Module 2 of 3Lesson 1 of 10 · Unit 12 · Simultaneous linear equations
Mathematics Form 3 · Term 2 · Unit 12

🔤 Simultaneous linear equations

AlgebraIGCSE path ✓JC ✓9Ae.059Ae.06
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Why this matters

One clue is not enough to find two unknowns. “A drink and a bun cost P12” could mean many things. Add a second clue and there is exactly one answer. Simultaneous equations are how shops, scientists and app designers find two unknowns from two facts.

🎯 By the end of this unit

  • I can explain why two unknowns need two equations.
  • I can solve simultaneous equations by elimination, adding or subtracting the equations.
  • I can solve simultaneous equations by substitution.
  • I can solve simultaneous equations by finding where two lines cross on a graph.
  • I can form simultaneous equations from a word problem, solve them and check my answer.
📖 Key words

simultaneous
happening at the same time: both equations must be true together
eliminate
get rid of one unknown, so only one is left
coefficient
the number in front of a letter, such as 3 in \(3x\)
substitute
replace a letter with an expression or a number
point of intersection
the point where two lines cross. It lies on both lines
solution
the pair of values, such as \(x=3\) and \(y=2,\) that makes both equations true

1 Learn it

Two unknowns need two clues

Two drinks and a bun cost P19. One drink and a bun cost P12. How much is one drink?

Compare the two rows. The difference is exactly one drink, and the difference in price is \(19-12=7.\)

So a drink costs P7. Then a bun costs \(12-7=5,\) so P5.

+= P192 drinks, 1 bun+= P121 drink, 1 bun= P7top row − middle row
Taking the middle row away from the top row leaves one drink and P7.
⭐ Key ideaOne equation with two unknowns has many answers. Two equations together usually pin down exactly one pair.

Elimination by subtracting

Write the shopping story with letters: \(d\) for a drink and \(b\) for a bun.

Both equations have exactly one \(b.\) Subtracting one equation from the other eliminates \(b.\)

  1. \(2d+b=19\) … (1)
  2. \(d+b=12\) … (2)
  3. (1) − (2): \(d=7\)
  4. Substitute into (2): \(7+b=12\)
  5. \(b=5\)
⭐ Key ideaSame coefficient, same sign: subtract the equations.

Elimination by adding

If the coefficients are the same but the signs are different, add the equations.

\(+2y\) and \(-2y\) add up to 0, just like 2 steps forward and 2 steps back.

  1. \(3x+2y=13\) … (1)
  2. \(5x-2y=11\) … (2)
  3. (1) + (2): \(8x=24\)
  4. \(x=3\)
  5. Substitute into (1): \(9+2y=13\)
  6. \(2y=4\)
  7. \(y=2\)
⭐ Key ideaDifferent signs: add. Same signs: subtract.

Making the coefficients match

Sometimes no letter has matching coefficients. Multiply one or both equations so that they do.

Multiplying a whole equation by a number keeps it true, as long as you multiply every term.

  1. \(2x+3y=12\) … (1)
  2. \(3x+2y=13\) … (2)
  3. (1) × 3: \(6x+9y=36\)
  4. (2) × 2: \(6x+4y=26\)
  5. Subtract: \(5y=10\)
  6. \(y=2\)
  7. Substitute into (1): \(2x+6=12\)
  8. \(x=3\)
⭐ Key ideaChoose multipliers that make one letter’s coefficients equal. Then add or subtract.

Substitution

When one equation already says what \(y\) (or \(x\)) equals, put that expression into the other equation.

This leaves one equation with one unknown.

  1. \(y=2x-1\) … (1)
  2. \(3x+y=14\) … (2)
  3. Put (1) into (2)
  4. \(3x+(2x-1)=14\)
  5. \(5x-1=14\)
  6. \(5x=15\)
  7. \(x=3\)
  8. Substitute into (1): \(y=6-1\)
  9. \(y=5\)
⭐ Key ideaSubstitution is quickest when one equation starts with \(y=\) or \(x=.\)

Solving with a graph

Every point on a line makes its equation true. The point where two lines cross is on both lines, so it makes both equations true.

Here \(y=x+1\) and \(x+y=5\) cross at \((2,3).\) So \(x=2\) and \(y=3.\)

−1123456−1123456xy0y = x + 1x + y = 5(2, 3)
The point of intersection (2, 3) is the solution.
⭐ Key ideaThe solution is the point of intersection. Read both coordinates. If the lines are parallel, they never meet and there is no solution.

