Question

ex.
\\(\

$$\begin{cases} x^2 - 4xy + 3y^2 = 0 & \\cdots \\text{①} \\\\ x^2 - 5xy + 6y^2 = 2 & \\cdots \\text{②} \\end{cases}$$

\\)

sol from ①, \\((x - y)(x - 3y) = 0\\) therefore \\(x = y\\) or \\(x = 3y\\)

\\(\

$$\begin{cases} x = y & \\cdots \\text{③} \\\\ x^2 - 5xy + 6y^2 = 2 & \\cdots \\text{②} \\end{cases}$$

\\)
substituting ③ into ②,
\\(y^2 - 5y^2 + 6y^2 = 2\\)
\\(y = \pm 1\\)
from ③, \\(x = \pm 1\\)

\\(\

$$\begin{cases} x = 3y & \\cdots \\text{③} \\\\ x^2 - 5xy + 6y^2 = 2 & \\cdots \\text{②} \\end{cases}$$

\\)
substituting ③ into ②,
\\(9y^2 - 15y^2 + 6y^2 = 2\\)
\\(0 \cdot y^2 = 2\\)
there is no value of \\(y\\) that would satisfy this. therefore, there is no solution.

ans. \\(\

$$\begin{cases} x = \\pm 1 \\\\ y = \\pm 1 \\end{cases}$$

\\)

(2) \\(\

$$\begin{cases} x^2 + 3xy + 2y^2 = 0 \\\\ x^2 - 2xy - 3y^2 = 5 \\end{cases}$$

\\)

Explanation:

Step1: Factor the first equation

$$ x^2 + 3xy + 2y^2 = 0 \implies (x + y)(x + 2y) = 0 $$

Step2: Identify the two cases

$$ x = -y \quad \text{or} \quad x = -2y $$

Step3: Substitute the first case

$$ (-y)^2 - 2(-y)y - 3y^2 = 5 \implies y^2 + 2y^2 - 3y^2 = 5 \implies 0 = 5 \quad \text{(No solution)} $$

Step4: Substitute the second case

$$ (-2y)^2 - 2(-2y)y - 3y^2 = 5 \implies 4y^2 + 4y^2 - 3y^2 = 5 \implies 5y^2 = 5 $$

Step5: Solve for y

$$ y^2 = 1 \implies y = \pm 1 $$

Step6: Solve for x

$$ x = -2y \implies x = \mp 2 $$

Answer:

$$
LATEXBLOCK0
$$

(which means \(x = 2, y = -1\) or \(x = -2, y = 1\))