To solve the linear-quadratic system using the substitution method, we can substitute the expression for \(y\) from the first equation into the second equation.
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We have the equations: \[ y = 2x - 3 \] \[ y = x^2 + 6x - 3 \]
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Substitute \(y\) in the second equation: \[ 2x - 3 = x^2 + 6x - 3 \]
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Simplifying this equation: \[ 2x - 3 = x^2 + 6x - 3 \] Add \(3\) to both sides: \[ 2x = x^2 + 6x \]
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Rearranging: \[ 0 = x^2 + 6x - 2x \] \[ 0 = x^2 + 4x \]
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Factoring the equation: \[ 0 = x(x + 4) \]
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Setting each factor to zero: \[ x = 0 \quad \text{or} \quad x + 4 = 0 \implies x = -4 \]
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Now substitute these values back into the first equation \(y = 2x - 3\) to find \(y\):
- For \(x = 0\): \[ y = 2(0) - 3 = -3 \] So, one solution is \((0, -3)\).
- For \(x = -4\): \[ y = 2(-4) - 3 = -8 - 3 = -11 \] So, the other solution is \((-4, -11)\).
Thus, the solutions to the system are \((0, -3)\) and \((-4, -11)\).
For the security systems question:
Let \(d\) be the number of days.
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Cost for Safety First Security: \[ C_1 = 2d + 150 \]
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Cost for Home Safety Security: \[ C_2 = 1.5d + 200 \]
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Set the costs equal: \[ 2d + 150 = 1.5d + 200 \]
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Rearranging: \[ 2d - 1.5d = 200 - 150 \] \[ 0.5d = 50 \]
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Solve for \(d\): \[ d = \frac{50}{0.5} = 100 \]
So, both companies would cost the same after 100 days.
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