Question 1155382


if the width {{{W}}} of a rectangular painting is {{{10in}}} less than the length {{{L}}}, we have

{{{W=L-10}}}......eq.1

if the area of the painting is {{{96in^2}}}, we have

{{{W*L=96}}}..........substitute {{{W}}} from eq.1

{{{(L-10)*L=96}}}.......solve for {{{L}}}

{{{L^2-10L=96}}}

{{{L^2-10L-96=0}}}.....factor completely

{{{L^2+6L-16L-96=0}}}

{{{(L^2+6L)-(16L+96)=0}}}

{{{L(L+6)-16(L+6)=0}}}

{{{(L - 16) (L + 6) = 0}}}

take only positive solution for length

{{{(L - 16)  = 0}}}=> {{{L =16in}}}


go to eq.1

{{{W=L-10}}}......eq.1, substitute {{{L}}}

{{{W=16-10}}}

{{{W=6in}}}


answer:

the length is {{{16in}}} and the width is {{{6in}}}