MathematicsExam code: 0580
Linear Equations and Inequalities
Algebra and Sequences
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