Question

j142b solve the following simultaneous equations

(3) \\(\

$$\begin{cases} y = x - 3 \\\\ x^2 - y^2 = 15 \\end{cases}$$

\\)

(4) \\(\

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

\\)

Explanation:

Step1: Substitute first equation into second for (3)

$$x^2 - (x - 3)^2 = 15$$

Step2: Expand and simplify the equation

$$x^2 - (x^2 - 6x + 9) = 15 \implies 6x - 9 = 15$$

Step3: Solve for x

$$6x = 24 \implies x = 4$$

Step4: Find corresponding y value

$$y = 4 - 3 = 1$$

Step5: Express y in terms of x for (4)

$$y = 2x - 3$$

Step6: Substitute into second equation

$$x^2 - (2x - 3)^2 = 3$$

Step7: Expand and simplify the equation

$$x^2 - (4x^2 - 12x + 9) = 3 \implies -3x^2 + 12x - 12 = 0$$

Step8: Solve the quadratic equation

$$x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0 \implies x = 2$$

Step9: Find corresponding y value

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

Answer:

(3) \(x = 4\), \(y = 1\)
(4) \(x = 2\), \(y = 1\) or \(x = 2\), \(y = 1\) (double root)