1. Solving linear equations
What makes an equation linear
- A linear equation contains an unknown whose highest power is 1, so there is no x² term
- Solving means isolating the unknown by undoing each operation applied to it
- Whatever is done to one side must be done to the other, which is what keeps the equation true
The method
- Undo additions and subtractions first, then multiplications and divisions
- Work one operation per line so each step is visible
Worked example
A two-step equation
Solve 5x − 8 = 22.
Solution:
- Adding 8 to each side clears the subtraction, leaving 5x = 30
- Since 5 multiplies the unknown, undo it by dividing each side by 5
- x = 6
Exam tip
A correct first line scores on its own
Your first correct move earns a mark on its own: 5x = 30 above keeps that mark even if the division then goes wrong. With a bracket or fractions, the line that collects the unknown on one side earns a second mark, which still counts when it follows correctly from a first line with a slip. Only the last mark needs the right answer, so write every move down rather than jumping to the answer.
2. Brackets, fractions and unknowns on both sides
Brackets
- Expand the bracket first, then solve as usual
- 3(2x + 5) = 39 expands to 6x + 15 = 39, giving 6x = 24 and x = 4
- Dividing through by the number outside also works, though it can create fractions
Fractions
- Scale the whole equation by the lowest common denominator, so no term is left divided
- Terms without a denominator must be multiplied too, which is the step most often missed
Worked example
Clearing fractions
Solve x/3 + x/4 = 7.
Solution:
- The lowest common denominator of 3 and 4 is 12
- Scaling every term by 12 gives 4x + 3x = 84
- Collect: 7x = 84
- x = 12
Unknowns on both sides
- Remove the smaller unknown term by applying its inverse to both sides, which avoids negatives
- 7x − 4 = 2x + 21 becomes 5x − 4 = 21 after subtracting 2x
- Then 5x = 25 and x = 5
- Fractional equations with algebraic denominators are handled the same way, and are covered under algebraic fractions
Worked example
Fractions with the unknown on both sides
Solve (3x − 1)/5 = (x + 3)/2.
Solution:
- Multiply both sides by 5 and by 2, keeping each numerator in a bracket: 2(3x − 1) = 5(x + 3)
- Expand: 6x − 2 = 5x + 15
- Collect the x terms on one side and the numbers on the other: 6x − 5x = 15 + 2
- x = 17
- Check: (3 × 17 − 1)/5 = 50/5 = 10 and (17 + 3)/2 = 20/2 = 10
- Without the bracket, 2 × 3x − 1 gives 6x − 1, and the answer comes out wrong
3. Inequalities
Reading the symbols
- An inequality states that one quantity is greater or less than another rather than equal to it
- < and > are strict, excluding the end value; ⩽ and ⩾ are inclusive
- x > 5 excludes 5, whereas x ⩾ 5 includes it
Listing integer solutions
- Test the two ends independently, since one may be included while the other is not
- For −2 < x ⩽ 3 the integers are −1, 0, 1, 2 and 3
- The strict end is excluded and the inclusive end is kept
Worked example
Listing the integers that satisfy an inequality
List the integers n for which −2 ⩽ n < 3.
Solution:
- Read each end symbol separately, because they are not the same
- ⩽ on the left means −2 is included
- < on the right means 3 is not included
- Working up from −2: −2, −1, 0, 1, 2
- The next integer would be 3, which the strict sign excludes
- The integers are −2, −1, 0, 1, 2
- Zero is an integer and is easy to leave out by accident
4. Solving inequalities
Almost the same as an equation
- Apply the same operations to both sides as when solving an equation
- 4x + 5 ⩽ 29 gives 4x ⩽ 24 and x ⩽ 6
- A double inequality is solved by doing the same thing to all three parts
- −5 < 2x − 1 ⩽ 9 gives −4 < 2x ⩽ 10 and −2 < x ⩽ 5
The one rule that differs
- A negative multiplier or divisor flips the direction of the inequality
- −3x > 12 gives x < −4, not x > −4
- Avoiding the reversal altogether is often simpler: move the unknown to the side that keeps it positive
Worked example
Solving two inequalities
Solve 5x − 3 > 12, and solve −3x ⩾ 12.
Solution:
- For the first, treat it exactly as an equation
- Add 3 to both sides: 5x > 15
- Divide both sides by 5: x > 3
- The sign does not change, because 5 is positive
- For the second, divide both sides by −3
- Dividing by a negative reverses the inequality sign
- 12 ÷ −3 = −4, and ⩾ becomes ⩽, giving x ⩽ −4
- Check with a value: x = −5 gives −3 × −5 = 15, and 15 ⩾ 12 is true
- Leaving the sign as ⩾ would claim x ⩾ −4, and x = 0 gives 0 ⩾ 12, which is false
5. Inequalities on a number line
The conventions
- The specification fixes how end points are drawn, and marks are given for using them correctly
- An open circle marks a strict end point, which is excluded
- A closed circle marks an inclusive end point, which is included
- A line joins two end points, and an arrow shows an unbounded direction
- The diagram above represents −3 < x ⩽ 4
Worked example
Reading a solution off a number line
A number line has a closed circle at −1 and an open circle at 4, with the line between them shaded. Write down the inequality it represents.
Solution:
- Take each end in turn, using the circle to fix the symbol
- The closed circle at −1 means −1 is included, so that end uses ⩽
- The open circle at 4 means 4 is excluded, so that end uses <
- The shading lies between them, so x is between the two values
- The inequality is −1 ⩽ x < 4
- Written the other way round it reads 4 > x ⩾ −1, which says the same thing