To solve the simultaneous equations given by:
- \( 2x + y = 4 \)
- \( x^2 + xy = -12 \)
we can start by expressing \( y \) from the first equation and then substitute it into the second equation.
From the first equation:
\[ y = 4 - 2x \]
Now, we will substitute \( y \) into the second equation:
\[ x^2 + x(4 - 2x) = -12 \]
Expanding this gives:
\[ x^2 + 4x - 2x^2 = -12 \]
Combining like terms leads to:
\[ -x^2 + 4x + 12 = 0 \]
Multiplying the entire equation by -1 to make it easier to handle:
\[ x^2 - 4x - 12 = 0 \]
Now we can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, c = -12 \):
Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4(1)(-12) = 16 + 48 = 64 \]
Now, using the quadratic formula:
\[ x = \frac{4 \pm \sqrt{64}}{2 \cdot 1} = \frac{4 \pm 8}{2} \]
This gives us two possible values for \( x \):
- \( x = \frac{12}{2} = 6 \)
- \( x = \frac{-4}{2} = -2 \)
Now, we can find the corresponding \( y \) values for each \( x \):
-
For \( x = 6 \): \[ y = 4 - 2(6) = 4 - 12 = -8 \] So one solution pair is \( (6, -8) \).
-
For \( x = -2 \): \[ y = 4 - 2(-2) = 4 + 4 = 8 \] So the second solution pair is \( (-2, 8) \).
Thus, the pairs of solutions to the equations are:
\[ \boxed{(6, -8) \text{ and } (-2, 8)} \]
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