Equations and Inequalities

Full lesson notes

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\( \text{Solve} \;\Rightarrow\; \text{find every value that makes it true} \)

This lesson covers the equation work you need in Grade 10 Term 1:
• quadratic equations solved by factorising
• changing the subject of a formula
• equations with algebraic fractions and their restrictions
• linear inequalities, number lines and interval notation
• simultaneous linear equations
• word problems

Each topic has worked examples and quick checks. Try each check on paper first, then open Show worked answer.

1. Quadratics

\( A \times B = 0 \;\Rightarrow\; A = 0 \text{ or } B = 0 \)

If two factors multiply to give zero, at least one of them must be zero. That is why a quadratic must be written as (factor)(factor) = 0 before you solve it.

Example. Solve \( x(2x-5) = 0 \).

Never divide both sides by \( x \): you would lose the answer \( x = 0 \).

\( 3x^2 - 2x - 8 = 0 \)

Step 1. Make sure one side is 0. It is.
Step 2. Factorise. We need two brackets whose outer and inner products add to \( -2x \):

Step 3. Set each factor equal to zero:

\( (x-3)(x+2) = 0 \qquad (2x+1)(x-3) = 0 \)

Solve for \( x \): (a) \( (x-3)(x+2) = 0 \)   (b) \( (2x+1)(x-3) = 0 \)

Show worked answer

(a) \( x - 3 = 0 \) or \( x + 2 = 0 \), so \( x = 3 \) or \( x = -2 \).
(b) \( 2x + 1 = 0 \) or \( x - 3 = 0 \), so \( x = -\frac{1}{2} \) or \( x = 3 \).

Both are already factorised and equal to zero, so you go straight to the zero-product step.

\( \text{Standard form} \;\downarrow \) \( \text{factorise} \;\downarrow \) \( \text{each factor} = 0 \)

• Get 0 on one side first.
• Factorise completely (common factor first, then trinomial).
• Set each factor equal to 0 and solve.
• A quadratic usually has two answers; never divide by \( x \).

2. Changing the subject

\( x = y + xy \)

Make \( x \) the subject. \( x \) appears in two terms, so collect them on one side and take \( x \) out as a common factor.

Treat the other letters exactly as if they were numbers. Every operation you do must be done to the whole of both sides.

\( \pi x^2 h = V \)

Make \( x \) the subject. Undo the operations in reverse order.

When you take a square root to solve, include both the positive and negative roots, unless the letter stands for a length (a length cannot be negative).

\( V = \frac{1}{3}\pi r^2 h \)

The volume of a cone is \( V = \frac{1}{3}\pi r^2 h \). Make \( r \) the subject.

Show worked answer

Multiply by 3: \( 3V = \pi r^2 h \)
Divide by \( \pi h \): \( r^2 = \dfrac{3V}{\pi h} \)
Square root: \( r = \sqrt{\dfrac{3V}{\pi h}} \)

Only the positive root is kept, because \( r \) is a radius (a length).

\( \text{collect} \;\downarrow \) \( \text{factorise} \;\downarrow \) \( \text{divide} \)

• If the subject appears more than once, collect those terms on one side and take it out as a common factor.
• Undo operations in reverse order.
• Square root gives \( \pm \); keep only \( + \) for lengths.

3. Algebraic fractions

\( \dfrac{3}{x+2} + \dfrac{5}{x-4} = 2 \)

A denominator can never be zero, so before solving write down the values \( x \) may not take:

Then multiply every term by the LCD \( (x+2)(x-4) \) to clear the fractions:

\( x = -1 \text{ or } x = 7 \) \( x \neq -2;\ 4 \)

Continue from
\( 3(x-4) + 5(x+2) \)
\( = 2(x+2)(x-4) \):

Compare with the restrictions \( x \neq -2;\ 4 \). Neither answer is excluded, so both are valid.

\( \dfrac{8x^3 - 1}{2x - 1} = 1 \)

Restriction: \( 2x - 1 \neq 0 \Rightarrow x \neq \frac{1}{2} \).

The numerator is a difference of cubes: \( 8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1) \).

Neither answer equals \( \frac{1}{2} \), so both are valid.

\( \dfrac{4x^2 - 3x - 1}{4x + 1} \) \( + \dfrac{x^3 + 1}{x^2 - x + 1} = 2 \)

Solve for \( x \). State any restriction. (Hint: factorise both numerators.)

Show worked answer

Restriction: \( 4x + 1 \neq 0 \Rightarrow x \neq -\frac{1}{4} \). (The trinomial \( x^2 - x + 1 \) is never zero for real \( x \).)


\( 4x^2 - 3x - 1 \)
\( = (4x + 1)(x - 1) \)
\( x^3 + 1 \)
\( = (x + 1)(x^2 - x + 1) \)
so

\( (x - 1) + (x + 1) = 2 \)
\( 2x = 2 \)
\( x = 1 \)

The only solution is \( x = 1 \), with \( x \neq -\frac{1}{4} \).

\( \text{restrictions} \;\downarrow \) \( \text{LCD} \;\downarrow \) \( \text{solve} \;\downarrow \) \( \text{check} \)

• Write the restrictions before you start.
• Factorise and simplify where possible, then multiply every term by the LCD.
• Solve the equation that remains.
• Reject any answer that breaks a restriction, and state the restrictions with your answer.