Forming the equations from a problem

Choose a letter for each unknown and say what it stands for.

Turn each fact in the question into an equation. Solve, answer the question in words, and check with the original facts.

⭐ Key ideaLetters → two equations → solve → answer in words → check.

2 Watch it work

EXAMPLE 1

Subtract to eliminate

Solve the simultaneous equations
\(3x+2y=18\) … (1)
\(x+2y=10\) … (2)
  1. Both have \(+2y\): same sign, so subtract
  2. (1) − (2): \(2x=8\)
  3. \(x=4\)
  4. Substitute into (2): \(4+2y=10\)
  5. \(2y=6\)
  6. \(y=3\)
  7. Check (1): \(12+6=18\) ✓
Answer \(x=4,\) \(y=3\)
EXAMPLE 2

Add to eliminate

Solve the simultaneous equations
\(4x+3y=25\) … (1)
\(2x-3y=-1\) … (2)
  1. \(+3y\) and \(-3y\): different signs, so add
  2. (1) + (2): \(6x=24\)
  3. \(x=4\)
  4. Substitute into (1): \(16+3y=25\)
  5. \(3y=9\)
  6. \(y=3\)
  7. Check (2): \(8-9=-1\) ✓
Answer \(x=4,\) \(y=3\)
EXAMPLE 3

Multiply first

Solve the simultaneous equations
\(5x+2y=16\) … (1)
\(3x-4y=20\) … (2)
  1. (1) × 2: \(10x+4y=32\)
  2. Now \(+4y\) and \(-4y\): add this to (2)
  3. \(13x=52\)
  4. \(x=4\)
  5. Substitute into (1): \(20+2y=16\)
  6. \(2y=-4\)
  7. \(y=-2\)
  8. Check (2): \(12+8=20\) ✓
Answer \(x=4,\) \(y=-2\)
EXAMPLE 4

Substitution

Solve the simultaneous equations
\(y=3x-5\) … (1)
\(2x+y=15\) … (2)
  1. Put \(y=3x-5\) into the second equation
  2. \(2x+3x-5=15\)
  3. \(5x=20\)
  4. \(x=4\)
  5. \(y=12-5\)
  6. \(y=7\)
Answer \(x=4,\) \(y=7\)
EXAMPLE 5

Using a graph

The graph shows \(y=2x-1\) and \(y=-x+5.\) Use it to solve the two equations.
−1123456−2−1123456xy0y = 2x − 1y = −x + 5(2, 3)
  1. The lines cross at \((2,3)\)
  2. Check in the first: \(2\times 2-1=3\) ✓
  3. Check in the second: \(-2+5=3\) ✓
Answer \(x=2,\) \(y=3\)
EXAMPLE 6

A shopping problem

At a Choppies store, 3 loaves of bread and 2 cartons of milk cost P43.00. 2 loaves and 3 cartons of milk cost P42.00. Find the price of each.
  1. Let a loaf cost P\(b\) and a carton of milk cost P\(m\)
  2. \(3b+2m=43\) … (1)
  3. \(2b+3m=42\) … (2)
  4. (1) × 3: \(9b+6m=129\)
  5. (2) × 2: \(4b+6m=84\)
  6. Subtract: \(5b=45\)
  7. \(b=9\)
  8. \(27+2m=43\) (substitute into (1))
  9. \(2m=16\)
  10. \(m=8\)
  11. Check (2): \(18+24=42\) ✓
Answer Bread P9.00, milk P8.00

3 Your turn

Try each question in your book first. Then tap the button to check every step.

