Pregunta
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Solve the non-linear system of equations. \( \begin{array}{l}(x-11)^{2}+(y)^{2}=25 \\ -2 x+y=-17\end{array} \) Present your answer in the form of \( (x, y) \). If there are more than one solutions, separate them by a comma. If there is no solution, enter DNE.

Ask by Spencer Rojas. in the United States
Mar 20,2025

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The solutions are \( (11, 5) \) and \( (7, -3) \).

Solución

Solve the equation \( (x-11)^{2}+(y)^{2}=25;-2x+y=-17 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\left(x-11\right)^{2}+y^{2}=25\\-2x+y=-17\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}\left(x-11\right)^{2}+y^{2}=25\\y=-17+2x\end{array}\right.\) - step2: Substitute the value of \(y:\) \(\left(x-11\right)^{2}+\left(-17+2x\right)^{2}=25\) - step3: Simplify: \(5x^{2}-90x+410=25\) - step4: Move the expression to the left side: \(5x^{2}-90x+410-25=0\) - step5: Subtract the numbers: \(5x^{2}-90x+385=0\) - step6: Factor the expression: \(5\left(x-11\right)\left(x-7\right)=0\) - step7: Divide the terms: \(\left(x-11\right)\left(x-7\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-11=0\\&x-7=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=11\\&x=7\end{align}\) - step10: Calculate: \(x=11\cup x=7\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=11\\y=-17+2x\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=-17+2x\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=11\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=-3\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=11\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=7\\y=-3\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(11,5\right)\cup \left(x,y\right) = \left(7,-3\right)\) The non-linear system of equations has two solutions: 1. \( (x, y) = (11, 5) \) 2. \( (x, y) = (7, -3) \) Therefore, the solutions to the system of equations are \( (11, 5) \) and \( (7, -3) \).

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To solve the non-linear system of equations, we begin with: 1. \((x-11)^{2}+(y)^{2}=25\) (This represents a circle centered at (11, 0) with a radius of 5) 2. \(-2x+y=-17\) (This can be rewritten as \(y=2x-17\)) Next, we substitute \(y\) from the second equation into the first equation: \((x-11)^{2} + (2x-17)^{2} = 25\) Expanding both sides: 1. \((x-11)^{2} = x^2 - 22x + 121\) 2. \((2x-17)^{2} = 4x^{2} - 68x + 289\) Now substitute these expansions into the equation: \[ x^2 - 22x + 121 + 4x^{2} - 68x + 289 = 25 \] Combine like terms: \[ 5x^{2} - 90x + 410 = 25 \] Subtract 25 from both sides: \[ 5x^{2} - 90x + 385 = 0 \] Now we divide everything by 5: \[ x^{2} - 18x + 77 = 0 \] Next, we can solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a = 1,\ b = -18,\ c = 77\) Calculating the discriminant: \[ D = b^2 - 4ac = (-18)^{2} - 4(1)(77) = 324 - 308 = 16 \] Since the discriminant is positive, there are two solutions: \[ x = \frac{18 \pm \sqrt{16}}{2(1)} \] So, \[ x = \frac{18 \pm 4}{2} \] This gives us: 1. \(x = \frac{22}{2} = 11\) 2. \(x = \frac{14}{2} = 7\) Next, substitute these \(x\) values back to find \(y\): 1. For \(x = 11\): \[ y = 2(11) - 17 = 22 - 17 = 5 \Rightarrow (11, 5) \] 2. For \(x = 7\): \[ y = 2(7) - 17 = 14 - 17 = -3 \Rightarrow (7, -3) \] Thus, the solutions to the non-linear system of equations are: **(11, 5), (7, -3)**

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