}} Add. So the original length of the rectangle is 34 cm




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Answer:



So the original dimensions of the rectangle are:


Length: 34 cm
Width: 22 cm