Question 957614
Length, L, width w.
<i>The length of a rectangle is 3 less than 5 times of its width</i>.
{{{L=-3+5w}}}.


p, perimeter of the rectangle.
{{{p=2L+2w}}}
{{{p=2(5w-3)+2w}}}
{{{p=10w-6+2w}}}
{{{highlight_green(p=12w-6)}}}




<i>if the rectangle width is tripled and its length is doubled the perimeter of new rectangle is 92 cm greater than original perimeter.</i>


Now the dimensions are changed to width {{{3w}}} and length  {{{2(5w-3)=10w-6}}}.  The new perimeter is given as {{{12w-6+92}}}.


The question is, find the area of the original rectangle, but you need to solve for w, and then for L so that the area can be calculated.  Return to the perimeter formula for using with the new perimeter situation.


{{{p=2*newLength+2*newWidth}}}
{{{12w-6+92=2(10w-6)+2(3w)}}}, with the expressions substituted.
{{{12w+86=10w-12+6w}}}
{{{86=4w-12}}}
{{{4w=92}}}
{{{highlight(w=23)}}}----------Use the given formula for L to find value of L.


{{{L=5w-3}}}
{{{L=5*23-3}}}
{{{highlight(L=112)}}}


Original rectangle's area, {{{highlight(w*L=23*112=highlight(2576))}}}.