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11. \( \left\{\begin{array}{l}y^{2}=x^{2}-9 \\ 2 y=x-3\end{array}\right. \)

Ask by Pope Lynch. in the United States
Jan 22,2025

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Answer

The solutions are \( (x,y) = (-5,-4) \) or \( (x,y) = (3,0) \).

Solution

Solve the system of equations \( y^{2}=x^{2}-9;2y=x-3 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y^{2}=x^{2}-9\\2y=x-3\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}y^{2}=x^{2}-9\\x=2y+3\end{array}\right.\) - step2: Substitute the value of \(x:\) \(y^{2}=\left(2y+3\right)^{2}-9\) - step3: Simplify: \(y^{2}=4y^{2}+12y\) - step4: Move the expression to the left side: \(y^{2}-\left(4y^{2}+12y\right)=0\) - step5: Subtract the terms: \(-3y^{2}-12y=0\) - step6: Factor the expression: \(-3y\left(y+4\right)=0\) - step7: Separate into possible cases: \(\begin{align}&-3y=0\\&y+4=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=0\\&y=-4\end{align}\) - step9: Calculate: \(y=0\cup y=-4\) - step10: Rearrange the terms: \(\left\{ \begin{array}{l}x=2y+3\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=2y+3\\y=-4\end{array}\right.\) - step11: Calculate: \(\left\{ \begin{array}{l}x=3\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=-5\\y=-4\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=-5\\y=-4\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=0\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=-5\\y=-4\end{array}\right.\cup \left\{ \begin{array}{l}x=3\\y=0\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(-5,-4\right)\cup \left(x,y\right) = \left(3,0\right)\) The solution to the system of equations is \( (x,y) = (-5,-4) \) or \( (x,y) = (3,0) \).

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Beyond the Answer

To solve this system of equations, let's first rewrite the equations for clarity: 1. \( y^2 = x^2 - 9 \) 2. \( 2y = x - 3 \) or rearranged, \( x = 2y + 3 \) Now, substitute \( x \) from the second equation into the first equation: \[ y^2 = (2y + 3)^2 - 9 \] Expanding the right side: \[ y^2 = 4y^2 + 12y + 9 - 9 \] This simplifies to: \[ y^2 = 4y^2 + 12y \] Now, rearranging gives: \[ 0 = 3y^2 + 12y \] Factor out \( 3y \): \[ 0 = 3y(y + 4) \] This gives us two potential solutions for \( y \): 1. \( y = 0 \) 2. \( y = -4 \) For each of these, we can substitute back to find \( x \): 1. For \( y = 0 \): \[ x = 2(0) + 3 = 3 \] Thus, one solution is \( (3, 0) \). 2. For \( y = -4 \): \[ x = 2(-4) + 3 = -8 + 3 = -5 \] Thus, the second solution is \( (-5, -4) \). The complete solution set to the system of equations is \( (3, 0) \) and \( (-5, -4) \).

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