1🟢 Warm-up
Solve the simultaneous equations
\(x+y=10\) … (1)
\(x-y=4\) … (2)
👀 Show the answer
Step by step
  1. Add: \(2x=14\)
  2. \(x=7\)
  3. \(7+y=10\)
  4. \(y=3\)
Answer \(x=7,\) \(y=3\)
2🟢 Warm-up
Solve the simultaneous equations
\(2x+y=11\) … (1)
\(x+y=7\) … (2)
👀 Show the answer
Step by step
  1. Subtract: \(x=4\)
  2. \(4+y=7\)
  3. \(y=3\)
Answer \(x=4,\) \(y=3\)
3🟢 Warm-up
Solve the simultaneous equations
\(y=x+2\) … (1)
\(x+y=8\) … (2)
👀 Show the answer
Step by step
  1. Substitute: \(x+x+2=8\)
  2. \(2x=6\)
  3. \(x=3\)
  4. \(y=3+2\)
  5. \(y=5\)
Answer \(x=3,\) \(y=5\)
4🟢 Warm-up
Use the graph to solve \(y=2x+1\) and \(y=-x+4.\)
−2−11234−2−1123456xy0y = 2x + 1y = −x + 4?
👀 Show the answer
Step by step
  1. The lines cross at \((1,3)\)
  2. Check: \(2\times 1+1=3\) ✓ and \(-1+4=3\) ✓
Answer \(x=1,\) \(y=3\)
5🔵 Core
Solve the simultaneous equations
\(3x+4y=26\) … (1)
\(3x+y=11\) … (2)
👀 Show the answer
Step by step
  1. Subtract: \(3y=15\)
  2. \(y=5\)
  3. \(3x+5=11\)
  4. \(3x=6\)
  5. \(x=2\)
Answer \(x=2,\) \(y=5\)
6🔵 Core
Solve the simultaneous equations
\(5x+2y=9\) … (1)
\(3x-2y=15\) … (2)
👀 Show the answer
Step by step
  1. Add: \(8x=24\)
  2. \(x=3\)
  3. \(15+2y=9\)
  4. \(2y=-6\)
  5. \(y=-3\)
Answer \(x=3,\) \(y=-3\)
7🔵 Core
Solve the simultaneous equations
\(2x+5y=1\) … (1)
\(3x+2y=7\) … (2)
👀 Show the answer
Step by step
  1. (1) × 3: \(6x+15y=3\)
  2. (2) × 2: \(6x+4y=14\)
  3. Subtract: \(11y=-11\)
  4. \(y=-1\)
  5. \(2x-5=1\)
  6. \(x=3\)
Answer \(x=3,\) \(y=-1\)
8🔵 Core
Solve the simultaneous equations
\(x=2y+1\) … (1)
\(3x-4y=9\) … (2)
👀 Show the answer
Step by step
  1. \(3(2y+1)-4y=9\) (substitute)
  2. \(6y+3-4y=9\)
  3. \(2y=6\)
  4. \(y=3\)
  5. \(x=6+1\)
  6. \(x=7\)
Answer \(x=7,\) \(y=3\)
9🔵 Core
The sum of two numbers is 42 and their difference is 8. Find the numbers.
👀 Show the answer
Step by step
  1. Let the numbers be \(x\) and \(y\)
  2. \(x+y=42\) and \(x-y=8\)
  3. Add: \(2x=50\)
  4. \(x=25\)
  5. \(y=42-25\)
  6. \(y=17\)
Answer 25 and 17
10🔵 Core
Tickets for a football match at the National Stadium in Gaborone: 2 adult tickets and 3 child tickets cost P110.00. 1 adult ticket and 2 child tickets cost P65.00. Find the price of each ticket.
👀 Show the answer
Step by step
  1. Let an adult ticket cost P\(a\) and a child ticket P\(c\)
  2. \(2a+3c=110\) … (1)
  3. \(a+2c=65\) … (2)
  4. (2) × 2: \(2a+4c=130\)
  5. Subtract (1): \(c=20\)
  6. \(a+40=65\)
  7. \(a=25\)
Answer Adult P25.00, child P20.00
11🔵 Core
A farmer in the Kgalagadi keeps chickens and goats. The animals have 30 heads and 84 legs altogether. How many of each are there?
👀 Show the answer
Step by step
  1. Let there be \(c\) chickens and \(g\) goats
  2. \(c+g=30\) (heads)
  3. \(2c+4g=84\) (legs)
  4. \(2c+2g=60\) (double the first)
  5. Subtract: \(2g=24\)
  6. \(g=12\)
  7. \(c=30-12\)
  8. \(c=18\)
Answer 18 chickens and 12 goats
12🟠 Stretch
Data plan A costs P20 plus P5 per GB. Plan B costs P50 plus P2 per GB. For how many GB do the plans cost the same? Which plan is cheaper for 15 GB?
👀 Show the answer
Step by step
  1. Cost \(y\) for \(g\) GB: \(y=20+5g\) and \(y=50+2g\)
  2. Same cost when
  3. \(20+5g=50+2g\)
  4. \(3g=30\)
  5. \(g=10,\) and the cost is \(20+50=70\) pula
  6. For 15 GB: A costs \(20+75=95,\) B costs \(50+30=80\)
Answer 10 GB (both P70.00). For 15 GB, plan B is cheaper.
13🟠 Stretch
The line \(y=mx+c\) passes through \((1,5)\) and \((3,11).\) Find \(m\) and \(c.\)
👀 Show the answer
Step by step
  1. Substitute \((1,5)\): \(m+c=5\)
  2. Substitute \((3,11)\): \(3m+c=11\)
  3. Subtract: \(2m=6\)
  4. \(m=3\)
  5. \(3+c=5\)
  6. \(c=2\)
Answer \(m=3,\) \(c=2,\) so \(y=3x+2\)
14🟠 Stretch
Solve \(\dfrac{x}{2}+\dfrac{y}{3}=4\) and \(x-y=3.\)
👀 Show the answer
Step by step
  1. Multiply the first by 6: \(3x+2y=24\)
  2. From the second: \(x=y+3\)
  3. \(3(y+3)+2y=24\) (substitute)
  4. \(5y+9=24\)
  5. \(y=3\)
  6. \(x=6\)
Answer \(x=6,\) \(y=3\)
15🟠 Stretch
Kabelo tries to solve \(y=2x+1\) and \(y=2x-3.\) Explain why there is no solution, using the graph and using algebra.
−224−4−2246xy0y = 2x + 1y = 2x − 3
👀 Show the answer
Step by step
  1. Both lines have gradient 2, so they are parallel and never cross
  2. Algebra: \(2x+1=2x-3\)
  3. Subtract \(2x\): \(1=-3,\) which is impossible
Answer There is no point on both lines, so there is no solution.

