Los op vir x: 1.1.1 $x^2 - x - 12 = 0$ 1.1.2 $x(x + 3) - 1 = 0$ (Laat jou antwoord in die eenvoudigste wortelvorm.) 1.1.3 $x(4 - x) < 0$ 1.1.4 $x = \frac{a^2 + a - 2}{a - 1}$ aas $a = 888888888$ 1.2 Los die volgende vergelykings gelyktydig op: $y + 7 = 2x$ en $x^2 - xy + 3y = 15$ 1.3 Bepaal die waardeverzameling van die funksie $y = x + \frac{1}{x}$, $x \neq 0$ en $x$ is reël. - NSC Mathematics - Question 1 - 2016 - Paper 1

Rearranging gives:

$x(x + 3) = 1$

Expanding this, we have: $x^2 + 3x - 1 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4(1)(-1)}}{2(1)}$

This simplifies to: $x = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}$

To solve the inequality, first, find the roots:

1. When $x = 0$, $4 - 0 > 0$.
2. When $x = 4$, $0 < 0$.

This means the critical points are $x = 0$ and $x = 4$. Using test intervals:

• For $x < 0$, choose $x = -1$: $(-1)(4 - (-1)) < 0$ (True)
• For $0 < x < 4$, choose $x = 2$: $(2)(4 - 2) > 0$ (False)
• For $x > 4$, choose $x = 5$: $(5)(4 - 5) < 0$ (True)

Thus, the solution is $x < 0$ or $x > 4$.

Substituting $a = 888888888$ into the expression:

$x = \frac{(888888888)^2 + 888888888 - 2}{888888888 - 1}$

This requires calculating the values:

1. First calculate $(888888888)^2$.
2. Then, add $888888888$ and subtract $2$.
3. Lastly, divide by $888888887$.

First, isolate $y$ from the first equation:

$y = 2x - 7$

Now substitute into the second equation:

$x^2 - x(2x - 7) + 3(2x - 7) = 15$

This expands to: $x^2 - 2x^2 + 7x + 6x - 21 = 15$

Combine the terms to form: $-x^2 + 13x - 36 = 0$

Using the quadratic formula: $x = \frac{-13 \pm \sqrt{(13)^2 - 4(-1)(-36)}}{-2}$

To find the range of $y = x + \frac{1}{x}$, analyze its behavior:

1. Differentiate to find critical points: $y' = 1 - \frac{1}{x^2}$
2. Setting $y' = 0$ results in critical points which can be analyzed for maxima and minima.
3. Calculate limits as $x \to 0^+$ and $x \to \infty$.
4. Determine that the function has a minimum value at $y = 2$ when $x = 1$ and approaches infinity otherwise.

Thus, the range is $y \geq 2$.