Question

Solve the pairs of simultaneous equations 2x+y=4,x^2+xy=-12

Answer 1

To solve the simultaneous equations given by:

1. \( 2x + y = 4 \)
2. \( x^2 + xy = -12 \)

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 \( x \):

1. \( x = \frac{12}{2} = 6 \)
2. \( x = \frac{-4}{2} = -2 \)

Now, we can find the corresponding \( y \) values for each \( x \):

1. For \( x = 6 \):
\[
y = 4 - 2(6) = 4 - 12 = -8
\]
So one solution pair is \( (6, -8) \).

2. 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:

\[
\boxed{(6, -8) \text{ and } (-2, 8)}
\]