4 Watch out!

✗ Wrong\(3x+2y=13\) and \(x+2y=7\): add to get \(4x=20\)
✓ RightSubtract: \(2x=6,\) so \(x=3\)

Adding gives \(4y,\) not 0. With the same sign in front of \(2y,\) you must subtract to eliminate \(y.\)

✗ Wrong\((5x-2y)-(3x-2y)\) \(=2x-4y\)
✓ Right\((5x-2y)-(3x-2y)\) \(=2x\)

Subtracting \(-2y\) is adding \(2y\): \(-2y-(-2y)=0.\) Take care with signs when you subtract.

✗ Wrong\(2x+y=7\) times 3 gives \(6x+y=21\)
✓ Right\(6x+3y=21\)

Multiply every term. Otherwise the equation is no longer true.

✗ WrongStopping at \(x=4\) and writing that as the answer
✓ Right\(x=4\) and \(y=3\)

The solution is a pair. Substitute \(x\) back to find \(y,\) then check both values in the other equation.

5 In your exams

🇧🇼 Junior Certificate (JCE)

Simultaneous equations are a regular JCE topic. In Paper 2 they are often a few marks in Section A (“Solve the simultaneous equations”) or part of a Section B real-life question where you form the equations first. You may also be asked to solve them by drawing two straight lines and reading the point of intersection. Paper 1 can ask which pair of values satisfies both equations: substitute each option to check.

🎓 Later in IGCSE

IGCSE 0580 Core papers ask you to solve two linear simultaneous equations by elimination or substitution, and to find the intersection of two lines on a graph. Extended papers add harder word problems and equations where one is linear and one is quadratic, which use the substitution method from this unit.

🏆 Challenge
A two-digit number has digits that add up to 11. When the digits are reversed, the number increases by 27. Find the number.
👀 Reveal the solution
Step by step
  1. Let the tens digit be \(t\) and the units digit \(u.\) The number is \(10t+u\)
  2. \(t+u=11\)
  3. Reversed number: \(10u+t\)
  4. \((10u+t)-(10t+u)\) \(=27\)
  5. \(9u-9t=27\)
  6. Divide by 9: \(u-t=3\)
  7. Add to \(t+u=11\): \(2u=14\)
  8. \(u=7\)
  9. \(t=4\)
  10. Check: \(74-47=27\) ✓
Answer 47
📝 Remember

  • Two unknowns need two equations. The solution is the pair of values that makes both true.
  • Elimination: make one letter’s coefficients equal. Same signs subtract, different signs add.
  • Multiply a whole equation (every term) to make coefficients match.
  • Substitution: replace \(y\) (or \(x\)) with what it equals in the other equation.
  • On a graph, the solution is the point where the two lines cross. Always check in both equations.

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