MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Substitution Method for a Pair of Linear Equations

Updated 14 October 2026 · Fact-checked

The substitution method solves a pair of linear equations by writing one variable in terms of the other from one equation and putting it into the second equation. This leaves one equation in one variable. Boards ask you to solve pairs this way, use the solution to find a constant, and solve word problems or case studies.

Rewrite one variable, substitute it into the other equation, solve for x, then put x back to find y and check both.

Illustration drawn for QuestionBuddy.

Understand Substitution Method

A pair of linear equations in two variables has two unknowns, $x$ and $y$ , and two conditions. You cannot solve one equation with two unknowns on its own, because it has infinitely many solutions. The substitution method fixes this. It turns the pair into a single equation in one variable, which you already know how to solve.

Here is the idea. From one equation, make one variable the subject, for example $y = 7 - 2 x$ . Then replace $y$ in the other equation with $7 - 2 x$ . Now only $x$ is left. Solve for $x$ , then put that value back into $y = 7 - 2 x$ to get $y$ .

Why it works. A solution of the pair is a point $(x, y)$ that makes both equations true at the same time. At that point, $y$ really does equal $7 - 2 x$ , so you are allowed to replace $y$ by $7 - 2 x$ in the second equation. You are not changing the problem, only rewriting it. Geometrically, you are finding the point of the first line that also lies on the second line.

Choosing well. Pick the variable whose coefficient is $1$ or $- 1$ . That keeps the working free of fractions. If no coefficient is $1$ , you can still substitute, but expect a fraction; clearing it by multiplying through by the LCM keeps the arithmetic safe.

When the variable disappears. Sometimes, after substituting, both variables cancel. If you get a true statement such as $0 = 0$ , the two equations describe the same line, so there are infinitely many solutions. If you get a false statement such as $0 = 6$ , the lines are parallel and there is no solution. This matches the ratio conditions and the graphs you study elsewhere in the chapter, so substitution can also tell you the nature of a pair.

Substitution and elimination always give the same answer, because both only make legal moves on true equations. Use substitution when one variable is easy to isolate, and always check your answer in the equation you did not use.

Substitution Method: formulas to remember

Making a variable the subject: $a_{1} x + y = c_{1} \Rightarrow y = c_{1} - a_{1} x$; Choose a variable with coefficient 1 or -1 to avoid fractions. Label this new equation (3).
Substituting into the other equation: $a_{2} x + b_{2} (c_{1} - a_{1} x) = c_{2}$; Always substitute into the equation you did not use to make the subject, or you will get 0 = 0 by mistake.
Reading a vanished variable: $0 = 0 \Rightarrow infinitely many solutions; 0 = k (k \neq = 0) \Rightarrow no solution$; A true statement means coincident lines; a false statement means parallel lines.
Condition for a unique solution: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$; For a1x + b1y = c1 and a2x + b2y = c2. The lines intersect at one point, and substitution ends with a single value of the variable, never 0 = 0 or 0 = k. The pair is consistent.
Condition for infinitely many solutions: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; The lines coincide, and substitution ends in a true statement such as 0 = 0. The pair is consistent.
Condition for no solution: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$; The lines are parallel, and substitution ends in a false statement such as 0 = k with k not zero. The pair is inconsistent.
Checking the solution: $LHS = RHS in the unused equation$; A one-line check catches sign errors before you move to a 'hence' part.

How to solve Substitution Method questions

This method works for any pair given in any form, including word problems once you have written the equations.

1Write both equations in the form $a x + b y = c$ and number them (1) and (2). Clear any fractions first.
2Choose the variable with coefficient $1$ or $- 1$ and make it the subject of its equation. Call this (3).
3Substitute (3) into the other equation, using brackets around the whole expression.
4Expand carefully, collect like terms and solve for the remaining variable.
5If the variable vanishes, read the result: $0 = 0$ means infinitely many solutions, a false statement means no solution.
6Put the value back into (3) to find the second variable.
7Check both values in the equation you did not use for substitution.
8Answer exactly what is asked: state $x$ and $y$ , then do any 'hence' part, with units in word problems.

Substitution 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.

Substitution method

Imagine your family plans a festival dinner and orders samosas and jalebis from a sweet shop. A samosa costs Rs 6 more than a plate of jalebi. The bill for 10 samosas and 8 plates of jalebi comes to Rs 312.

