Question

1.2 Solve for 𝑥 and 𝑦 simultaneously:
2𝑥 + 𝑦 = 1 and 2𝑥
2 − 𝑥𝑦 + 𝑦
2 = 4

Answer 1

To solve the system of equations

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

we will first express \( y \) in terms of \( x \) using Equation 1, and then substitute it into Equation 2.

From Equation 1, we can solve for \( y \):
\[
y = 1 - 2x
\]

Next, we substitute this expression for \( y \) into Equation 2:

Substituting \( y = 1 - 2x \) into Equation 2:

\[
2x^2 - x(1 - 2x) + (1 - 2x)^2 = 4
\]

Now, we expand the equation:

1. First, calculate \( -x(1 - 2x) \):
\[
-x(1 - 2x) = -x + 2x^2
\]

2. Next, calculate \( (1 - 2x)^2 \):
\[
(1 - 2x)^2 = 1 - 4x + 4x^2
\]

Now substituting back into the equation:

\[
2x^2 - x + 2x^2 + 1 - 4x + 4x^2 = 4
\]

Combine like terms:

\[
(2x^2 + 2x^2 + 4x^2) - x - 4x + 1 = 4
\]

This simplifies to:

\[
8x^2 - 5x + 1 = 4
\]

Next, we move 4 to the left side:

\[
8x^2 - 5x + 1 - 4 = 0
\]

This simplifies to:

\[
8x^2 - 5x - 3 = 0
\]

Now, we use the quadratic formula to solve for \( x \):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 8 \), \( b = -5 \), and \( c = -3 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-5)^2 - 4(8)(-3) = 25 + 96 = 121
\]

Now substituting into the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{121}}{2 \cdot 8}
\]
\[
x = \frac{5 \pm 11}{16}
\]

Calculating the two possible values for \( x \):

1. \( x = \frac{5 + 11}{16} = \frac{16}{16} = 1 \)
2. \( x = \frac{5 - 11}{16} = \frac{-6}{16} = \frac{-3}{8} \)

Now we'll substitute these values of \( x \) back into Equation 1 to find the corresponding \( y \) values.

**For \( x = 1 \)**:
\[
y = 1 - 2(1) = 1 - 2 = -1
\]

So, one solution is \( (1, -1) \).

**For \( x = -\frac{3}{8} \)**:
\[
y = 1 - 2\left(-\frac{3}{8}\right) = 1 + \frac{6}{8} = 1 + \frac{3}{4} = \frac{7}{4}
\]

So, the second solution is \( \left(-\frac{3}{8}, \frac{7}{4}\right) \).

In conclusion, the solutions for the system of equations are:
1. \( (1, -1) \)
2. \( \left(-\frac{3}{8}, \frac{7}{4}\right) \)