Question

j144b solve the following simultaneous equations

(2) \\(\

$$\begin{cases} xy+x+y=5 \\\\ 2xy+x+3y=9 \\end{cases}$$

\\)

(3) \\(\

$$\begin{cases} xy+2x+2y+1=0 \\\\ 3xy+3x+5y+3=0 \\end{cases}$$

\\)

Explanation:

Step1: Write down equations for (2)

$$
LATEXBLOCK0
$$

Step2: Eliminate the \(xy\) term

Multiply (1) by 2 and subtract from (2):

$$(2xy + x + 3y) - 2(xy + x + y) = 9 - 2(5)$$

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

Step3: Substitute \(y\) into (1)

$$x(x - 1) + x + (x - 1) = 5$$

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

Step4: Solve the quadratic equation

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

Step5: Find corresponding \(y\) values

If \(x = 2\), then \(y = 2 - 1 = 1\).
If \(x = -3\), then \(y = -3 - 1 = -4\).
Thus, the solutions are \((2, 1)\) and \((-3, -4)\).

Step6: Write down equations for (3)

$$
LATEXBLOCK1
$$

Step7: Eliminate the \(xy\) term

Multiply (3) by 3 and subtract from (4):

$$(3xy + 3x + 5y + 3) - 3(xy + 2x + 2y + 1) = 0$$

$$-3x - y = 0 \implies y = -3x$$

Step8: Substitute \(y\) into (3)

$$x(-3x) + 2x + 2(-3x) + 1 = 0$$

$$-3x^2 - 4x + 1 = 0 \implies 3x^2 + 4x - 1 = 0$$

Step9: Solve the quadratic equation

Using the quadratic formula:

$$x = \frac{-4 \pm \sqrt{16 - 4(3)(-1)}}{2(3)} = \frac{-4 \pm \sqrt{28}}{6} = \frac{-2 \pm \sqrt{7}}{3}$$

Step10: Find corresponding \(y\) values

If \(x = \frac{-2 + \sqrt{7}}{3}\), then \(y = 2 - \sqrt{7}\).
If \(x = \frac{-2 - \sqrt{7}}{3}\), then \(y = 2 + \sqrt{7}\).

Answer:

For (2):
\(x = 1, y = 2\) or \(x = -4, y = \frac{1}{3}\)

For (3):
\(x = -1, y = 1\) or \(x = -3, y = -1\)