Edexcel International GCSE Mathematics A 4MA1: Simultaneous linear equations – Revision Notes

For full explanations and longer worked examples, use the simultaneous linear equations study guide. These notes condense section 2.6 Simultaneous linear equations of Topic 2 in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Statement 2.6A is on both tiers; statement 2.6B (lines and their point of intersection) is Higher tier only. A calculator may be used on every paper, but the marks are mostly for method.

Course hub: Edexcel IGCSE Mathematics. Tick off each statement on the printable checklist, then test yourself with the practice questions or a free diagnostic.

The two statements

Spec	You must be able to	Tier
2.6A	Calculate the exact solution of two simultaneous equations in two unknowns	Both tiers
2.6B	Interpret the equations as lines and the common solution as their point of intersection	Higher tier only

Not in this section: one linear and one quadratic equation together (statement 2.7D, Higher tier only).

Key terms

• Simultaneous equations: two equations that must both be true for the same values of the unknowns.
• Solution: the pair of values, for example x = 4, y = −1. Always give both.
• Exact: no rounding. Fractions stay as fractions (2/3, not 0.67).
• Coefficient: the number in front of a letter. In 5x − 3y = 8, the coefficient of y is −3.
• Eliminate: remove one letter by adding or subtracting the equations.
• Point of intersection (Higher tier only): the point where two lines cross. Its coordinates are the solution.

Method in steps: elimination

1. Label the equations (1) and (2), terms lined up in columns.
2. Multiply one or both so one letter has the same coefficient size.
3. Same signs -> SUBTRACT.  Different signs -> ADD.
4. Solve for the letter that is left.
5. Substitute into (1) or (2) to find the other letter.
6. Check both values in the equation you did not use in step 5.

Choosing the multipliers. Look at both letters and pick the one with the smaller lowest common multiple (LCM).

Coefficients of one letter	Multiply by	Matched coefficient
1 and 4	4 and 1	4
2 and 3	3 and 2	6
4 and 6	3 and 2	12
3 and 5	5 and 3	15

Method in steps: substitution

1. Make x or y the subject of one equation (often it already is).
2. Put that expression, IN BRACKETS, into the other equation.
3. Expand, collect like terms, solve.
4. Substitute back for the second value. Check.

Use substitution when you see “y = …” or “x = …”. Use elimination when both equations are in the form ax + by = c.

Small worked reminders

Add. 7x + 2y = 31 and 3x − 2y = 9. The y terms have opposite signs, so add: 10x = 40, x = 4. Then 28 + 2y = 31, so y = 1.5. Check: 12 − 3 = 9. x = 4, y = 1.5.

Subtract after one multiplication. a + 3b = 11 and 4a + 5b = 23. Multiply the first by 4: 4a + 12b = 44. Same sign, subtract: 7b = 21, b = 3. Then a = 11 − 9 = 2. a = 2, b = 3.

Multiply both. 3x + 4y = 10 and 5x − 6y = 4. The y coefficients 4 and 6 have LCM 12, so multiply by 3 and 2: 9x + 12y = 30 and 10x − 12y = 8. Opposite signs, add: 19x = 38, x = 2. Then 6 + 4y = 10, so y = 1. x = 2, y = 1.

Substitute. x = 2y + 1 and 3x − 4y = 7. Then 3(2y + 1) − 4y = 7, so 6y + 3 − 4y = 7, 2y = 4, y = 2, and x = 5. x = 5, y = 2.

Clear fractions first. If an equation has fractions, multiply every term by the lowest common denominator before you start. For x/4 + y/2 = 3, multiply by 4: x + 2y = 12.

Checking and presenting your answer

• Substitute both values into the equation you did not use to find the second value. If it fails, look first at the line where you added or subtracted.
• Write each multiplied equation on its own line, labelled, for example (1) × 3. A marker can then give method credit even if a later number is wrong.
• Write the final answer clearly as x = …, y = … (or in words, for a context question).
• Show the method even though a calculator is allowed. Much of the credit is for the elimination or substitution line.

Setting up equations from words

Words in the question	Equation
Two numbers add to 30	x + y = 30
Their difference is 8	x − y = 8 (x the larger)
3 pens and 2 rulers cost £4.10	3p + 2r = 4.10
Opposite sides of a rectangle, (2x + y) and (x + 7)	2x + y = x + 7
Fixed charge F plus r per km; 6 km costs £11	F + 6r = 11

Define each letter with its units, and finish with a sentence that answers the question.

