MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Elimination Method for a Pair of Linear Equations

Updated 13 October 2026 · Fact-checked

The elimination method solves two linear equations in two variables by multiplying them so that one variable has equal coefficients, then adding or subtracting the equations to cancel it. The board asks you to solve pairs by elimination, use the solution to find an unknown, and solve word problems and case studies with it.

Make the y coefficients equal in size (4 and −4), add to cancel y, solve for x, then substitute back to find y.

Illustration drawn for QuestionBuddy.

Understand Elimination Method

You have two linear equations in the same two unknowns, $x$ and $y$ . Each one alone has infinitely many solutions. You want the one pair $(x, y)$ that makes both true at once. The elimination method gets there by removing one variable completely. That leaves one equation in one unknown, which you can solve in a line.

The idea rests on two facts. First, if two equations are both true, their sum and their difference are also true. Second, multiplying an equation by a non-zero number gives the same line. For example, $2 x + y = 5$ and $6 x + 3 y = 15$ are the same line, so the solution does not change. So you may scale the equations freely, then add or subtract them without losing the solution.

To eliminate a variable, make its coefficients equal in size. Take $3 x + 4 y = 10$ and $2 x - 2 y = 2$ . Multiply the second by $2$ to get $4 x - 4 y = 4$ . Now $y$ has $+ 4$ and $- 4$ . The signs are opposite, so add: $7 x = 14$ , giving $x = 2$ . If the signs had been the same, you would subtract. Use the LCM of the two coefficients as the target so the numbers stay small.

Elimination is often faster than substitution when no coefficient is $1$ or $- 1$ . It avoids fractions in the middle of the working. It also has a neat shortcut. When the coefficients are swapped, as in $37 x + 43 y = 123$ and $43 x + 37 y = 117$ , adding and subtracting give $x + y$ and $x - y$ directly.

Sometimes both variables cancel together. If you are left with $0 = 0$ , the equations are multiples of each other. The lines coincide, so there are infinitely many solutions. If you get $0 = 2$ or another false statement, the lines are parallel and there is no solution. This matches the ratio conditions you learn in this chapter: both variables cancel exactly when $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ .

The method works for any pair of linear equations. First write both in the same form, such as $a x + b y = c$ , with like terms lined up. Clear decimals or fractions first by multiplying through. Then check your answer in the equation you did not use last.

Elimination Method: formulas to remember

Elimination principle: $E_{1} = 0, E_{2} = 0 \Rightarrow m E_{1} \pm n E_{2} = 0 (m, n \neq = 0)$; Scaling an equation by a non-zero number and adding or subtracting equations keeps the solution unchanged.
Choosing multipliers: $a_{1} x + b_{1} y = c_{1}, a_{2} x + b_{2} y = c_{2} : multiply by \frac{L}{a _{1}}, \frac{L}{a _{2}}, L = LCM (a_{1}, a_{2})$; Gives both x-coefficients equal to L. Use the same idea with b-coefficients to eliminate y.
Add or subtract rule: $same signs \Rightarrow subtract; opposite signs \Rightarrow add$; Apply it to the coefficients of the variable you want to remove.
Swapped-coefficient shortcut: $a x + b y = c, b x + a y = d \Rightarrow x + y = \frac{c + d}{a + b}, x - y = \frac{c - d}{a - b}$; Needs a ≠ b and a + b ≠ 0. Then add and subtract these two results to get x and y.
Reading a total cancellation: $0 = 0 \Rightarrow infinitely many solutions; 0 = k (k \neq = 0) \Rightarrow no solution$; Both variables cancel only when a1/a2 = b1/b2.

How to solve Elimination Method questions

Use this routine for any elimination question, from a 2-mark solve to a Section E case.

1Write both equations as $a x + b y = c$ with $x$ , $y$ and constant terms lined up. Number them (1) and (2).
2Clear decimals or fractions by multiplying each equation by a suitable number, such as 10 or the LCM of the denominators.
3Choose the variable to eliminate. Pick the one whose coefficients have the smaller LCM, or opposite signs, because then you can add.
4Multiply each equation so that this variable has the same coefficient in both. Write the new equations as (3) and (4).
5Add if the signs of that coefficient are opposite. Subtract if they are the same. Put every term of the subtracted equation in brackets.
6Solve for the remaining variable. Substitute it into the simplest original equation to find the other variable.
7Check both values in the equation you did not use. Then answer any 'hence' part, or state the answer in words with units.
8If both variables cancel, read the result: $0 = 0$ means infinitely many solutions, and a false statement means no solution.