Find the price of one samosa and one plate of jalebi. Your uncle then wants to buy 15 samosas and 5 plates of jalebi with a Rs 500 note. How much change should he get?

Try it first, then show the solutionSolution

Let one samosa cost Rs $s$ and one plate of jalebi cost Rs $j$ .
First condition: $s = j + 6$ ... (1)
Second condition: $10 s + 8 j = 312$ ... (2)
Substitute (1) in (2): $10 (j + 6) + 8 j = 312$
$10 j + 60 + 8 j = 312$ , so $18 j = 252$
$j = 14$
From (1): $s = 14 + 6 = 20$
Check in (2): $10 (20) + 8 (14) = 200 + 112 = 312$ , correct.
Cost of 15 samosas and 5 jalebi plates $= 15 (20) + 5 (14) = 300 + 70 = 370$
Change $= 500 - 370 = 130$

Answer: Samosa Rs 20, jalebi Rs 14 per plate; change Rs 130.

Memory hook: 'Rs 6 more than' already gives you one variable in terms of the other: substitute it straight in.

Substitution method

Suppose your school's mobile data pack for a project costs a fixed monthly charge plus a rate per GB used beyond the free limit. In March your class used 5 extra GB and paid Rs 230. In April it used 9 extra GB and paid Rs 310.

Find the fixed charge and the rate per extra GB. If the class can spend at most Rs 400 in May, what is the largest number of whole extra GB it can use?

Try it first, then show the solutionSolution

Let the fixed charge be Rs $f$ and the rate be Rs $r$ per extra GB.
March: $f + 5 r = 230$ ... (1)
April: $f + 9 r = 310$ ... (2)
From (1): $f = 230 - 5 r$ ... (3)
Substitute (3) in (2): $230 - 5 r + 9 r = 310$
$4 r = 80$ , so $r = 20$
From (3): $f = 230 - 100 = 130$
Check in (2): $130 + 9 (20) = 310$ , correct.
May: $130 + 20 n \leq 400$ , so $20 n \leq 270$ , $n \leq 13.5$
Largest whole number of extra GB is $13$ .

Answer: Fixed charge Rs 130, rate Rs 20 per extra GB; at most 13 extra GB in May.

Memory hook: Fixed charge has coefficient 1: isolate it, substitute, and only the rate is left to solve.

Tricks to remember Substitution Method

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

Pick the 'lonely' variable
Look for the variable standing alone with coefficient $1$ or $- 1$ . Isolate that one first.
Use it when: Choosing which variable to make the subject so no fractions appear.
Out of one, into the other
Take the expression OUT of equation (1) and put it INTO equation (2). Never back into (1).
Use it when: Avoiding the fake '0 = 0' you get by substituting into the same equation.
Bracket it before you break it
Whenever you substitute, write the whole expression inside brackets first, then expand.
Use it when: Prevents sign errors like $- 3 (2 x - 5)$ becoming $- 6 x - 15$ .
True means many, false means none
If the letters vanish: true statement, infinitely many; false statement, no solution.
Use it when: Reading the nature of a pair directly from substitution.

Substitution Method: solved examples

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

Example 1: Solve by substitution and check · 2 marks · Very short answerBoard level

Solve the following pair by the substitution method: $4 x - y = 9$ and $3 x + 2 y = 15$ .

Try it first, then show the solutionSolution

From $4 x - y = 9$ : $y = 4 x - 9$ ... (3)
Substitute in $3 x + 2 y = 15$ : $3 x + 2 (4 x - 9) = 15$ (½ mark)
$3 x + 8 x - 18 = 15$ , so $11 x = 33$ , $x = 3$ (½ mark)
From (3): $y = 4 (3) - 9 = 3$ (½ mark)
Check: $3 (3) + 2 (3) = 15$ , correct. (½ mark)

Answer: $x = 3$ , $y = 3$

Example 2: Solve and hence find m · 3 marks · Short answerBoard level

Solve $2 x + 3 y = 11$ and $x - 2 y = - 12$ by substitution. Hence find the value of $m$ for which $y = m x + 3$ .

