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The width of a rectangular window frame is 12 inches less than its length. if you add 30 inches to both the length and the width, you double the perimeter. find the length and the width of the original rectangle?

Respuesta :

Let's assume

length of rectangle is x

width of rectangle is y

we are given

The width of a rectangular window frame is 12 inches less than its length

so, we get

[tex] y=x-12 [/tex]

now, we are given

if you add 30 inches to both the length and the width, you double the perimeter

perimeter of rectangle =2*length+2*width

perimeter of rectangle =[tex] 2x+2y [/tex]

After adding 30 inches,

perimeter will become =[tex] 2(x+30)+2(y+30) [/tex]

and it is double of original

so, we get

[tex] 2(x+30)+2(y+30)=2(2x+2y) [/tex]

now, we can simplify it

[tex] 2x+60+2y+60=4x+4y [/tex]

[tex] 2x+2y=120 [/tex]

[tex] x+y=60 [/tex]

so, we will get system of equations as

first equation:

[tex] y=x-12 [/tex]

Second equation:

[tex] x+y=60 [/tex]

now, we can solve it by using substitution

we can plug first equation into second equation

[tex] x+x-12=60 [/tex]

[tex] 2x=72 [/tex]

[tex] x=36 [/tex]

now, we can find y

[tex] y=36-12 [/tex]

[tex] y=24 [/tex]

so, dimensions of rectangle are

length is 36 inches

width is 24 inches..............Answer

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