Question

j 146a kumon
simultaneous equations

solve the following equations.

ex.
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$$\begin{cases} (x-y)(x+2y) = 0 & \\cdots\\cdots (1) \\\\ x^2 - xy + 2y^2 = 16 & \\cdots\\cdots (2) \\end{cases}$$

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sol from (1), \\(x = y\\) or \\(x = -2y\\)
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$$\begin{cases} x = y & \\cdots (3) \\\\ x^2 - xy + 2y^2 = 16 & \\cdots (2) \\end{cases}$$

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substituting (3) into (2),
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y^2 - y^2 + 2y^2 = 16
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y^2 = 8
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y = \pm 2\sqrt{2}
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from (3), \\(x = \pm 2\sqrt{2}
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$$\begin{cases} x = -2y & \\cdots (3) \\\\ x^2 - xy + 2y^2 = 16 & \\cdots (2) \\end{cases}$$

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substituting (3) into (2),
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4y^2 + 2y^2 + 2y^2 = 16
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y^2 = 2
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y = \pm \sqrt{2}
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from (3), \\(x = \mp 2\sqrt{2}
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ans. \\(\

$$\begin{cases} x = \\pm 2\\sqrt{2} \\\\ y = \\pm 2\\sqrt{2} \\end{cases}$$

\\) \\(\

$$\begin{cases} x = \\pm 2\\sqrt{2} \\\\ y = \\mp \\sqrt{2} \\end{cases}$$

\\)

(1) \\(\

$$\begin{cases} (x-3y)(x-2y) = 0 \\\\ x^2 - 2xy + 6y^2 = 6 \\end{cases}$$

\\)

Explanation:

Step1: Identify linear relations

From the first equation \((x-3y)(x-2y) = 0\), we obtain:

$$ x = 3y \quad \text{or} \quad x = 2y $$

Step2: Substitute first relation

Substitute \(x = 3y\) into the second equation \(x^2 - 2xy + 6y^2 = 6\):

$$ (3y)^2 - 2(3y)y + 6y^2 = 6 \implies 9y^2 - 6y^2 + 6y^2 = 6 \implies 9y^2 = 6 $$

Step3: Solve for y

Solve the simplified equation for \(y\):

$$ y^2 = \frac{2}{3} \implies y = \pm \frac{\sqrt{6}}{3} $$

Step4: Find corresponding x

Substitute \(y\) back into \(x = 3y\):

$$ x = \pm \sqrt{6} $$

Step5: Substitute second relation

Substitute \(x = 2y\) into the second equation \(x^2 - 2xy + 6y^2 = 6\):

$$ (2y)^2 - 2(2y)y + 6y^2 = 6 \implies 4y^2 - 4y^2 + 6y^2 = 6 \implies 6y^2 = 6 $$

Step6: Solve for y

Solve the simplified equation for \(y\):

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

Step7: Find corresponding x

Substitute \(y\) back into \(x = 2y\):

$$ x = \pm 2 $$

Answer:

$$
LATEXBLOCK0
\quad \text{and} \quad
LATEXBLOCK1
$$