Find two unknown values that satisfy both equations. Learn how adding or subtracting the equations removes one unknown, then practise the method yourself.
Your learning plan
What you’ll do in this lesson
Turn two conditions into equations, solve them and check what the answers mean.
2 notebooks + 1 pen₹70
1 notebook + 1 pen₹45
Both purchases contain one pen. The first has one extra notebook. What does that extra notebook cost?
Understand the idea and what the unknown values mean.
Follow a worked example, one calculation at a time.
Explore how changing the values affects both equations.
Answer practice questions, use hints if needed, then try different questions.
Before you begin
What you need to know: One-variable equations and signed arithmetic.
This lesson covers: Start with modelling, elimination and solution types. Substitution and drawing graphs will come in later lessons.
Opens a focused learning view. Closing keeps your place and answers on this page. Reloading clears them.
Use this reference when you want all the explanations and questions together. Guided rehearsal is not an independent test; write the changed questions before revealing their answers.
Before you start
You will need: One-variable equations and signed arithmetic.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
Three ideas to keep in view
A notebook and a pen costing ₹45 could mean ₹25 + ₹20, or ₹30 + ₹15. Both add to ₹45. So that total alone does not tell us the two prices. We need to use the other purchase too.
Let n be the price of one notebook and p the price of one pen, both in rupees. The equation n + p = 45 means the two prices add up to ₹45. The equation 2n + p = 70 describes two notebooks and one pen. Here 2n means twice the notebook price; it does not mean two extra rupees.
We need the same values of n and p to make both totals correct. For example, n = 30 and p = 15 fit the ₹45 purchase, but two notebooks and a pen would then cost ₹75, not ₹70. So that pair cannot be our answer, even though it fits one equation.
Try the worked example first
reference-M10-LIN-W1Translate pricesFocused practice
Two notebooks and a pen cost ₹70. One notebook and a pen cost ₹45. Find each price.
Show your working
Need a starting hint?
Use n for notebook price and p for pen price.
Still stuck? Reveal the setup
Write 2n+p=70 and n+p=45.
Show worked solution+
1
Compare the purchases before doing algebra
Both purchases include one pen. The ₹70 purchase has one extra notebook compared with the ₹45 purchase. That extra notebook accounts for the extra ₹25, so one notebook costs ₹25.
2
See why subtracting works
Write the comparison as (2n + p) − (n + p) = 70 − 45. Subtract the whole second purchase: 2n − n leaves n, and p − p leaves 0. On the right, 70 − 45 is 25. We are left with n = 25. Removing the matching pen term is called elimination.
3
Use the notebook price to find the pen price
The smaller purchase costs ₹45. We now know ₹25 of that is the notebook, so the pen costs 45 − 25 = ₹20. Replace n with 25 in n + p = 45: 25 + p = 45, which gives p = 20.
4
Check both purchases
Two notebooks and a pen cost 2 × 25 + 20 = ₹70. One notebook and a pen cost 25 + 20 = ₹45. Both totals match, so the two prices work together as a solution.
5
Carry the idea into the next question
Look for a term you can remove by comparing the whole equations. Sometimes subtraction does it, as here. With x + y = 9 and x − y = 3, addition removes y because y + (−y) = 0. The aim is the same: make one unknown disappear so you can find the other.
Answer: Notebook ₹25; pen ₹20
Before moving on: can you explain why your method works, as well as give the answer?
Your turn: three different checks
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
reference-M10-LIN-Q1EliminateFocused practice
Solve x+y=9 and x−y=3.
Show your working
Need a starting hint?
Addition cancels y.
Still stuck? Reveal the setup
2x=12.
Show worked solution+
1
Add equations
x=6.
2
Substitute
6+y=9 gives y=3.
Answer: x=6, y=3
Before moving on: can you explain why your method works, as well as give the answer?
reference-M10-LIN-Q2No common solutionFocused practice
Do x+2y=5 and 2x+4y=12 have a solution?
Show your working
Need a starting hint?
Double the first equation.
Still stuck? Reveal the setup
It becomes 2x+4y=10.
Show worked solution+
1
Compare the whole equations
Doubling x + 2y = 5 gives 2x + 4y = 10. The other equation says 2x + 4y = 12, so the same expression would have to equal both 10 and 12.
2
Explain the contradiction
No values of x and y can do that. The conditions disagree, so the pair has no solution; it is not just a calculation that we have not finished.
Answer: No solution
Before moving on: can you explain why your method works, as well as give the answer?
reference-M10-LIN-Q3Same lineFocused practice
How many solutions do 3x−y=4 and 6x−2y=8 have?
Show your working
Need a starting hint?
Compare the entire equations.
Still stuck? Reveal the setup
The second is twice the first.
Show worked solution+
1
Look for a new condition
The second equation is exactly twice the whole first equation, including its right side. It repeats the same condition rather than adding a new one.
2
Try more than one pair
Both (x, y) = (0, −4) and (1, −1) satisfy the first equation, and therefore the second. There are infinitely many such pairs. On a graph, the equations draw the same line, so every point on it is shared.
Answer: Infinitely many
Before moving on: can you explain why your method works, as well as give the answer?
Catch a likely mistake
A tempting claim: “Multiply only one term when scaling an equation.”
That changes the equality instead of producing an equivalent equation.
Independent check
Close the examples and try again
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
reference-M10-LIN-R1New pairIndependent retest
Solve x+y=13 and x−y=5.
Show your working
Need a starting hint?
Add the equations.
Still stuck? Reveal the setup
2x=18.
Show worked solution+
1
Solve x
x=9.
2
Solve y
y=13−9=4.
Answer: x=9, y=4
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Three tickets and a snack cost ₹170. Two tickets and a snack cost ₹125. Find both prices.
Show your working
Need a starting hint?
Subtract the conditions.
Still stuck? Reveal the setup
The difference is one ticket.
Show worked solution+
1
Ticket
170−125=45.
2
Snack
125−2(45)=35.
Answer: Ticket ₹45; snack ₹35
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Choose the next useful step
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
Optional TutorMax original practice: five questions on linear equations, with explanations and a fresh retest. These checks sample a few skills; they are not official paper questions or a whole-subject readiness score. Existing explanations and official paper downloads stay available on this page.
