Question

j 142a kumon
simultaneous equations

solve the following simultaneous equations.

(1) \\(\

$$\begin{cases} y = 2x + 1 \\\\ x^2 - y^2 = -21 \\end{cases}$$

\\)

(2) \\(\

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

\\)

Explanation:

Step1: Substitute \(y\) in the first system

$$x^2 - (2x + 1)^2 = -21$$

Step2: Expand and simplify the equation

$$-3x^2 - 4x + 20 = 0$$

Step3: Solve the quadratic equation for \(x\)

$$(3x + 22)(x - 2) = 0 \implies x = 2 \text{ or } x = -\frac{22}{3}$$

Step4: Find corresponding \(y\) values for (1)

$$x = 2 \implies y = 5; \quad x = -\frac{22}{3} \implies y = -\frac{41}{3}$$

Step5: Substitute \(y\) in the second system

$$x^2 - 2(2x - 1)^2 = -1$$

Step6: Expand and simplify the equation

$$-7x^2 + 8x - 1 = 0$$

Step7: Solve the quadratic equation for \(x\)

$$(7x + 1)(x - 1) = 0 \implies x = 1 \text{ or } x = -\frac{1}{7}$$

Step8: Find corresponding \(y\) values for (2)

$$x = 1 \implies y = 1; \quad x = -\frac{1}{7} \implies y = -\frac{9}{7}$$

Answer:

(1) \(x = 2, y = 5\) or \(x = -\frac{22}{3}, y = -\frac{41}{3}\)
(2) \(x = 1, y = 1\) or \(x = -\frac{1}{7}, y = -\frac{9}{7}\)