To solve the system of equations:
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Set the equations equal to each other: \[ 2x + 4 = x^2 - 4 \]
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Rearrange to form a standard quadratic equation: \[ x^2 - 2x - 8 = 0 \]
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Factor the quadratic equation: To factor \(x^2 - 2x - 8\), we look for two numbers that multiply to \(-8\) and add to \(-2\): \[ (x - 4)(x + 2) = 0 \]
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Solve for \(x\): Setting each factor equal to zero gives us: \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
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Find the corresponding \(y\) values: For \(x = 4\): \[ y = 2(4) + 4 = 8 + 4 = 12 \] So, one solution is \((4, 12)\).
For \(x = -2\): \[ y = 2(-2) + 4 = -4 + 4 = 0 \] So, another solution is \((-2, 0)\).
The solutions to the system of equations are: \[ (4, 12) \quad \text{and} \quad (-2, 0) \]
Verification Graphically
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Graph the first equation \(y = 2x + 4\) (a straight line):
- This line has a slope of 2 and a y-intercept of 4.
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Graph the second equation \(y = x^2 - 4\) (a parabola):
- This parabola opens upwards with a vertex at \((0, -4)\).
Systems of Equations Graph
To visualize this, plot the following points:
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For \(y = 2x + 4\):
- When \(x = 0\), \(y = 4\) → point \((0, 4)\)
- When \(x = 2\), \(y = 8\) → point \((2, 8)\)
- When \(x = 4\), \(y = 12\) → point \((4, 12)\)
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For \(y = x^2 - 4\):
- When \(x = -2\), \(y = 0\) → point \((-2, 0)\)
- When \(x = 0\), \(y = -4\) → point \((0, -4)\)
- When \(x = 2\), \(y = 0\) → point \((2, 0)\)
- When \(x = 4\), \(y = 12\) → point \((4, 12)\)
Plot these points on the same graph, and you will see:
- The line \(y = 2x + 4\) crosses the parabola \(y = x^2 - 4\) at the points \((4, 12)\) and \((-2, 0)\), confirming that the solutions are correct.
The solutions \((4, 12)\) and \((-2, 0)\) have been verified.