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Learn · Mathematics · Grade 9 · Algebra

Systems of two linear equations

Substitution and elimination, what the two lines look like when there is one solution, none or infinitely many, and word problems that turn into two equations.

High school Lesson 5 of 31 Interactive figure Practice

First, the rules

Two unknowns need two equations

★ Substitution
Example 1
Solve one equation for one unknown (y = …), put that expression into the other equation, solve, then go back for the second unknown.
★ Elimination (addition)
Examples 1, 3
Multiply one or both equations so that one unknown has opposite coefficients, add the equations, solve, substitute back.
Always check
Put the pair (x, y) back into BOTH original equations.
★ What the picture says
Example 2
Each equation is a line. One crossing point: exactly one solution. Parallel lines: no solution. The same line: infinitely many solutions.
Recognising the three cases from the coefficients
Example 2
a₁a₂ ≠ b₁b₂ one solution · a₁a₂ = b₁b₂ ≠ c₁c₂ none · all three equal infinitely many
Word problems
Example 3
Name the two unknowns, write one equation per sentence of information, solve, then answer the actual question in words.

See it

Two lines, one crossing

Change the coefficients. The green dot is the solution. Make the lines parallel (same ratio a : b, different c) and the solution disappears; make them identical and every point works.

Step by step

Worked examples

1

Elimination

match a coefficient

Solve 2x + 3y = 12 and 3x − 2y = 5.

Ask yourself: Which unknown is easier to cancel? Multiply the first by 2 and the second by 3 to get 6y and −6y.

Show the solution
  1. First × 2: 4x + 6y = 24. Second × 3: 9x − 6y = 15.
  2. Add: 13x = 39 ⇒ x = 3.
  3. Back into the first: 6 + 3y = 12 ⇒ y = 2. Check in the second: 9 − 4 = 5 ✓.

Answer (x, y) = (3, 2)

2

When there is no solution

parallel lines

Solve x − 2y = 4 and −3x + 6y = 9.

Ask yourself: Multiply the first equation by −3. What do you get on the left? Compare with the second.

Show the solution
  1. First × (−3): −3x + 6y = −12.
  2. Second says the same left side equals 9. Subtracting: 0 = 21 — impossible.
  3. The ratios 1 / (−3) = (−2) / 6 are equal but 4 / 9 is different: parallel lines, never meet.

Answer no solution

3

Tickets

word problem

A cinema sold 150 tickets for a total of 1,050. Adult tickets cost 9 and child tickets cost 5. How many of each were sold?

Ask yourself: Two unknowns: adults a and children c. One sentence about the count, one about the money.

Show the solution
  1. Count: a + c = 150. Money: 9a + 5c = 1050.
  2. Substitution: c = 150 − a, so 9a + 5(150 − a) = 1050 ⇒ 4a + 750 = 1050 ⇒ a = 75.
  3. c = 75. Check money: 9 · 75 + 5 · 75 = 675 + 375 = 1050 ✓.

Answer 75 adult and 75 child tickets

Practice: solve the system

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About this lesson

Systems of two linear equations in two unknowns: the substitution method, the elimination (addition) method, what the picture of two lines says — one crossing point, parallel lines (no solution) or the same line (infinitely many) — and how to set up a system from a word problem about prices or ages. Move the lines yourself, then practise.

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