Question

Solve the simultaneous equation below.
X² +xy=4
Y-x=2

Answer 1

To solve the simultaneous equations:

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

First, we can express \( y \) in terms of \( x \) from Equation 2:

\[
y = x + 2
\]

Now, we can substitute this expression for \( y \) back into Equation 1:

\[
x^2 + x(x + 2) = 4
\]

Simplifying this, we get:

\[
x^2 + x^2 + 2x = 4
\]
\[
2x^2 + 2x - 4 = 0
\]

Next, we can simplify this equation by dividing everything by 2:

\[
x^2 + x - 2 = 0
\]

Now, we can factor the quadratic equation:

\[
(x + 2)(x - 1) = 0
\]

Setting each factor to zero gives us:

1. \( x + 2 = 0 \) → \( x = -2 \)
2. \( x - 1 = 0 \) → \( x = 1 \)

Now we can find the corresponding \( y \) values using \( y = x + 2 \):

1. For \( x = -2 \):
\[ y = -2 + 2 = 0 \]
So, one solution is \( (x, y) = (-2, 0) \).

2. For \( x = 1 \):
\[ y = 1 + 2 = 3 \]
So, another solution is \( (x, y) = (1, 3) \).

The final solutions to the simultaneous equations are:

\[
(x, y) = (-2, 0) \quad \text{and} \quad (x, y) = (1, 3)
\]