Elimination Method in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Elimination method

Suppose your school canteen sells samosas and cups of tea at fixed prices. One day you buy 3 samosas and 2 teas for Rs 65. The canteen owner says that if you had taken 2 more samosas and 1 more tea, your bill would have been Rs 40 more. For a class outing you have Rs 200, and your teacher's rule is at least one tea for every two samosas.

Form a pair of linear equations and find the price of a samosa and of a tea by elimination. Then find the largest number of samosas you can buy within Rs 200 while following the teacher's rule.

Try it first, then show the solutionSolution

Let one samosa cost Rs $s$ and one tea cost Rs $t$ .
First bill: $3 s + 2 t = 65$ ... (1).
The extra 2 samosas and 1 tea cost Rs 40: $2 s + t = 40$ ... (2).
(2) × 2: $4 s + 2 t = 80$ ... (3).
(3) − (1): $s = 15$ .
From (2): $30 + t = 40$ , so $t = 10$ .
Check in (1): $45 + 20 = 65$ . Correct.
To buy as many samosas as possible, buy the fewest teas the rule allows: half as many teas as samosas, rounded up.
10 samosas need 5 teas: $150 + 50 = 200$ , which fits the Rs 200 budget.
11 samosas need at least 6 teas: $165 + 60 = 225$ , which is over Rs 200. More samosas cost even more.

Answer: A samosa costs Rs 15 and a tea costs Rs 10. The largest number of samosas is 10 (with 5 teas), costing exactly Rs 200.

Memory hook: Match the tea count, subtract the bills: what is left is pure samosa.

Tricks to remember Elimination Method

Short memory aids. Say each one aloud once, then use it on a question the same day.

SAME-Subtract, OPPOSITE-Add
Same Signs Subtract. Opposite signs add. Just remember the double S.
Use it when: Decide the operation the moment the coefficients match.
LCM, not product
Aim for the LCM of the two coefficients, not their product. For $6$ and $4$ , aim for $12$ , not $24$ .
Use it when: Keeps numbers small and cuts arithmetic errors.
Swapped? Add and subtract
If the numbers on $x$ and $y$ swap places between the equations, add for $x + y$ and subtract for $x - y$ .
Use it when: Pairs like $47 x + 31 y$ and $31 x + 47 y$ with big coefficients.
Bracket before you subtract
Write $(\dots) - (\dots)$ around the whole equation you subtract, so every sign flips.
Use it when: Stops the most common sign error with negative terms.

Elimination Method: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A direct elimination with opposite signs · 2 marks · Very short answerBoard level

Solve by the elimination method: $3 x + 4 y = 10$ and $2 x - 2 y = 2$ .

Try it first, then show the solutionSolution

$3 x + 4 y = 10$ ... (1), $2 x - 2 y = 2$ ... (2).
(2) × 2: $4 x - 4 y = 4$ ... (3). The $y$ -coefficients are $+ 4$ and $- 4$ , so add. (½ mark)
(1) + (3): $7 x = 14$ , so $x = 2$ . (½ mark)
From (2): $4 - 2 y = 2$ , so $y = 1$ . (½ mark)
Check in (1): $6 + 4 = 10$ . Correct. (½ mark)

Answer: $x = 2$ , $y = 1$

Example 2: Swapped-coefficient shortcut · 3 marks · Short answerHarder board level

Solve $37 x + 43 y = 123$ and $43 x + 37 y = 117$ . Hence find the value of $x^{2} + y^{2}$ .