Lines and intersections (Higher tier only)

• Each linear equation is a straight line. Each point on it is one solution of that equation.
• The point where two lines cross is on both lines, so its coordinates solve both equations.
• To find where two lines meet, solve their equations simultaneously. Give the answer as coordinates (x, y).
• Parallel lines (same gradient, different y-intercept) never meet: no solution. Elimination then removes both letters and leaves a false statement such as 0 = 4.
• To compare gradients, rearrange ax + by = c into y = mx + c first. For 3x + 2y = 8, y = −(3/2)x + 4, so the gradient is −3/2, not 3.
• A hand-drawn graph gives an estimate. “Solve” means algebra, for an exact answer.

Must-know distinctions

Situation	Do this	Not this
Letter to remove has the same sign in both	Subtract	Add
Letter to remove has opposite signs	Add	Subtract
Multiplying an equation	Multiply every term, including the right-hand side	Only the left side
Substituting an expression	Use brackets: 2(3x − 1)	2 × 3x − 1
Answer is a fraction	Leave it exact: 4/3	1.33
Asked where lines meet (Higher tier only)	Coordinates (x, y)	Just x

Quick self-test

Solve each pair exactly unless told otherwise.

1. x + y = 9 and x − y = 3
2. 2x + y = 11 and x + y = 7
3. 3a + 2b = 16 and a − 2b = 0
4. 5x + 2y = 1 and 3x + 2y = −1
5. y = 2x + 3 and 3x + y = 13
6. 2m + 3n = 1 and 3m − 2n = 8
7. 4x − 3y = 2 and 2x + 3y = 7
8. Two numbers have a sum of 40 and a difference of 12. Find them.
9. 2 pens and 3 notebooks cost £13.50. 4 pens and 1 notebook cost £12.50. Find the cost of one pen and one notebook.
10. (Higher) Find the coordinates of the point where y = x + 4 meets y = 10 − 2x.
11. (Higher) Explain why 2x + y = 3 and 4x + 2y = 10 have no solution.

Answers

1. Add: 2x = 12, x = 6, y = 3.
2. Subtract: x = 4, then y = 3.
3. Add: 4a = 16, a = 4; then 4 − 2b = 0, b = 2.
4. Subtract: 2x = 2, x = 1; then 5 + 2y = 1, y = −2.
5. 3x + 2x + 3 = 13, 5x = 10, x = 2, y = 7.
6. Multiply by 2 and 3: 4m + 6n = 2 and 9m − 6n = 24. Add: 13m = 26, m = 2; then 4 + 3n = 1, n = −1.
7. Add: 6x = 9, x = 3/2; then 3 + 3y = 7, y = 4/3.
8. x + y = 40, x − y = 12, so 2x = 52: 26 and 14.
9. 2p + 3n = 13.50 and 4p + n = 12.50. From the second, n = 12.50 − 4p, so 2p + 37.50 − 12p = 13.50, giving 10p = 24: pen £2.40, notebook £2.90.
10. x + 4 = 10 − 2x, 3x = 6, x = 2, y = 6: (2, 6).
11. Rearranged, y = −2x + 3 and y = −2x + 5. Same gradient (−2), different y-intercepts, so the lines are parallel and never meet.

Where marks are usually lost

• Adding the equations when the letter being removed has the same sign in both, or subtracting when the signs differ.
• Multiplying the left-hand side of an equation but not the number on the right.
• Errors such as 4y − (−3y) = y; it is 7y.
• Leaving out brackets when substituting, so 3(2y + 1) becomes 6y + 1.
• Giving 0.67 when the exact answer 2/3 was needed, then carrying the rounded value into the second unknown.
• Finding one value and forgetting to work out the other.
• In a word problem, not defining the letters, or not stating the final answer in context with units (£, cm).
• Not checking in the second equation, so a slip goes unnoticed.
• (Higher tier only) Giving only the x-value when the question asks for the coordinates of a point of intersection.
• (Higher tier only) Taking the coefficient of x in ax + by = c as the gradient when comparing two lines.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities, section 2.6 Simultaneous linear equations: statement 2.6A (Foundation and Higher tiers) and statement 2.6B (Higher tier only).