Sample Grade 11 Mathematics worksheet on Equations and Inequalities from StudyChamp.co.za.

Here are the step-by-step solutions for the visible questions on the worksheet.

Question 1: Solve for $x$ using factorisation or simplification

1.1 $3x^2 - 27 = 0$
* Step 1: Divide the entire equation by 3 to make it simpler.
$$x^2 - 9 = 0$$
* Step 2: Recognize this as a difference of two squares ($a^2 - b^2$).
$$(x - 3)(x + 3) = 0$$
* Step 3: Set each bracket to zero to solve for $x$.
$$x - 3 = 0 \Rightarrow x = 3$$
$$x + 3 = 0 \Rightarrow x = -3$$

1.3 $7x - x^2 - 6 = 0$
* Step 1: Rearrange the terms into standard form ($ax^2 + bx + c = 0$). It is easier to work with a positive $x^2$, so multiply by $-1$.
$$x^2 - 7x + 6 = 0$$
* Step 2: Find two numbers that multiply to $+6$ and add up to $-7$. These numbers are $-1$ and $-6$.
$$(x - 1)(x - 6) = 0$$
* Step 3: Solve for $x$.
$$x = 1 \text{ or } x = 6$$

1.5 $x(x - 1) = 4(3x - 10)$
* Step 1: Expand both sides (remove the brackets).
$$x^2 - x = 12x - 40$$
* Step 2: Move all terms to one side to set the equation to zero.
$$x^2 - x - 12x + 40 = 0$$
$$x^2 - 13x + 40 = 0$$
* Step 3: Factorise. Find two numbers that multiply to $40$ and add to $-13$. These are $-5$ and $-8$.
$$(x - 5)(x - 8) = 0$$
* Step 4: Solve for $x$.
$$x = 5 \text{ or } x = 8$$

1.7 $(2x - 5)(3x - 2) + 5x = 22$
* Step 1: Expand the brackets first using FOIL (First, Outer, Inner, Last).
$$(6x^2 - 4x - 15x + 10) + 5x = 22$$
$$6x^2 - 19x + 10 + 5x = 22$$
* Step 2: Simplify the left side.
$$6x^2 - 14x + 10 = 22$$
* Step 3: Subtract 22 from both sides to get zero on the right.
$$6x^2 - 14x - 12 = 0$$
* Step 4: Divide by 2 to simplify.
$$3x^2 - 7x - 6 = 0$$
* Step 5: Factorise. We need numbers that multiply to $3 \times -6 = -18$ and add to $-7$. These are $-9$ and $2$.
$$3x^2 - 9x + 2x - 6 = 0$$
$$3x(x - 3) + 2(x - 3) = 0$$
$$(3x + 2)(x - 3) = 0$$
* Step 6: Solve for $x$.
$$3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$$
$$x - 3 = 0 \Rightarrow x = 3$$

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Question 2: Fractions – solve for $x$

2.1 $\frac{10}{3(x-1)} + \frac{1}{3} = \frac{4}{x-2}$
* Step 1: Determine restrictions. The denominators cannot be zero, so $x \neq 1$ and $x \neq 2$.
* Step 2: Find the Lowest Common Denominator (LCD), which is $3(x-1)(x-2)$. Multiply every term by the LCD to remove fractions.
$$10(x-2) + 1(x-1)(x-2) = 4[3(x-1)]$$
* Step 3: Expand the terms.
$$10x - 20 + (x^2 - 3x + 2) = 12(x - 1)$$
$$x^2 + 7x - 18 = 12x - 12$$
* Step 4: Bring everything to one side.
$$x^2 + 7x - 12x - 18 + 12 = 0$$
$$x^2 - 5x - 6 = 0$$
* Step 5: Factorise. Numbers multiplying to $-6$ and adding to $-5$ are $-6$ and $1$.
$$(x - 6)(x + 1) = 0$$
* Step 6: Solve for $x$.
$$x = 6 \text{ or } x = -1$$
*(Check restrictions: Neither is 1 or 2, so both are valid.)*

2.2 $\frac{2}{x-2} = \frac{3x+6}{x^2-x-2} + \frac{2}{x^2+4x+3}$
* Step 1: Factorise the quadratic denominators to find the LCD.
* $x^2 - x - 2 = (x - 2)(x + 1)$
* $x^2 + 4x + 3 = (x + 3)(x + 1)$
* LCD $= (x - 2)(x + 1)(x + 3)$
* Step 2: Restrictions: $x \neq 2, -1, -3$.
* Step 3: Multiply the whole equation by the LCD.
$$2(x+1)(x+3) = (3x+6)(x+3) + 2(x-2)$$
*Note: $3x+6$ can be factored as $3(x+2)$, but let's just expand normally.*
* Step 4: Expand.
Left Side: $2(x^2 + 4x + 3) = 2x^2 + 8x + 6$
Right Side Term 1: $(3x+6)(x+3) = 3x^2 + 9x + 6x + 18 = 3x^2 + 15x + 18$
Right Side Term 2: $2(x-2) = 2x - 4$

Equation: $2x^2 + 8x + 6 = 3x^2 + 15x + 18 + 2x - 4$
$2x^2 + 8x + 6 = 3x^2 + 17x + 14$
* Step 5: Simplify to a quadratic equation equal to zero.
$$0 = x^2 + 9x + 8$$
* Step 6: Factorise.
$$(x + 8)(x + 1) = 0$$
* Step 7: Solve for $x$.
$$x = -8 \text{ or } x = -1$$
* Step 8: Check restrictions. Since $x \neq -1$, we discard $-1$.
Final Answer: $x = -8$

