GCSE · Algebra

Simultaneous Equations

Two equations, two unknowns, one answer that makes both true at the same time. On a graph, that answer is where two lines (or a line and a curve) cross. This page takes the KS3 methods further: scaling both equations, setting up equations from real situations, and (Higher) solving a straight line with a quadratic or a circle.

GCSE subject content: solve two simultaneous equations in two variables (linear/linear) algebraically; find approximate solutions using a graph; translate situations into simultaneous equations and interpret the solution; Higher: solve linear/quadratic simultaneous equations algebraically and graphically

Recap. If you haven't met simultaneous equations before, start with Simultaneous equations (KS3): it explains elimination with matching coefficients and simple substitution. You'll also need to draw lines from their equations (Linear graphs), and, for the Higher sections, to solve quadratics (Solving quadratic equations).

1 · One answer, two lines

An equation like x + y = 7 has infinitely many solutions: (1, 6), (2, 5), (3.5, 3.5)… Plotted, they make a straight line. A second equation, 2x − y = 2, is a second line. Simultaneous means "at the same time": we want the one pair (x, y) that is on both lines, which is where they cross. Change the equations and watch the crossing point.

x + y = x + y =

Reading the crossing off a graph is only accurate when it lands on grid points. Algebra gives the exact answer every time.

2 · Elimination when both equations need scaling

Elimination means adding or subtracting the equations so that one unknown disappears. That only works when one unknown has the same size coefficient in both equations (the coefficient is the number in front of the letter). In 3x + 4y = 18 and 5x − 3y = 1 nothing matches, so we multiply both equations: to make the y's match, aim for the lowest common multiple of 4 and 3, which is 12.

Two big ideas the graph shows. Multiplying an equation by a number doesn't move its line (6x + 8y = 36 is the same line as 3x + 4y = 18). And adding or subtracting the equations makes a new line through the same crossing point: when y has been eliminated it is a vertical line, x = number.

Pair eliminate

Add or subtract? Look at the signs of the matching terms. Same signs: subtract (+12y − (+12y) = 0). Different signs: add (+12y + (−12y) = 0). When subtracting, subtract every term, including the right-hand side: 9x − 20x = −11x and 54 − 4 = 50.

Common mistake: multiplying only the left-hand side. ×3 applies to every term: 3x + 4y = 18 becomes 9x + 12y = 54.

3 · Substitution

When one equation already says "y = …" (or "x = …"), substitution is usually quicker: replace that letter in the other equation with the whole expression, in a bracket. The second equation then has only one unknown. This is also the only method that works for the Higher curve and circle questions later on, so it's worth getting fluent.

Pair

Use a bracket. Substituting y = 2x − 3 into 3x + 2y = 15 gives 3x + 2(2x − 3) = 15. Without the bracket, students write 3x + 4x − 3 and get the wrong answer.

4 · Setting up equations from a situation

Exam questions often describe a situation in words. The method is always the same: choose letters for the two unknowns (say what each stands for, with units), turn each fact into one equation, solve, then answer the question that was asked in words, and check against the original facts.

Situation

Units trap: if one price is in pounds and another in pence, convert first. Writing 4p + 3r = 3.30 and 2p + 5r = 340 mixes units and gives nonsense. Work in pence throughout, then convert the answer back.

5 · A line and a quadratic curve Higher

Now one equation is a quadratic, like y = x² − 2x − 3, and the other is a straight line, y = mx + k. A line can cut a U-shaped curve twice, touch it once, or miss it. Elimination won't work here (there is an x² in only one equation), so use substitution: both equations say "y = …", so set the right-hand sides equal. You get a quadratic equation in x, and its discriminant b² − 4ac tells you how many crossing points there are.

curve y = m = k =

6 · A line and a circle Higher

The equation x² + y² = r² is a circle with centre (0, 0) and radius r (it is Pythagoras' theorem for every point on the circle). To find where the line y = mx + k meets it, substitute the line into the circle: replace y with (mx + k), expand the bracket squared carefully, and solve the quadratic. Each x then gives its y from the line equation. When there is exactly one point, the line is a tangent to the circle.

circle x² + y² = m = k =

Two classic slips. (1) Expanding (x + 1)² as x² + 1: it is x² + 2x + 1. (2) Finding the x values but then putting them into the circle to get y: that gives ± two y values for each x, and only one of them is on the line. Always use the linear equation to find y, and give your answers as pairs.