Try it first, then show the solutionSolution

$37 x + 43 y = 123$ ... (1), $43 x + 37 y = 117$ ... (2).
(1) + (2): $80 x + 80 y = 240$ , so $x + y = 3$ ... (3). (1 mark)
(2) − (1): $6 x - 6 y = - 6$ , so $x - y = - 1$ ... (4). (½ mark)
(3) + (4): $2 x = 2$ , so $x = 1$ . Then $y = 2$ . (½ mark)
Check in (1): $37 + 86 = 123$ . Correct. (½ mark)
$x^{2} + y^{2} = 1 + 4 = 5$ . (½ mark)

Answer: $x = 1$ , $y = 2$ , and $x^{2} + y^{2} = 5$

Example 3: Clearing fractions first, then a hence part · 3 marks · Short answerHarder board level

Solve by elimination: $\frac{x}{2} + \frac{2 y}{3} = - 1$ and $x - \frac{y}{3} = 3$ . Hence find $p$ if the solution satisfies $y = p x - 7$ .

Try it first, then show the solutionSolution

Multiply the first equation by 6: $3 x + 4 y = - 6$ ... (1). (½ mark)
Multiply the second equation by 3: $3 x - y = 9$ ... (2). (½ mark)
Same sign on $3 x$ , so subtract. (1) − (2): $5 y = - 15$ , so $y = - 3$ . (½ mark)
From (2): $3 x + 3 = 9$ , so $x = 2$ . (½ mark)
Check: $\frac{2}{2} + \frac{2 ( - 3 )}{3} = 1 - 2 = - 1$ . Correct.
Substitute into $y = p x - 7$ : $- 3 = 2 p - 7$ , so $2 p = 4$ . (½ mark)
$p = 2$ . (½ mark)

Answer: $x = 2$ , $y = - 3$ , and $p = 2$

Example 4: Ticket sales word problem · 5 marks · Long answerHarder board level

A school fair sells adult and child tickets at fixed prices. On Saturday, 7 adult and 5 child tickets bring in Rs 1500. On Sunday, 4 adult and 9 child tickets bring in Rs 1410. Using the elimination method, find the price of each ticket. Hence find the amount a family pays for 10 adult and 6 child tickets.

Try it first, then show the solutionSolution

Let an adult ticket cost Rs $a$ and a child ticket cost Rs $c$ . (½ mark)
$7 a + 5 c = 1500$ ... (1), $4 a + 9 c = 1410$ ... (2). (1 mark)
(1) × 4: $28 a + 20 c = 6000$ ... (3). (2) × 7: $28 a + 63 c = 9870$ ... (4). (1 mark)
(4) − (3): $43 c = 3870$ , so $c = 90$ . (1 mark)
From (1): $7 a + 450 = 1500$ , so $7 a = 1050$ and $a = 150$ . (½ mark)
Check in (2): $600 + 810 = 1410$ . Correct. (½ mark)
Cost $= 10 \times 150 + 6 \times 90 = 1500 + 540 = 2040$ . (½ mark)

Answer: An adult ticket costs Rs 150 and a child ticket Rs 90. The family pays Rs 2040.

Example 5: Case study: checking a classmate's elimination · 4 marks · Case-basedHOTS / competency

Arjun is asked to solve $5 x - 3 y = 9$ and $2 x - 3 y = 0$ . He writes: '(1) − (2): $3 x - 6 y = 9$ .'
(i) Which operation removes $y$ correctly, and why? (1 mark)
(ii) Find the correct value of $x$ . (1 mark)
(iii) Find $y$ . Then show, using elimination, that the pair $5 x - 3 y = 9$ and $10 x - 6 y = 20$ has no solution. (2 marks)

Try it first, then show the solutionSolution

(i) Both $y$ -terms are $- 3 y$ , with the same sign, so subtract. Arjun subtracted, but he wrote $- 3 y - 3 y$ . The correct result is $- 3 y - (- 3 y) = 0$ . (1 mark)
(ii) $(5 x - 3 y) - (2 x - 3 y) = 9 - 0$ gives $3 x = 9$ , so $x = 3$ . (1 mark)
(iii) From $2 x - 3 y = 0$ : $6 = 3 y$ , so $y = 2$ . Check: $15 - 6 = 9$ . Correct. (½ mark)
Multiply $5 x - 3 y = 9$ by 2: $10 x - 6 y = 18$ . (½ mark)
Subtract this from $10 x - 6 y = 20$ : $0 = 2$ , which is false. (½ mark)
So no pair $(x, y)$ satisfies both equations. The lines are parallel and the pair is inconsistent. (½ mark)

Answer: (i) Subtract, because the signs are the same. Arjun's sign error left $- 6 y$ . (ii) $x = 3$ (iii) $y = 2$ . The second pair leads to $0 = 2$ , so it has no solution.