Try it first, then show the solutionSolution

From $x - 2 y = - 12$ : $x = 2 y - 12$ ... (3) (½ mark)
Substitute in $2 x + 3 y = 11$ : $2 (2 y - 12) + 3 y = 11$ (½ mark)
$4 y - 24 + 3 y = 11$ , so $7 y = 35$ , $y = 5$ (½ mark)
From (3): $x = 10 - 12 = - 2$ (½ mark)
Check: $2 (- 2) + 3 (5) = - 4 + 15 = 11$ , correct.
Put $x = - 2$ , $y = 5$ in $y = m x + 3$ : $5 = - 2 m + 3$ (½ mark)
$- 2 m = 2$ , so $m = - 1$ (½ mark)

Answer: $x = - 2$ , $y = 5$ and $m = - 1$

Example 3: Equations with fractions · 3 marks · Short answerBoard level

Solve by substitution: $\frac{x}{2} + \frac{y}{3} = 4$ and $\frac{x}{3} + \frac{y}{2} = \frac{7}{2}$ .

Try it first, then show the solutionSolution

Multiply each equation by the LCM 6: $3 x + 2 y = 24$ ... (1) and $2 x + 3 y = 21$ ... (2) (½ mark)
No coefficient is 1, so from (1): $y = \frac{24 - 3 x}{2}$ ... (3) (½ mark)
Substitute (3) in (2): $2 x + \frac{3 ( 24 - 3 x )}{2} = 21$ (½ mark)
Multiply by 2: $4 x + 72 - 9 x = 42$ , so $- 5 x = - 30$ , $x = 6$ (½ mark)
From (3): $y = \frac{24 - 18}{2} = 3$ (½ mark)
Check in the second original equation: $\frac{6}{3} + \frac{3}{2} = 2 + \frac{3}{2} = \frac{7}{2}$ , correct. (½ mark)

Answer: $x = 6$ , $y = 3$

Example 4: Finding two unknown constants · 3 marks · Short answerHarder board level

The pair $a x + b y = 7$ and $b x - a y = 4$ has solution $x = 2$ , $y = 1$ . Find $a$ and $b$ using the substitution method, and hence write the two equations.

Try it first, then show the solutionSolution

Put $x = 2$ , $y = 1$ in the first equation: $2 a + b = 7$ ... (1) (½ mark)
Put $x = 2$ , $y = 1$ in the second: $2 b - a = 4$ ... (2) (½ mark)
From (1): $b = 7 - 2 a$ ... (3) (½ mark)
Substitute in (2): $2 (7 - 2 a) - a = 4$ , so $14 - 5 a = 4$ (½ mark)
$a = 2$ ; from (3): $b = 3$ (½ mark)
Equations: $2 x + 3 y = 7$ and $3 x - 2 y = 4$ . Check: $4 + 3 = 7$ , $6 - 2 = 4$ , correct. (½ mark)

Answer: $a = 2$ , $b = 3$ ; the equations are $2 x + 3 y = 7$ and $3 x - 2 y = 4$ .

Example 5: Case study: school canteen · 4 marks · Case-basedHOTS / competency

A school canteen sells veg rolls and fruit juice. One day it sold 120 items in total and collected Rs 3300. A veg roll costs Rs 30 and a juice costs Rs 25.
(i) Taking the number of rolls as $x$ and juices as $y$ , write the pair of equations. (1 mark)
(ii) Using substitution, find how many rolls were sold. (1 mark)
(iii) The next day the canteen again sold 120 items, but it collected Rs 3400 and sold exactly 20 more rolls than juices. Show, by substitution, that these two facts cannot both be true. (2 marks)

Try it first, then show the solutionSolution

(i) $x + y = 120$ and $30 x + 25 y = 3300$ (1 mark)
(ii) From the first: $y = 120 - x$ . Then $30 x + 25 (120 - x) = 3300$ , so $5 x + 3000 = 3300$ , $x = 60$ rolls (and $y = 60$ ). (1 mark)
(iii) Conditions: $x + y = 120$ and $x = y + 20$ . Substituting: $2 y + 20 = 120$ , so $y = 50$ , $x = 70$ . (1 mark)
Money collected would be $30 (70) + 25 (50) = 2100 + 1250 = 3350$ , not Rs 3400, so the three facts contradict each other. (1 mark)

Answer: (i) $x + y = 120$ , $30 x + 25 y = 3300$ (ii) 60 rolls (iii) 120 items with 20 more rolls than juices gives Rs 3350, not Rs 3400, so the claim is inconsistent.