7 · Practice: linear pairs

Solve each pair (use elimination or substitution) and enter x and y. The feedback checks your answer against each equation separately, so you can see where it went wrong.

x = y =

Worked examples

Solve 3x + 4y = 18 and 5x − 3y = 1

Plan: eliminate y. LCM of 4 and 3 is 12, so multiply the first equation by 3 and the second by 4.

Scale: 9x + 12y = 54 and 20x − 12y = 4.

Different signs on 12y, so add: 29x = 58, so x = 2.

Substitute back into 3x + 4y = 18: 6 + 4y = 18, 4y = 12, y = 3.

Check in the other equation: 5(2) − 3(3) = 10 − 9 = 1 ✓. x = 2, y = 3.

Solve y = 2x − 3 and 3x + 2y = 15 by substitution

Substitute the first into the second, in a bracket: 3x + 2(2x − 3) = 15.

Expand and solve: 3x + 4x − 6 = 15, 7x = 21, x = 3.

Find y from the first equation: y = 2(3) − 3 = 3.

Check: 3(3) + 2(3) = 15 ✓. x = 3, y = 3.

Exam-style: 3 adult and 2 child tickets cost £36.50; 2 adult and 5 child tickets cost £39. Find the price of each.

Letters: let a = price of an adult ticket (£), c = price of a child ticket (£).

Equations: 3a + 2c = 36.5 and 2a + 5c = 39.

Eliminate a: × 2 and × 3: 6a + 4c = 73 and 6a + 15c = 117. Same signs, subtract: 11c = 44, c = 4.

Back-substitute: 3a + 8 = 36.5, 3a = 28.5, a = 9.5.

Answer in words: adult £9.50, child £4.00. Check: 2 × 9.50 + 5 × 4 = 19 + 20 = 39 ✓

Solve y = x² − 3x + 1 and y = 2x − 3 Higher

Both are "y = …", so set them equal: x² − 3x + 1 = 2x − 3.

Rearrange to = 0: x² − 5x + 4 = 0.

Factorise: (x − 1)(x − 4) = 0, so x = 1 or x = 4.

y from the line: x = 1 gives y = −1; x = 4 gives y = 5.

Answer as pairs: (1, −1) and (4, 5). Check (4, 5) in the curve: 16 − 12 + 1 = 5 ✓

Solve x² + y² = 25 and y = x + 1 Higher

Substitute the line into the circle: x² + (x + 1)² = 25.

Expand carefully: x² + x² + 2x + 1 = 25, so 2x² + 2x − 24 = 0.

Divide by 2: x² + x − 12 = 0, (x + 4)(x − 3) = 0, so x = −4 or x = 3.

y from the line: x = 3 → y = 4; x = −4 → y = −3.

Answer: (3, 4) and (−4, −3). Check: 3² + 4² = 25 ✓, (−4)² + (−3)² = 25 ✓

Show that the line y = 2x + 5 does not meet the curve y = x² + 7 Higher

Set equal: x² + 7 = 2x + 5, so x² − 2x + 2 = 0.

Discriminant: b² − 4ac = (−2)² − 4 × 1 × 2 = 4 − 8 = −4.

Conclude: −4 < 0, so the equation has no real solutions, so the line and the curve never meet.

Quick check

What you've learned

The solution of a pair of simultaneous equations is the point that lies on both graphs: where they cross.
Elimination: scale one or both equations so that one unknown has matching coefficients (use the LCM), then add if the signs differ, subtract if they're the same. Scaling an equation doesn't move its line.
Substitution: replace a letter with the expression it equals, in a bracket. It's best when an equation is already "y = …".
In context: define your letters with units, one fact gives one equation, answer in words and check against the story.
Higher For a line and a quadratic or a circle, substitute the line into the other equation, rearrange to a quadratic = 0 and solve. Use the line to find each y. Two, one or no solutions match two, one or no intersections (and one means the line is a tangent).