Where students lose marks in Elimination Method

The slips examiners see most often, and what to write instead.

Mistake: Sign error while subtracting a negative term
Only the first term of the second equation gets subtracted, not every term.
Wrong working
$(4 x + 3 y = 17) - (4 x - 5 y = 1)$ gives $- 2 y = 16$ .
Right working
$3 y - (- 5 y) = 8 y$ and $17 - 1 = 16$ , so $8 y = 16$ and $y = 2$ .
Do this: Bracket the whole subtracted equation and flip every sign.
Usually costs: Usually all marks after that line.
Mistake: Multiplying only the left side
You rush to match coefficients and forget the constant.
Wrong working
$2 x + 5 y = 1$ times 3 written as $6 x + 15 y = 1$ .
Right working
$6 x + 15 y = 3$ .
Do this: Multiply every term, including the right-hand side, and write the new equation in full.
Usually costs: The final answer is wrong, so you keep only method marks.
Mistake: Eliminating before lining up the terms
One equation is given as $9 x = 2 y + 16$ , so the $y$ -terms are not on the same side.
Wrong working
$9 x + 2 y = 16$ taken as is, and then subtracted from $9 x - 15 y = 3$ .
Right working
Rewrite as $9 x - 2 y = 16$ first, then subtract to get $13 y = 13$ .
Do this: Rewrite both equations in the form $a x + b y = c$ before anything else.
Usually costs: 1 to 2 marks in a 3-mark question.
Mistake: Misreading $0 = 0$ or $0 = k$
You think that because both variables vanished, the solution must be $x = 0$ , $y = 0$ .
Wrong working
$0 = 2$ , so $x = 0$ and $y = 0$ .
Right working
$0 = 2$ is false, so there is no solution. The lines are parallel.
Do this: A statement with no variables tells you how many solutions there are, not their values.
Usually costs: The full 1 mark in an MCQ or assertion-reason question.

Exam tips

If a question says 'by elimination', show the multiplied equations clearly. An answer reached another way may not get full method marks.
Number your equations (1), (2), (3) and write operations such as '(1) × 3 − (2) × 2'. Examiners award step marks for these.
Large coefficients that swap between the equations are a signal to use the add-and-subtract shortcut.
Always check in the unused equation, and answer the 'hence' part. That part often carries the last mark.
In word problems and cases, define the variables with units and end with a sentence answer.

