Section A · The Idea
What Is a Solution?
Why should I care? Simultaneous equations are how you find two mystery numbers from two clues. That happens all the time:
- Shopping: 2 coffees and 1 cake cost $8; 1 coffee and 1 cake cost $5. One clue alone cannot tell you the price of a coffee — both together can.
- Choosing a phone plan: plan A is $10 a month + $2 per GB, plan B is $25 a month + $0.50 per GB. At how many GB do they cost the same? Where the two lines cross — try the phone plans button in the Graph Explorer (Section C).
- Break-even: a T-shirt shop spends $500 plus $3 per shirt and sells each for $8. How many shirts before it makes a profit? Where the cost line meets the income line.
- Later on: the same idea finds where supply meets demand in economics, when one car catches up with another in physics, and it powers the software that solves systems with thousands of unknowns.
One equation is not enough. x + y = 10 has lots of answers: x = 1 and y = 9, x = 2 and y = 8, x = 7.5 and y = 2.5, and so on for ever. Two unknowns need two clues.
A second equation narrows it down. Add the clue x − y = 4. Of all the pairs that add up to 10, only x = 7 and y = 3 also have a difference of 4. Two equations that must be true at the same time are called simultaneous equations, and their solution is one pair of values: x = 7, y = 3, often written as the point (7, 3).
There are three ways to find that one pair. This sheet practises all three, then mixes them:
| Method | Use it when | The first move |
|---|---|---|
| Substitution | both equations start with y = (Questions B) | set the two right-hand sides equal to each other |
| Graphs | you have a grid to draw on (Questions C) | draw both lines; the crossing point is the answer |
| Elimination | both equations look like ax + by = c (Questions D) | add or subtract the equations so one letter disappears |
Testing a pair: put the values into both equations. A pair is only a solution if both sides match in both equations. This is also how you check every answer at the end.
Q1
Which pair of values makes both x + y = 9 and x − y = 3 true?
1 mark
Q2
Is the point (2, 7) the solution of y = 2x + 3 and y = x + 5?
1 mark
Q3
The point (3, 5) fits the equation y = x + 2. Which second equation would make (3, 5) the solution of the pair?
1 mark
WHAT IS A SOLUTION? — SECTION SCORE
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Section B · Workbook Q1–Q2
The Method of Substitution
Why it works: if y = 3x − 2 and also y = x + 4, then 3x − 2 and x + 4 are the same number (both are y). So you can write them equal to each other and get an equation with only x in it.
The four steps (the same boxes as the workbook):
- Work out x — write the right-hand sides equal: 3x − 2 = x + 4. Take x from both sides and add 2 to both sides: 3x − x = 4 + 2, so 2x = 6 and x = 3.
- Work out y — substitute x = 3 into the first equation: y = 3 × 3 − 2 = 7.
- Check — substitute into the other equation: y = 3 + 4 = 7. Both give 7, so the answer is right.
- Write the answers: x = 3 and y = 7.
Watch the signs when a number changes side: a number that is subtracted on one side is added when it moves to the other side, and the other way round. Doing the same thing to both sides keeps the equation balanced.
Q4
Solve y = 2x + 3 and y = x + 7. Step 1: 2x + 3 = x + 7, so 2x − x = 7 − 3. What is x?
1 mark
Q5
Same equations, y = 2x + 3 and y = x + 7. Step 2: substitute your x into y = 2x + 3. What is y?
1 mark
Q6
Solve y = 5x − 4 and y = 2x + 8. What is x?
1 mark
Q7
Same equations, y = 5x − 4 and y = 2x + 8. What is y?
1 mark
Q8
Solve y = 4x + 1 and y = x − 8. Give both values.
2 marks
x =y =
Q9
Sam solves y = 3x + 2 and y = x + 10. He writes: 3x − x = 10 + 2, so 2x = 12 and x = 6. What went wrong?
1 mark
SUBSTITUTION — SECTION SCORE
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Section C · Workbook Q3 & Q8
The Graphical Method
Why it works: every point on the line of y = 2x + 1 is a pair (x, y) that makes that equation true. Every point on the other line makes the other equation true. The only point that is on both lines is where they cross — so the crossing point is the solution.
The steps:
- Make a table of values for each equation (for example x = 0, 2, 4).
- Plot the points and draw each straight line with a ruler.
- Read the coordinates of the intersection (the crossing point): (x, y).
- Check the values in both equations.
Limits of graphs: a graph is only as exact as your drawing and the grid. Use the explorer below to see what happens when the crossing is not on a gridline, and when the two lines never meet.
