Question

Pre Calc Help?? Much appreciated

Answer 1

$\dfrac{18x^2-68x+24}{x(x-2)(x-4)}=\dfrac ax+\dfrac b{x-2}+\dfrac c{x-4}$

$18x^2-68x+24=a(x-2)(x-4)+bx(x-4)+cx(x-2)$

$x=0\implies24=8a\implies a=3$

$x=2\implies-40=-4b\implies b=10$

$x=4\implies40=8c\implies c=5$

So we have

$\dfrac{18x^2-68x+24}{x(x-2)(x-4)}=\dfrac3x+\dfrac{10}{x-2}+\dfrac5{x-4}$

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$xy=12\implies y=\dfrac{12}x$

$x^2+\left(\dfrac{12}x\right)^2=40$

$x^2+\dfrac{144}{x^2}=40$

$x^4-40x^2+144=0$

$(x^2-36)(x^2-4)=0$

$x^2-36=0\implies x=\pm6$

$x^2-4=0\implies x=\pm2$

The solutions are then

$x=-6\implies y=-2$

$x=-2\implies y=-6$

$x=2\implies y=6$

$x=6\implies y=2$