If the perimeter of a rectangle is $5x^2+2xy-7y^2+16$ , and the length is $-2x^2+6xy-y^2+15$ , what is the width?

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Answer 1 · 2016-10-27T06:37:07

$rArrW=9/2x^2-5xy-5/2y^2-7$

Name $P$ the perimeter of the rectangle.
Name $L$ its length and $W$ its width.

Given:
$P=5x^2+2xy-7y^2+16$

$L=-2x^2+6xy-y^2+15$

$P_(Rec)=2*(L+W)$

$rArr5x^2+2xy-7y^2+16=2*(-2x^2+6xy-y^2+15+W)$

$rArr5x^2+2xy-7y^2+16=-4x^2+12xy-2y^2+30+2W$

$rArr5x^2+2xy-7y^2+16+4x^2-12xy+2y^2-30=+2W$

$rArr5x^2+4x^2+2xy-12xy-7y^2+2y^2+16-30=+2W$

$rArr9x^2-10xy-5y^2-14=+2W$

$rArr(9x^2-10xy-5y^2-14)/2=W$

$rArr9/2x^2-5xy-5/2y^2-7=W$

If the perimeter of a rectangle is 5x^2+2xy-7y^2+165x^2+2xy-7y^2+16, and the length is -2x^2+6xy-y^2+15-2x^2+6xy-y^2+15, what is the width?