Question

j141b solve the simultaneous equations

(2) \\(\

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

\\)

(3) \\(\

$$\begin{cases} x = y + 1 \\\\ x^2 + y^2 = 5 \\end{cases}$$

\\)

Explanation:

Step1: Substitute first equation into second for (2)

$$x^2 + (x + 3)^2 = 5$$

Step2: Expand and simplify the equation

$$2x^2 + 6x + 4 = 0$$

Step3: Divide by 2 and factorize

$$(x + 1)(x + 2) = 0$$

Step4: Solve for x

$$x = -1 \quad \text{or} \quad x = -2$$

Step5: Find corresponding y values

$$y = -1 + 3 = 2 \quad \text{or} \quad y = -2 + 3 = 1$$

Step6: Substitute first equation into second for (3)

$$(y + 1)^2 + y^2 = 5$$

Step7: Expand and simplify the equation

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

Step8: Divide by 2 and factorize

$$(y - 1)(y + 2) = 0$$

Step9: Solve for y

$$y = 1 \quad \text{or} \quad y = -2$$

Step10: Find corresponding x values

$$x = 1 + 1 = 2 \quad \text{or} \quad x = -2 + 1 = -1$$

Answer:

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