Graph Explorer — change the two lines and watch where they cross
Line 1: y =x +
Line 2: y =x +
Q10
Table of values for y = 3x − 1. What is y when x = 4?
1 mark
Q11
Table of values for y = x + 3. What is y when x = 4?
1 mark
Q12
Here are both tables. At which x do the two lines meet?
1 mark
| x | 0 | 2 | 4 |
|---|---|---|---|
| y = 3x − 1 | −1 | 5 | 11 |
| y = x + 3 | 3 | 5 | 7 |
Q13
The graphs of y = x + 1 and y = 7 − 2x are drawn below. Read off the crossing point and write the solution.
2 marks
x =y =
Q14
Mia draws y = 2x + 1 and y = 5 − x. Their crossing point is at x = 1⅓, y = 3⅔. Why is the graph a poor way to give her answer?
1 mark
GRAPHS — SECTION SCORE
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Section D · Workbook Q4 & Q9
The Method of Elimination
Why it works: both equations are true, so you can add them together (left side + left side = right side + right side) and the result is still true. If you choose well, one letter cancels out (is eliminated) and only one letter is left.
Stack the equations, like the workbook. Write one equation under the other, with x under x, y under y and the numbers under the numbers. Then look at the letter you want to get rid of, and work down each column:
x + y=12+x − y=22x + 0y=14
+y and −y → add. y + (−y) = 0y, so 2x = 14 and x = 7.
x + 5y=23−x + 2y=110x + 3y=12
+x and +x → subtract every column: x − x, 5y − 2y, 23 − 11. So 3y = 12 and y = 4.
Then finish like substitution: put the value you found into the first equation to find the other letter (in the first stack: 7 + y = 12, so y = 5), check in the second equation (7 − 5 = 2 ✓), and write both answers.
Q15
Solve x + y = 15 and x − y = 7 by adding the equations. What is x?
1 mark
x + y=15+x − y=72x + 0y=?
Q16
Same equations, x + y = 15 and x − y = 7. What is y?
1 mark
Q17
Complete the stack: what do you get when you subtract x + 2y = 13 from x + 4y = 23?
1 mark
x + 4y=23−x + 2y=13?=?
Q18
Same equations, x + 4y = 23 and x + 2y = 13. Find y, then work out x. What is x?
1 mark
Q19
Solve 2x + 3y = 24 and 4x − 3y = 12. Stack them and choose: add or subtract?
2 marks
2x + 3y=24?4x − 3y=12?=?
x =y =
Q20
Which first step solves 5x + 2y = 29 and 3x + 2y = 19?
1 mark
ELIMINATION — SECTION SCORE
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Section E · Workbook Q5–Q10
Mixed Practice
Choose your method first. Both equations start with y = → substitution. Both look like ax + by = c with a matching letter → elimination. One is y = … and the other is not → substitute the y = … expression into the other equation. Give both values, and check them in the equation you did not use last.
Q21
Solve y = 4x and y = x + 9.
2 marks
x =y =
Q22
Solve y = x + 10 and y = −3x + 2.
2 marks
x =y =
Q23
Solve x + y = 25 and x − y = 9.
2 marks
x =y =
Q24
Solve 2x + y = 17 and x − y = 4.
2 marks
x =y =
Q25
Solve x + 3y = 17 and x + y = 9.
2 marks
x =y =
Q26
Solve y = 2x and x + y = 18.
2 marks
x =y =
Q27
Solve y = 3(x − 1) and y = x + 5.
2 marks
x =y =
MIXED PRACTICE — SECTION SCORE
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Section F · Workbook Q11–Q13
Problems in Words
Turning words into equations:
- Choose a letter for each unknown and write down what it means (c = cost of one cake in $).
- Turn each sentence with a total into one equation. “2 cakes and 3 coffees cost $9” becomes 2c + 3f = 9.
- Solve the pair with the method that fits.
- Answer the question in words, with units — and check the values make sense in the story.
Q28
3 pens and 2 notebooks cost $13. 1 pen and 2 notebooks cost $7. Let p = the cost of a pen and n = the cost of a notebook. Which pair of equations fits?
1 mark
Q29
Solve 3p + 2n = 13 and p + 2n = 7. How much does one pen cost?
1 mark
$
Q30
Same equations. How much does one notebook cost?
1 mark
$
Q31
Two numbers add up to 30. Their difference is 8. What is the larger number?
1 mark
Q32
x + y = 15 and x − y = 3. Work out the value of 3x + 2y.
2 marks
Q33
At the cinema, 2 adult tickets and 3 child tickets cost $36. 2 adult tickets and 1 child ticket cost $24. What does one adult ticket cost?
2 marks
$
PROBLEMS IN WORDS — SECTION SCORE
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