B. Substitute to solve the system of equations.
To solve the system of equations:
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From the first equation, solve for \(y\): \[ y = x + 5 \]
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Now substitute this expression for \(y\) into the second equation: \[ x + 5 = (x - 3)^2 + 2 \]
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Expand the right-hand side: \[ x + 5 = (x^2 - 6x + 9) + 2 \] \[ x + 5 = x^2 - 6x + 11 \]
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Rearranging gives: \[ 0 = x^2 - 7x + 6 \]
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Factor the quadratic: \[ 0 = (x - 6)(x - 1) \]
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Thus, \(x = 6\) or \(x = 1\).
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Substitute these values back to find \(y\):
- For \(x = 6\): \[ y = 6 + 5 = 11 \]
- For \(x = 1\): \[ y = 1 + 5 = 6 \]
So, the solutions are \((6, 11)\) and \((1, 6)\).
C. Graph the equations.
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Graph the linear equation:
- The line represented by \(-x + y = 5\) can be rewritten as \(y = x + 5\), which has a slope of 1 and a y-intercept at (0, 5).
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Graph the quadratic equation:
- The equation \(y = (x - 3)^2 + 2\) is a parabola that opens upwards, with the vertex at (3, 2).
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Label the points of intersection:
- The points of intersection, which are the solutions, are \((6, 11)\) and \((1, 6)\).
D. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions?
No, a system of two equations consisting of a linear equation and a quadratic equation cannot have an infinite number of solutions because the linear equation represents a straight line, while the quadratic equation represents a parabola; they can intersect at most twice in the Cartesian plane.
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