# 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?

Oct 27, 2016

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

#### Explanation:

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

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

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

${P}_{R e c} = 2 \cdot \left(L + W\right)$

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

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

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

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

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

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

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