Question
Use a graphing utility to solve each system of equations. Express the solution(s) rounded to two decimal places.$$\left\{\begin{array}{r}x^{2}+y^{3}=2 \\x^{3} y=4\end{array}\right.$$
Step 1
Starting with the first equation, we subtract $x^{2}$ from both sides to get $y^{3} = 2 - x^{2}$. Then, we take the cube root of both sides to solve for $y$, which gives us $y = \sqrt[3]{2 - x^{2}}$. Show more…
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