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

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Answer 1 · 2016-10-27T06:37:07

$\Rightarrow W = \frac{9}{2} x^{2} - 5 x y - \frac{5}{2} y^{2} - 7$ $rArrW=9/2x^2-5xy-5/2y^2-7$

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

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

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

$P_{R e c} = 2 \cdot (L + W)$ $P_(Rec)=2*(L+W)$

$\Rightarrow 5 x^{2} + 2 x y - 7 y^{2} + 16 = 2 \cdot (- 2 x^{2} + 6 x y - y^{2} + 15 + W)$ $rArr5x^2+2xy-7y^2+16=2*(-2x^2+6xy-y^2+15+W)$

$\Rightarrow 5 x^{2} + 2 x y - 7 y^{2} + 16 = - 4 x^{2} + 12 x y - 2 y^{2} + 30 + 2 W$ $rArr5x^2+2xy-7y^2+16=-4x^2+12xy-2y^2+30+2W$

$\Rightarrow 5 x^{2} + 2 x y - 7 y^{2} + 16 + 4 x^{2} - 12 x y + 2 y^{2} - 30 = + 2 W$ $rArr5x^2+2xy-7y^2+16+4x^2-12xy+2y^2-30=+2W$

$\Rightarrow 5 x^{2} + 4 x^{2} + 2 x y - 12 x y - 7 y^{2} + 2 y^{2} + 16 - 30 = + 2 W$ $rArr5x^2+4x^2+2xy-12xy-7y^2+2y^2+16-30=+2W$

$\Rightarrow 9 x^{2} - 10 x y - 5 y^{2} - 14 = + 2 W$ $rArr9x^2-10xy-5y^2-14=+2W$

$\Rightarrow \frac{9 x^{2} - 10 x y - 5 y^{2} - 14}{2} = W$ $rArr(9x^2-10xy-5y^2-14)/2=W$

$\Rightarrow \frac{9}{2} x^{2} - 5 x y - \frac{5}{2} y^{2} - 7 = W$ $rArr9/2x^2-5xy-5/2y^2-7=W$

If the perimeter of a rectangle is 5x^2+2xy-7y^2+165x2+2xy−7y2+165x^2+2xy-7y^2+16, and the length is -2x^2+6xy-y^2+15−2x2+6xy−y2+15-2x^2+6xy-y^2+15, what is the width?