Question 1185996

The length of a rectangle is {{{5}}} cm more than {{{3}}} times its width:

{{{L=3W+5}}}.......eq.1

 The area of the rectangle is {{{1428}}} square cm.

{{{A=L*W}}}

{{{1428=L*W}}}........substitute {{{L}}} from eq.1

{{{1428=(3W+5)*W}}}

{{{1428=3W^2+5W}}}

{{{3W^2+5W-1428=0}}}....factor

{{{3W^2-63W+68W-1428=0}}}

{{{(3W^2-63W )+(68W-1428)=0}}}

{{{3W(W-21 )+68(W-21)=0}}}

{{{(W - 21) (3W + 68) = 0}}}


=>  {{{(W - 21)   = 0}}} =>{{{W = 21cm}}}

go to  {{{L=3W+5}}}.......eq.1, substitute {{{W}}}

{{{L=3*21+5}}}

{{{L=68cm}}}