Santiago Stephens
05/01/2023 · High School
Find the solution to the system of equations: \(y = x^2\) and \(y = 2x + 3\).
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Step-by-step Solution
Solve the system of equations \( y=x^2; y=2x+3 \).
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
\(\left\{ \begin{array}{l}y=x^{2}\\y=2x+3\end{array}\right.\)
- step1: Substitute the value of \(y:\)
\(x^{2}=2x+3\)
- step2: Move the expression to the left side:
\(x^{2}-\left(2x+3\right)=0\)
- step3: Remove the parentheses:
\(x^{2}-2x-3=0\)
- step4: Factor the expression:
\(\left(x-3\right)\left(x+1\right)=0\)
- step5: Separate into possible cases:
\(\begin{align}&x-3=0\\&x+1=0\end{align}\)
- step6: Solve the equation:
\(\begin{align}&x=3\\&x=-1\end{align}\)
- step7: Calculate:
\(x=3\cup x=-1\)
- step8: Rearrange the terms:
\(\left\{ \begin{array}{l}x=3\\y=x^{2}\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=x^{2}\end{array}\right.\)
- step9: Calculate:
\(\left\{ \begin{array}{l}x=3\\y=9\end{array}\right.\cup \left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\)
- step10: Calculate:
\(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=9\end{array}\right.\)
- step11: Check the solution:
\(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=9\end{array}\right.\)
- step12: Rewrite:
\(\left(x,y\right) = \left(-1,1\right)\cup \left(x,y\right) = \left(3,9\right)\)
The solution to the system of equations \(y = x^2\) and \(y = 2x + 3\) is \((x, y) = (-1, 1)\) or \((x, y) = (3, 9)\).
Quick Answer
A solução do sistema de equações \(y = x^2\) e \(y = 2x + 3\) é \((x, y) = (-1, 1)\) ou \((x, y) = (3, 9)\).
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