Question 1

Pre Calc Help?? Much appreciated

Question 2

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

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

$(18x^2-68x+24)/(x(x-2)(x-4))=\frac3x+(10)/(x-2)+\frac5{x-4}$

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

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

x^2+(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

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

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

$x=2\implies y=6$

$x=6\implies y=2$

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