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#