Simultaneous Linear Equations — Grade 10 Maths (CAPS)

Solving two equations in two unknowns by substitution and by elimination, what the solution means graphically, and the word problems that lead to a pair of equations.

One equation with two unknowns has endless solutions. Two equations pin them both down — and there are exactly two methods you need.

1What the answer means

y = x + 1 and y = −x + 5 cross at (2 ; 3). That single point satisfies both equations at once.

Each equation is a straight line, and a solution is a point lying on both lines — where they cross. Every example in this lesson has exactly one such point, which is the usual exam case. In general two lines can also be parallel (no solution at all) or be the same line written twice (infinitely many).

2Method 1: substitution

Best when one equation already has x or y on its own.

Worked ExampleSolve: y = x + 1 and y = −x + 5

1
Both equal y, so set them equal: x + 1 = −x + 5.
Substitute one expression for y into the other.
2
2x = 4, so x = 2.
Collect the x terms.
3
Substitute back: y = 2 + 1 = 3.
Use the simpler equation.
4
Solution: x = 2 and y = 3, i.e. the point (2 ; 3).
Check in the OTHER equation: −2 + 5 = 3 ✓

3Method 2: elimination

Best when both equations are in the form ax + by = c. Add or subtract the equations so one letter disappears.

Worked ExampleSolve: 2x + y = 7 and 3x − y = 8

1
The y terms are +y and −y, so ADD the two equations.
They cancel each other.
2
(2x + 3x) + (y − y) = 7 + 8 → 5x = 15.
3
x = 3.
4
Substitute into the first: 2(3) + y = 7, so y = 1.
Solution: (3 ; 1). Check in the second: 9 − 1 = 8 ✓

Worked ExampleSolve: 3x + 2y = 16 and 5x − 4y = 1

1
Multiply the first equation by 2: 6x + 4y = 32.
Now the y terms are +4y and −4y.
2
Add: 11x = 33, so x = 3.
3
Substitute: 3(3) + 2y = 16 → 2y = 7 → y = 3,5.
Check: 15 − 14 = 1 ✓

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Always check in the equation you did NOT use for the substitution. If it balances, both values are right.

4Word problems

Worked ExampleTwo pens and three books cost R74. Four pens and one book cost R38. Find each price.

1
Let a pen be x and a book be y.
Naming the unknowns earns a mark.
2
2x + 3y = 74 and 4x + y = 38.
Translate each sentence.
3
Multiply the second by 3: 12x + 3y = 114. Subtract the first: 10x = 40, so x = 4.
Elimination.
4
Then 4(4) + y = 38, so y = 22.
A pen costs R4 and a book R22. Check: 8 + 66 = 74 ✓

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Solve: y = 2x and x + y = 9
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x + 2x = 9 → x = 3, y = 6 → (3 ; 6)
2
Solve: y = x + 4 and y = 2x + 1
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x + 4 = 2x + 1 → x = 3, y = 7
3
Solve: x + y = 10 and x − y = 2
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Add: 2x = 12 → x = 6, y = 4
4
Solve: 2x + y = 7 and 3x − y = 8
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Add: 5x = 15 → x = 3, y = 1
5
Solve: 3x + 2y = 12 and x − 2y = 4
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Add: 4x = 16 → x = 4, y = 0
6
Solve: 2x + 3y = 13 and 4x − y = 5
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x = 2, y = 3
7
What does the solution represent graphically?
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The point where the two lines cross.
8
Three apples and two pears cost R16; one apple and one pear cost R6. Find each price.
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3a + 2p = 16, a + p = 6 → apple R4, pear R2