Question 946397
w and L, the original width (or breadth) and length.



First part of description means {{{(L+3)(w+4)=3wL}}}.


The second part means (L-1) and (w+1) become the new dimensions and
{{{L-1=w+1}}}.


Non-linear system in w and L, and look for the intersection point or points.



{{{system((L+3)(w+4)=3wL,L-1=w+1)}}}
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{{{wL+3w+4L+12=3wL}}}
{{{2wL-3w-4L-12=0}}}
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Use the linear equation to allow for substitution of one of the variables.
{{{L=w+1+1}}}
{{{L=w+2}}}
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{{{2w(w+2)-3w-4(w+2)-12=0}}}
{{{2w^2+4-3w-4w-8-12=0}}}
{{{2w^2-7w-4-12=0}}}
{{{highlight_green(2w^2-7w-16=0)}}}


Discriminant for prep to use general solution method:
{{{(-7)^2-4*2*(-16)}}}
{{{49+8*16}}}
{{{49+128}}}
{{{highlight_green(177)}}}
{{{3*59}}}, a prime factorization, still only using none repeated.


{{{w=(7+- sqrt(177))/4}}}
The meaningful value for w is {{{highlight(w=(7+sqrt(177))/4)}}}


Return to the use of {{{highlight_green(L=w+2)}}}  to get the corresponding value for L.