Question 994258

Let the width of the matte be {{{x}}}cm.

the outside dimensions of the matte are {{{40}}} by {{{60}}} cm

the length of the picture will be {{{60 -2x }}} ... ( {{{x}}} cm on each side, so {{{2x}}})
the width of the picture will be {{{40-2x}}}
.

the area of the {{{picture + matte = 60*40 = 2400cm^2}}}

the area of the {{{picture = (60-2x)(40-2x)}}}
.
the area of the{{{( picture + matte)  - (area_ of_ picture) = 2(60-2x)(40-2x)}}}
.
=>{{{2400-(60-2x)(40-2x)= 2(60-2x)(40-2x)}}}

{{{2400= 3(60-2x)(40-2x)}}}

{{{800= 2400-120x-80x+4x^2}}}

{{{4x^2-200x+1600=0}}}

{{{x^2-50x+400=0}}}

{{{x^2-40x-10x+400=0}}}

{{{x(x-40)-10(x-40)=0}}}

{{{(x-10)(x-40) =0}}}

{{{x= 10}}} or {{{x=40}}}

{{{x=10}}} satisfies the equation, so the width of matte is {{{10cm}}}