Question

j148b solve the following simultaneous equations

(2) \\(\

$$\begin{cases} xy = 3 & \\dots \\dots \\text{①} \\\\ x^2 + y^2 = 10 & \\dots \\dots \\text{②} \\end{cases}$$

\\)
there are 4 solutions.

sol from ②,
\\((x+y)^2 - 2xy = 10 \dots \dots \text{③}\\)
substituting ① into ③,
\\((x+y)^2 - 6 = 10\\)
\\((x+y)^2 = 16\\)
\\(x+y = \\)

(3) \\(\

$$\begin{cases} xy = 2 \\\\ x^2 + y^2 = 4 \\end{cases}$$

\\)

Explanation:

Step1: Solve for \(x+y\) in (2)

$$(x+y)^2 = 16 \implies x+y = \pm 4$$

Step2: Form quadratic equations for (2)

$$t^2 \mp 4t + 3 = 0$$

Step3: Solve for \(t\) to find solutions

$$t = 1, 3 \text{ or } t = -1, -3$$

Step4: Express solutions for (2)

$$(x, y) = (1, 3), (3, 1), (-1, -3), (-3, -1)$$

Step5: Rewrite system (3) using identity

$$(x+y)^2 - 2xy = 4 \implies (x+y)^2 - 4 = 4 \implies (x+y)^2 = 8$$

Step6: Solve for \(x+y\) in (3)

$$x+y = \pm 2\sqrt{2}$$

Step7: Form quadratic equations for (3)

$$t^2 \mp 2\sqrt{2}t + 2 = 0$$

Step8: Solve for \(t\) to find solutions

$$(t \mp \sqrt{2})^2 = 0 \implies t = \pm \sqrt{2}$$

Step9: Express solutions for (3)

$$(x, y) = (\sqrt{2}, \sqrt{2}), (-\sqrt{2}, -\sqrt{2})$$

Answer:

For (2):

$$(x, y) = (1, 3), (3, 1), (-1, -3), (-3, -1)$$

For (3):

$$(x, y) = (\sqrt{2}, \sqrt{2}), (-\sqrt{2}, -\sqrt{2})$$