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(D) The first pair of lines is $xy+4x-3y-12=0$,which factors as $(x-3)(y+4)=0$. Thus,the lines are $x=3$ and $y=-4$.The second pair of lines is $xy-3x

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$x^2-y^2+x-y=0$

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(D) The first pair of lines is $xy+4x-3y-12=0$,which factors as $(x-3)(y+4)=0$. Thus,the lines are $x=3$ and $y=-4$.The second pair of lines is $xy-3x+4y-12=0$,which factors as $(x+4)(y-3)=0$. Thus,the lines are $x=-4$ and $y=3$.The four lines forming the square are $x=3, x=-4, y=-4, y=3$.The vertices of the square are $(3, 3), (3, -4), (-4, -4), (-4, 3)$.The diagonals connect $(3, 3)$ to $(-4, -4)$ and $(3, -4)$ to $(-4, 3)$.The equation of the diagonal passing through $(3, 3)$ and $(-4, -4)$ is $y-3 = \frac{-4-3}{-4-3}(x-3)$,which simplifies to $y-3 = x-3$,or $x-y=0$.The equation of the diagonal passing through $(3, -4)$ and $(-4, 3)$ is $y-(-4) = \frac{3-(-4)}{-4-3}(x-3)$,which simplifies to $y+4 = -1(x-3)$,or $x+y+1=0$.The combined equation is $(x-y)(x+y+1) = x^2+xy+x-xy-y^2-y = x^2-y^2+x-y=0$.

The combined equation of the diagonals of the square formed by the two pairs of straight lines given by $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ is

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