Practice questions on Elimination Method

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Solve by the elimination method: $2 x + 5 y = 1$ and $3 x + 2 y = 7$ .
Hint
Make the $x$ -coefficients 6 in both equations, then subtract.
Answer
Multiply the first by 3 and the second by 2: $6 x + 15 y = 3$ and $6 x + 4 y = 14$ . Subtracting gives $11 y = - 11$ , so $y = - 1$ . Then $2 x - 5 = 1$ , so $x = 3$ . Answer: $x = 3$ , $y = - 1$ .
Q2 · 3 marks · Short answerBoard level
Solve $101 x + 99 y = 499$ and $99 x + 101 y = 501$ . Hence find $x y$ .
Hint
The coefficients are swapped. Add the equations, then subtract them.
Answer
Adding: $200 (x + y) = 1000$ , so $x + y = 5$ . Subtracting the second from the first: $2 x - 2 y = - 2$ , so $x - y = - 1$ . This gives $x = 2$ and $y = 3$ . Check: $202 + 297 = 499$ . So $x y = 6$ .
Q3 · 3 marks · Short answerBoard level
Solve by elimination: $3 x - 5 y = 1$ and $9 x = 2 y + 16$ . Hence find $m$ if $y = m x - 3$ .
Hint
First rewrite the second equation as $9 x - 2 y = 16$ .
Answer
Multiply the first by 3: $9 x - 15 y = 3$ . Subtract this from $9 x - 2 y = 16$ : $13 y = 13$ , so $y = 1$ . Then $3 x = 6$ , so $x = 2$ . Substituting: $1 = 2 m - 3$ , so $m = 2$ .
Q4 · 5 marks · Long answerHarder board level
Solve $2 x + 3 y = 12$ and $3 x - 2 y = 5$ by elimination. Find the area of the triangle formed by these two lines and the $y$ -axis. Also find $k$ for which $4 x + 6 y = k$ and $2 x + 3 y = 12$ have infinitely many solutions.
Hint
Eliminate $y$ using 6 as the LCM. The base of the triangle is on the $y$ -axis, and its height is the $x$ -coordinate of the intersection.
Answer
Multiply to get $4 x + 6 y = 24$ and $9 x - 6 y = 15$ . Adding gives $13 x = 39$ , so $x = 3$ and $y = 2$ . The lines meet the $y$ -axis at $(0, 4)$ and $(0, - \frac{5}{2})$ , so the base is $\frac{13}{2}$ and the height is 3. Area $= \frac{1}{2} \times \frac{13}{2} \times 3 = \frac{39}{4} = 9.75$ square units. For infinitely many solutions, $4 x + 6 y = k$ must be 2 times the first equation, so $k = 24$ .
Q5 · 4 marks · Case-basedHOTS / competency
Suppose a mobile plan charges a fixed monthly fee of Rs $F$ plus Rs $r$ for each extra GB of data. Meera uses 3 extra GB and pays Rs 335. Kabir uses 7 extra GB and pays Rs 415.
(i) Write the pair of equations. (1 mark)
(ii) Find $F$ and $r$ by elimination. (2 marks)
(iii) Another plan costs Rs 395 a month with unlimited data. For how many extra GB do the two plans cost the same? (1 mark)
Hint
Both equations have the same coefficient of $F$ , so subtract them straight away.
Answer
(i) $F + 3 r = 335$ and $F + 7 r = 415$ . (ii) Subtracting gives $4 r = 80$ , so $r = 20$ and $F = 275$ . (iii) $275 + 20 g = 395$ gives $g = 6$ . The plans cost the same at 6 extra GB.
Q6 · 3 marks · Short answerHarder board level
The pair $a x + 3 y = 15$ and $2 x + 5 y = 21$ has a solution in which $x = y$ . Find $a$ . Then verify your answer by solving the pair by elimination.
Hint
Put $x = y$ in the equation that has no unknown coefficient first.
Answer
With $x = y$ , the second equation gives $7 y = 21$ , so $x = y = 3$ . In the first: $3 a + 9 = 15$ , so $a = 2$ . Verify: $2 x + 3 y = 15$ and $2 x + 5 y = 21$ . Subtracting gives $2 y = 6$ , so $y = 3$ , and then $2 x = 6$ , so $x = 3$ . Indeed $x = y$ . Answer: $a = 2$ .
Q7 · 5 marks · Long answerHarder board level
Suppose a motorboat moves at $u$ km/h in still water and the river flows at $v$ km/h. On Monday it travels downstream for 2 hours and then upstream for 3 hours, covering 66 km in all. On Tuesday it travels downstream for 4 hours and upstream for 1 hour, covering 82 km in all. Form a pair of linear equations, find $u$ and $v$ by elimination, and hence find the total time for a trip of 90 km downstream and 90 km back upstream.
Hint
Downstream speed is $u + v$ and upstream speed is $u - v$ . Distance = speed × time; expand and collect terms in $u$ and $v$ .
Answer
