I've been stuck on these two. If you could help me and show the work. Thanks.
8-sqrt(y-1)=6
x^2/3-10x^1/3+9=0
$\displaystyle 8 - \sqrt{y - 1} = 6$
$\displaystyle \Rightarrow - \sqrt {y - 1} = -2$
now, square both sides and continue
note that we havex^2/3-10x^1/3+9=0
$\displaystyle \left( x^{1/3} \right)^2 - 10 \left( x^{1/3} \right) + 9 = 0$
this is quadratic in $\displaystyle x^{1/3}$. this becomes as clear as day if we say let $\displaystyle x^{1/3} = y$, then we get
$\displaystyle y^2 - 10y + 9 = 0$
now solve for y. when done, replace y with $\displaystyle x^{1/3}$ and then solve for x