4. Inequalities

\( -6 \le 2x - 2 \lt 10 \)

Solve an inequality like an equation, but do the same thing to all three parts.

Closed dot for \( \le \) (included), open dot for \( \lt \) (not included):Interval notation: \( x \in [-2;\ 6) \). A square bracket means included, a round bracket means not included.

\( -11 \lt -2x + 1 \lt -9 \)

When you multiply or divide an inequality by a negative number, every inequality sign flips.Interval notation: \( x \in (5;\ 6) \). Both ends are open.

\( 5 \lt \sqrt{29} \lt 6 \)

Does \( x = \sqrt{29} \) satisfy \( 5 \lt x \lt 6 \)?

So yes: \( \sqrt{29} \approx 5{,}39 \) lies inside the interval. Squeeze the number between two perfect squares.

\( -9 \le 2x + 3 \lt 5 \)

Solve, show the answer on a number line and write it in interval notation.

Show worked answer

Subtract 3: \( -12 \le 2x \lt 2 \)
Divide by 2: \( -6 \le x \lt 1 \)Interval notation: \( x \in [-6;\ 1) \).

\( -4 \le 3x - 1 \le 5 \)

Solve, show the answer on a number line and write it in interval notation.

Show worked answer

Add 1: \( -3 \le 3x \le 6 \)
Divide by 3: \( -1 \le x \le 2 \)Interval notation: \( x \in [-1;\ 2] \). Both ends are closed.

\( \times \text{ or } \div \text{ by a negative:} \) \( \text{flip the sign} \)

• Do the same operation to every part of the inequality.
• Multiplying or dividing by a negative reverses the signs.
• Number line: closed dot for \( \le, \ge \); open dot for \( \lt, \gt \).
• Interval notation: [ ] included, ( ) not included.

5. Simultaneous

\( 2x - y - 3 = 0 \qquad 3x + 2y - 8 = 0 \)

Two unknowns need two equations. Make one variable the subject of one equation, then substitute it into the other.

Solution: \( x = 2;\ y = 1 \). Check in the second equation: \( 3(2) + 2(1) = 8 \).

\( 4a - b = 5 \qquad -3a + 4b = 19 \)

Multiply so that one variable has opposite coefficients, then add the equations.

Solution: \( a = 3;\ b = 7 \).

\( 4x - 10 + 3y = 0 \qquad y + 2x - 6 = 0 \)

Solve for \( x \) and \( y \).

Show worked answer

From the second: \( y = 6 - 2x \)
Substitute: \( 4x - 10 + 3(6 - 2x) = 0 \)
\( -2x + 8 = 0 \Rightarrow x = 4 \)
\( y = 6 - 2(4) = -2 \)

Solution: \( x = 4;\ y = -2 \).

\( 2x - y + 1 = 0 \qquad x + 2y - 12 = 0 \)

Solve for \( x \) and \( y \).

Show worked answer

From the first: \( y = 2x + 1 \)
Substitute: \( x + 2(2x + 1) - 12 = 0 \)
\( 5x - 10 = 0 \Rightarrow x = 2 \)
\( y = 2(2) + 1 = 5 \)

Solution: \( x = 2;\ y = 5 \).

\( \text{one subject} \;\downarrow \) \( \text{substitute} \;\downarrow \) \( \text{solve} \;\downarrow \) \( \text{back-substitute} \)

• Substitution: best when a variable has coefficient 1 or \( -1 \).
• Elimination: multiply to get opposite coefficients, then add.
• Always find both values and check them in the equation you did not use.

6. Word problems

\( \dfrac{x + (2x - 4) + (4x + 2)}{3} = 25 \)

The mean of three numbers is 25. The first number is \( x \), the second is \( 2x - 4 \) and the third is \( 4x + 2 \). Find \( x \).

• Name the unknown and write the other quantities in terms of it.
• Translate the words into an equation: mean = sum ÷ number of values.

\( x = 11 \)

Solve the equation from the previous step and find the three numbers.

Show worked answer

\( x + 2x - 4 + 4x + 2 = 75 \)
\( 7x - 2 = 75 \)
\( 7x = 77 \Rightarrow x = 11 \)

The numbers are \( 11 \), \( 2(11) - 4 = 18 \) and \( 4(11) + 2 = 46 \).
Check: \( \frac{11 + 18 + 46}{3} = \frac{75}{3} = 25 \).

\( \text{words} \;\downarrow \) \( \text{equation} \;\downarrow \) \( \text{solve} \;\downarrow \) \( \text{check} \)

• Let \( x \) be the unknown; write everything else in terms of \( x \).
• Build one equation from the information.
• Solve and answer the question that was asked.
• Check your answer in the original words.

\( \text{factorise} \cdot \text{restrict} \) \( \text{flip} \cdot \text{substitute} \)

• Quadratics: 0 on one side, factorise, each factor = 0.
• Changing the subject: collect, factorise, divide; \( \pm \) when you square root.
• Algebraic fractions: restrictions first, multiply by the LCD, reject excluded answers.
• Inequalities: flip the sign when you multiply or divide by a negative; show a number line and interval notation.
• Simultaneous: substitution or elimination; find both values.
• Word problems: define \( x \), build the equation, check in the words.