Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$

I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?

$$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$$
$$ \sqrt[3]{5 - 3x} = 1-\sqrt{3x - 4}$$
$$ 5 - 3x = (1-\sqrt{3x - 4})^3$$
$$ 5 - 3x = (1-3x)\sqrt{3x-4}+9x-11$$
$$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$
$$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$
$$ -4\sqrt{3x-4}\cdot \sqrt{3x-4}= (1-3x)\sqrt{3x-4}$$
so $x=4/3$ or: $$ -4 \sqrt{3x-4}=1-3x$$
which is easier. You can finish?
