\( \left\{ \left. \begin{array} { l } { 3 x - y = - 10 } \\ { x ^ { 2 } + y = 10 } \end{array} \right. \Rightarrow \right. \)

Система уравнений: \[ \begin{cases} 3x - y = -10 \\ x^2 + y = 10 \end{cases} \] Нам нужно найти решение этой системы уравнений. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x-y=-10\\x^{2}+y=10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y=10+3x\\x^{2}+y=10\end{array}\right.\) - step2: Substitute the value of \(y:\) \(x^{2}+10+3x=10\) - step3: Cancel equal terms: \(x^{2}+3x=0\) - step4: Factor the expression: \(x\left(x+3\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x=0\\&x+3=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=0\\&x=-3\end{align}\) - step7: Calculate: \(x=0\cup x=-3\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=0\\y=10+3x\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=10+3x\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=0\\y=10\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=1\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-3\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=10\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-3\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=10\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-3,1\right)\cup \left(x,y\right) = \left(0,10\right)\) Решение системы уравнений: \[ \begin{cases} 3x - y = -10 \\ x^2 + y = 10 \end{cases} \] Решение системы уравнений: \[ (x, y) = (-3, 1) \cup (x, y) = (0, 10) \] Таким образом, система уравнений имеет два решения: \((-3, 1)\) и \((0, 10)\).