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Question 3: Solve by completing the square (leave answers in surd form)

3.1 $x^2 + 2x = 24$
* Step 1: Take half of the coefficient of $x$ (which is 2), divide by 2 to get 1, and square it ($1^2 = 1$). Add this to both sides.
$$x^2 + 2x + 1 = 24 + 1$$
* Step 2: Write the left side as a perfect square.
$$(x + 1)^2 = 25$$
* Step 3: Take the square root of both sides.
$$x + 1 = \pm\sqrt{25}$$
$$x + 1 = \pm 5$$
* Step 4: Solve for $x$.
$$x = -1 + 5 = 4$$
$$x = -1 - 5 = -6$$

3.2 $x^2 + 5x - 3 = 0$
* Step 1: Move the constant to the right side.
$$x^2 + 5x = 3$$
* Step 2: Half of 5 is $\frac{5}{2}$. Square it to get $\frac{25}{4}$. Add to both sides.
$$x^2 + 5x + \frac{25}{4} = 3 + \frac{25}{4}$$
* Step 3: Convert 3 to $\frac{12}{4}$ to add the fractions.
$$(x + \frac{5}{2})^2 = \frac{12}{4} + \frac{25}{4} = \frac{37}{4}$$
* Step 4: Square root both sides.
$$x + \frac{5}{2} = \pm\sqrt{\frac{37}{4}}$$
$$x + \frac{5}{2} = \pm\frac{\sqrt{37}}{2}$$
* Step 5: Isolate $x$.
$$x = -\frac{5}{2} \pm \frac{\sqrt{37}}{2}$$
$$x = \frac{-5 \pm \sqrt{37}}{2}$$

3.4 $4x^2 - 9x = 12$
* Step 1: Make the coefficient of $x^2$ equal to 1 by dividing everything by 4.
$$x^2 - \frac{9}{4}x = 3$$
* Step 2: Half of $-\frac{9}{4}$ is $-\frac{9}{8}$. Square it to get $\frac{81}{64}$. Add to both sides.
$$x^2 - \frac{9}{4}x + \frac{81}{64} = 3 + \frac{81}{64}$$
* Step 3: Convert 3 to $\frac{192}{64}$ ($3 \times 64 = 192$).
$$(x - \frac{9}{8})^2 = \frac{192 + 81}{64} = \frac{273}{64}$$
* Step 4: Square root both sides.
$$x - \frac{9}{8} = \pm\sqrt{\frac{273}{64}}$$
$$x - \frac{9}{8} = \pm\frac{\sqrt{273}}{8}$$
* Step 5: Solve for $x$.
$$x = \frac{9 \pm \sqrt{273}}{8}$$

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Question 4: Solve for $x$ by squaring the root

4.1 $\sqrt{5 - x} = x + 1$
* Step 1: Square both sides to remove the square root.
$$5 - x = (x + 1)^2$$
$$5 - x = x^2 + 2x + 1$$
* Step 2: Rearrange into a quadratic equation equal to zero.
$$0 = x^2 + 3x - 4$$
* Step 3: Factorise.
$$(x + 4)(x - 1) = 0$$
$$x = -4 \text{ or } x = 1$$
* Step 4: Crucial Step: Check for extraneous solutions (solutions that don't work in the original equation because you can't have a negative result from a principal square root).
* If $x = -4$: LHS $= \sqrt{5 - (-4)} = \sqrt{9} = 3$. RHS $= -4 + 1 = -3$. ($3 \neq -3$, so reject).
* If $x = 1$: LHS $= \sqrt{5 - 1} = \sqrt{4} = 2$. RHS $= 1 + 1 = 2$. ($2 = 2$, so accept).
* Answer: $x = 1$

4.2 $\sqrt{3x - 1} - 1 = \frac{2}{\sqrt{3x - 1}}$
* Step 1: Let $u = \sqrt{3x - 1}$. The equation becomes:
$$u - 1 = \frac{2}{u}$$
* Step 2: Multiply by $u$ to clear the fraction.
$$u(u - 1) = 2$$
$$u^2 - u - 2 = 0$$
* Step 3: Factorise for $u$.
$$(u - 2)(u + 1) = 0$$
$$u = 2 \text{ or } u = -1$$
* Step 4: Substitute back $u = \sqrt{3x - 1}$.
* Case A: $\sqrt{3x - 1} = -1$. This has no solution because a square root cannot be negative.
* Case B: $\sqrt{3x - 1} = 2$. Square both sides.
$$3x - 1 = 4$$
$$3x = 5$$
$$x = \frac{5}{3}$$

Final Answer:
1.1 $x = 3; x = -3$
1.3 $x = 1; x = 6$
1.5 $x = 5; x = 8$
1.7 $x = -\frac{2}{3}; x = 3$
2.1 $x = 6; x = -1$
2.2 $x = -8$
3.1 $x = 4; x = -6$
3.2 $x = \frac{-5 \pm \sqrt{37}}{2}$
3.4 $x = \frac{9 \pm \sqrt{273}}{8}$
4.1 $x = 1$
4.2 $x = \frac{5}{3}$