Monday: $2 (u + v) + 3 (u - v) = 66$ , so $5 u - v = 66$ ... (1). Tuesday: $4 (u + v) + (u - v) = 82$ , so $5 u + 3 v = 82$ ... (2). (2) − (1): $4 v = 16$ , so $v = 4$ . Then $5 u = 70$ , so $u = 14$ . Check: downstream speed 18 km/h, upstream 10 km/h; $36 + 30 = 66$ and $72 + 10 = 82$ . Trip time $= \frac{90}{18} + \frac{90}{10} = 5 + 9 = 14$ hours. Answer: $u = 14$ km/h, $v = 4$ km/h, total time 14 hours.
Q8 · 3 marks · Short answerHarder board level
Five years ago, a mother was three times as old as her daughter. Ten years from now, she will be twice as old as her daughter. Form a pair of linear equations and find their present ages by elimination.
Hint
Let the present ages be $M$ and $D$ years. Simplify each condition to the form $M + b D = c$ .
Answer
$M - 5 = 3 (D - 5)$ gives $M - 3 D = - 10$ ... (1). $M + 10 = 2 (D + 10)$ gives $M - 2 D = 10$ ... (2). (2) − (1): $D = 20$ , so $M = 50$ . Check: $45 = 3 \times 15$ and $60 = 2 \times 30$ . The mother is 50 years old and the daughter 20 years old.
Q9 · 3 marks · Short answerHarder board level
Find $a$ and $b$ for which the pair $x + 2 y = 5$ and $(a - 1) x + (a + b) y = 15$ has infinitely many solutions.
Hint
Multiply the first equation by 3 so the constants match, then subtract and demand $0 = 0$ .
Answer
(1) × 3: $3 x + 6 y = 15$ . Subtracting this from the second: $(a - 4) x + (a + b - 6) y = 0$ . This must be $0 = 0$ for every $x$ and $y$ , so $a - 4 = 0$ and $a + b - 6 = 0$ . Hence $a = 4$ and $b = 2$ . Check: the second equation becomes $3 x + 6 y = 15$ , which is 3 times the first.
Q10 · 1 mark · MCQBoard level
If $5 x + 2 y = 16$ and $3 x - 4 y = - 6$ , then $x + y$ equals:
(a) $3$
(b) $4$
(c) $5$
(d) $6$
Hint
Multiply the first equation by 2 and add to remove $y$ .
Answer
(c) $5$ . Doubling the first gives $10 x + 4 y = 32$ ; adding the second gives $13 x = 26$ , so $x = 2$ and $y = 3$ . Then $x + y = 5$ .
Q11 · 1 mark · MCQHOTS / competency
While solving $4 x + 6 y = 10$ and $6 x + 9 y = 15$ by elimination, Riya multiplies the first equation by 3 and the second by 2 and subtracts. She says, 'Both variables vanished, so the pair has no solution.' Which statement is correct?
(a) Riya is right; the lines are parallel.
(b) Riya is wrong; she gets $0 = 0$ , so there are infinitely many solutions.
(c) Riya is wrong; the pair has the unique solution $x = 0$ , $y = 0$ .
(d) Riya is wrong; she gets $0 = 5$ , so there is no solution.
Hint
Write out both multiplied equations in full, including the constants.
Answer
(b). Both become $12 x + 18 y = 30$ , so subtracting gives $0 = 0$ . The lines coincide and there are infinitely many solutions. Vanishing variables alone do not mean no solution; the constant decides.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When do you add the equations in elimination?Show answer
When the matched coefficients have opposite signs.
2.Why does multiplying an equation by 3 not change the solution?Show answer
Because it is still the same line, so every solution is unchanged.
3.What does $0 = 0$ mean after elimination?Show answer
Infinitely many solutions, because the lines coincide.
4.What does $0 = 5$ mean after elimination?Show answer
No solution, because the lines are parallel.
5.Shortcut for $a x + b y = c$ and $b x + a y = d$ ?Show answer
Add to get $x + y$ and subtract to get $x - y$ .

Elimination Method: frequently asked questions

When should I choose elimination over substitution?

Choose elimination when no variable has coefficient 1 or minus 1, or when the coefficients are swapped. It avoids fractions in the middle of the working. If the question names a method, use that method.

Do I have to eliminate x, or can I eliminate y?

Either one works and gives the same answer. Pick the variable whose coefficients have the smaller LCM, or opposite signs, to keep the arithmetic simple.

Is the cross-multiplication method still in the syllabus?

No. Cross-multiplication is not in the 2026-27 syllabus, so use only substitution, elimination and the graphical method.

How do I handle decimals or fractions?

Multiply each equation by a power of 10 or by the LCM of its denominators so that all coefficients become whole numbers. Then eliminate as usual.