Example 6: When the variable vanishes: 0 = 0 · 3 marks · Short answerBoard level

Use the substitution method to find the nature of the pair $3 x - 6 y = 9$ and $x - 2 y = 3$ . State what this means graphically and write any two solutions.

Try it first, then show the solutionSolution

From $x - 2 y = 3$ : $x = 2 y + 3$ ... (3) (½ mark)
Substitute (3) in $3 x - 6 y = 9$ : $3 (2 y + 3) - 6 y = 9$ (½ mark)
$6 y + 9 - 6 y = 9$ gives $9 = 9$ , a true statement for every $y$ (½ mark)
So the pair has infinitely many solutions: the two lines are coincident and the pair is consistent. (½ mark)
Ratio check: $\frac{3}{1} = \frac{- 6}{- 2} = \frac{9}{3} = 3$ , which agrees. (½ mark)
From (3): $y = 0$ gives $x = 3$ , and $y = 1$ gives $x = 5$ . Check $(5, 1)$ : $15 - 6 = 9$ , correct. (½ mark)

Answer: Infinitely many solutions; the lines coincide, so the pair is consistent. Two solutions: $(3, 0)$ and $(5, 1)$ .

Example 7: No coefficient equal to 1 · 3 marks · Short answerBoard level

Solve by substitution: $3 x + 4 y = 10$ and $2 x - 5 y = - 1$ .

Try it first, then show the solutionSolution

No coefficient is 1. From $3 x + 4 y = 10$ : $3 x = 10 - 4 y$ , so $x = \frac{10 - 4 y}{3}$ ... (3) (½ mark)
Substitute (3) in $2 x - 5 y = - 1$ : $\frac{2 ( 10 - 4 y )}{3} - 5 y = - 1$ (½ mark)
Multiply by 3 to clear the fraction: $2 (10 - 4 y) - 15 y = - 3$ (½ mark)
$20 - 8 y - 15 y = - 3$ , so $- 23 y = - 23$ , $y = 1$ (½ mark)
From (3): $x = \frac{10 - 4}{3} = 2$ (½ mark)
Check in the second equation: $2 (2) - 5 (1) = - 1$ , correct. (½ mark)

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

Example 8: 5-mark word problem: finding a fraction · 5 marks · Long answerHarder board level

A fraction becomes $\frac{4}{3}$ if 3 is added to its numerator and 2 is subtracted from its denominator. It becomes $\frac{4}{7}$ if 1 is subtracted from both its numerator and its denominator. Form a pair of linear equations and find the fraction using the substitution method.

Try it first, then show the solutionSolution

Let the fraction be $\frac{x}{y}$ . (½ mark)
First condition: $\frac{x + 3}{y - 2} = \frac{4}{3}$ , so $3 x + 9 = 4 y - 8$ , giving $3 x - 4 y = - 17$ ... (1) (1 mark)
Second condition: $\frac{x - 1}{y - 1} = \frac{4}{7}$ , so $7 x - 7 = 4 y - 4$ , giving $7 x - 4 y = 3$ ... (2) (1 mark)
From (1): $4 y = 3 x + 17$ ... (3) (½ mark)
Substitute (3) in (2): $7 x - (3 x + 17) = 3$ (½ mark)
$4 x - 17 = 3$ , so $4 x = 20$ , $x = 5$ (½ mark)
From (3): $4 y = 15 + 17 = 32$ , so $y = 8$ (½ mark)
Check: $\frac{5 + 3}{8 - 2} = \frac{8}{6} = \frac{4}{3}$ and $\frac{5 - 1}{8 - 1} = \frac{4}{7}$ , both correct. The fraction is $\frac{5}{8}$ . (½ mark)

Answer: The fraction is $\frac{5}{8}$ .

Example 9: An age problem by substitution · 5 marks · Long answerHarder board level

Ten years from now, a mother will be twice as old as her daughter. Five years ago, she was five times as old as her daughter. Form a pair of linear equations and find their present ages by substitution. In how many years will the ratio of their ages be $5 : 3$ ?

