To solve the system of equations:
- \(-x + y = 5\)
- \(y = \frac{(x - 3)^2}{2} + 2\)
We can first express \(y\) from the first equation:
From equation 1: \[ y = x + 5 \]
Now, we can substitute \(y\) into equation 2:
\[ x + 5 = \frac{(x - 3)^2}{2} + 2 \]
Now, let's simplify and solve this equation:
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Multiply both sides by 2 to eliminate the fraction: \[ 2(x + 5) = (x - 3)^2 + 4 \]
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Distribute on the left side: \[ 2x + 10 = (x - 3)^2 + 4 \]
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Expand the right side: \[ 2x + 10 = x^2 - 6x + 9 + 4 \]
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Combine like terms on the right: \[ 2x + 10 = x^2 - 6x + 13 \]
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Rearrange the equation to one side: \[ 0 = x^2 - 6x - 2x + 13 - 10 \] \[ 0 = x^2 - 8x + 3 \]
Now we have a quadratic equation: \[ x^2 - 8x + 3 = 0 \]
We can solve this using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = -8\), and \(c = 3\).
Calculating the discriminant: \[ b^2 - 4ac = (-8)^2 - 4(1)(3) = 64 - 12 = 52 \]
Now applying the quadratic formula: \[ x = \frac{-(-8) \pm \sqrt{52}}{2(1)} = \frac{8 \pm \sqrt{52}}{2} \]
\(\sqrt{52}\) can be simplified: \[ \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} \]
Thus, we have: \[ x = \frac{8 \pm 2\sqrt{13}}{2} = 4 \pm \sqrt{13} \]
This gives us two values for \(x\): \[ x_1 = 4 + \sqrt{13}, \quad x_2 = 4 - \sqrt{13} \]
Now, we can find corresponding \(y\) values for each \(x\):
Using \(y = x + 5\):
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For \(x_1 = 4 + \sqrt{13}\): \[ y_1 = (4 + \sqrt{13}) + 5 = 9 + \sqrt{13} \]
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For \(x_2 = 4 - \sqrt{13}\): \[ y_2 = (4 - \sqrt{13}) + 5 = 9 - \sqrt{13} \]
Thus, the solutions to the system of equations are: \[ (4 + \sqrt{13}, 9 + \sqrt{13}) \quad \text{and} \quad (4 - \sqrt{13}, 9 - \sqrt{13}) \]