Question

j 150a kumon
simultaneous equations

solve the following equations.

(1) \\(\

$$\begin{cases} x - y = 3 \\\\ x^2 + y^2 = 9 \\end{cases}$$

\\)

(2) \\(\

$$\begin{cases} x^2 + y^2 - 2x + 2y = 3 \\\\ x^2 + y^2 - 4x + 4y = 1 \\end{cases}$$

\\)

Explanation:

Step1: Express \(x\) in terms of \(y\) for (1)

$$x = y + 3$$

Step2: Substitute \(x\) into the second equation of (1)

$$(y + 3)^2 + y^2 = 9$$

Step3: Expand and simplify the equation

$$2y^2 + 6y = 0$$

Step4: Solve for \(y\)

$$2y(y + 3) = 0 \implies y = 0 \text{ or } y = -3$$

Step5: Find corresponding \(x\) values

$$\text{If } y = 0, x = 3; \text{ if } y = -3, x = 0$$

Step6: Subtract the second equation from the first in (2)

$$2x - 2y = 2 \implies x - y = 1$$

Step7: Express \(x\) in terms of \(y\) for (2)

$$x = y + 1$$

Step8: Substitute \(x\) into the first equation of (2)

$$(y + 1)^2 + y^2 - 2(y + 1) + 2y = 3$$

Step9: Simplify the quadratic equation

$$2y^2 + 2y - 4 = 0 \implies y^2 + y - 2 = 0$$

Step10: Solve for \(y\)

$$(y + 2)(y - 1) = 0 \implies y = 1 \text{ or } y = -2$$

Step11: Find corresponding \(x\) values

$$\text{If } y = 1, x = 2; \text{ if } y = -2, x = -1$$

Answer:

(1) \(x = 3, y = 0\) or \(x = 0, y = -3\)
(2) \(x = 2, y = -1\) or \(x = 1, y = -2\)