x^{2}-y^{3}=10

x^{4}-y^{6}=100

x^2+y^3=

asked by guest
on Nov 19, 2024 at 2:36 am



You asked:

Given the following equations: \[\begin{aligned}{x}^{2} - {y}^{3} &= 10\\ {x}^{4} - {y}^{6} &= 100\end{aligned}\] Evaluate \({x}^{2} + {y}^{3}\).

MathBot Answer:

The system of equations has \(2\) solutions.

\[x = \sqrt{10}, y = 0\]\[x = - \sqrt{10}, y = 0\]

By substituting these values into the expression, we find that: $$\begin{aligned}x^{2} + y^{3} &= 10\end{aligned}$$

Solve \(x^{2} - y^{3} = 10\) for \(x\). \[x = - \sqrt{y^{3} + 10}, x = \sqrt{y^{3} + 10}\]Substitute \(- \sqrt{y^{3} + 10}\) for \(x\) in \(x^{4} - y^{6} = 100\) and simplify. $$\begin{aligned}x^{4} - y^{6} &amp= 100 \\ \left(- \sqrt{y^{3} + 10}\right)^{4} - y^{6} &= 100 \\ y^{3} &= 0 \\y^{3} &= 0 \\ y = 0\end{aligned}$$Substitute \(0\) into \(x^{2} - y^{3} = 10\) to solve for \(x\). $$\begin{aligned}x^{2} - 0^{3} &= 10 \\ x^{2} &= 10 \\ \sqrt{x^{2}} &= \sqrt{10} \\ x = - \sqrt{10}&, x = \sqrt{10}\end{aligned}$$This yields the following solutions. $$\begin{aligned}x = - \sqrt{10},\,y = 0\\ x = \sqrt{10},\,y = 0\end{aligned}$$

By substituting these values into the expression, we find that: $$\begin{aligned}x^{2} + y^{3} &= 10\end{aligned}$$

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