Try it first, then show the solutionSolution

Let the mother be $x$ years and the daughter $y$ years old now. (½ mark)
Ten years hence: $x + 10 = 2 (y + 10)$ , so $x = 2 y + 10$ ... (1). Five years ago: $x - 5 = 5 (y - 5)$ , so $x - 5 y + 20 = 0$ ... (2). (1 mark)
Substitute (1) in (2): $2 y + 10 - 5 y + 20 = 0$ , so $3 y = 30$ and $y = 10$ . Then $x = 30$ . (1 mark)
Check: in 10 years, 40 and 20 (twice); 5 years ago, 25 and 5 (five times). (½ mark)
After $n$ years: $\frac{30 + n}{10 + n} = \frac{5}{3}$ , so $90 + 3 n = 50 + 5 n$ and $n = 20$ . Check: $50 : 30 = 5 : 3$ . (2 marks)

Answer: Mother 30 years, daughter 10 years; the ratio will be $5 : 3$ in 20 years.

Example 10: Finding k for no solution or infinitely many · 3 marks · Short answerHarder board level

For the pair $k x + 3 y = k - 3$ and $12 x + k y = k$ , find the value of $k$ for which the pair has (i) no solution (ii) infinitely many solutions. Confirm each answer by substitution.

Try it first, then show the solutionSolution

For no solution or infinitely many, the lines cannot meet in exactly one point, so $\frac{k}{12} = \frac{3}{k}$ (½ mark)
$k^{2} = 36$ , so $k = 6$ or $k = - 6$ (½ mark)
$k = 6$ : the pair is $6 x + 3 y = 3$ and $12 x + 6 y = 6$ . From the first, $y = 1 - 2 x$ ; substituting, $12 x + 6 (1 - 2 x) = 6$ (½ mark)
$12 x + 6 - 12 x = 6$ gives $6 = 6$ , a true statement, so infinitely many solutions when $k = 6$ (½ mark)
$k = - 6$ : the pair is $- 6 x + 3 y = - 9$ and $12 x - 6 y = - 6$ . From the first, $y = 2 x - 3$ ; substituting, $12 x - 6 (2 x - 3) = - 6$ (½ mark)
$12 x - 12 x + 18 = - 6$ gives $18 = - 6$ , a false statement, so no solution when $k = - 6$ (½ mark)

Answer: (i) No solution for $k = - 6$ (ii) Infinitely many solutions for $k = 6$

Example 11: MCQ: reading a true statement · 1 mark · MCQBoard level

When the pair $x - 2 y = 3$ and $3 x - 6 y = k$ is solved by substituting $x = 2 y + 3$ into the second equation, the result is a true statement for every value of $y$ . The value of $k$ is
(a) $3$
(b) $6$
(c) $9$
(d) $- 9$

Try it first, then show the solutionSolution

Substitute: $3 (2 y + 3) - 6 y = k$
$6 y + 9 - 6 y = k$ , so the equation becomes $9 = k$
This is true for every $y$ only when $k = 9$ ; the lines then coincide. (1 mark)

Answer: (c) $9$

Example 12: Assertion-reason on a parameter · 1 mark · Assertion-reasonHarder board level

Assertion (A): The pair $3 x - 2 y = 5$ and $6 x - 4 y = k$ has infinitely many solutions only when $k = 10$ .
Reason (R): When $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ , the pair has infinitely many solutions if $\frac{c _{1}}{c _{2}}$ is also equal to that ratio.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Substitution: from the first, $y = \frac{3 x - 5}{2}$ . Then $6 x - 4 \cdot \frac{3 x - 5}{2} = k$ gives $6 x - 6 x + 10 = k$ , i.e. $10 = k$ . This is true for every $x$ only when $k = 10$ ; for any other $k$ it is false and there is no solution. So A is true.
R is the ratio rule: $\frac{3}{6} = \frac{- 2}{- 4} = \frac{1}{2}$ , and $\frac{- 5}{- k} = \frac{1}{2}$ gives $k = 10$ . R is true and gives exactly the condition in A. (1 mark)

Answer: (a) Both A and R are true and R is the correct explanation of A.

Where students lose marks in Substitution Method

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

Mistake: Substituting back into the same equation
The expression came from that equation, so it always gives a true statement and no information.
Wrong working
From $x + y = 6$ , $y = 6 - x$ ; then $x + (6 - x) = 6 \Rightarrow 6 = 6$ , 'infinitely many solutions'.
Right working
Substitute $y = 6 - x$ into the other equation, for example $2 x - y = 3$ : $2 x - 6 + x = 3$ , so $x = 3$ , $y = 3$ .
Do this: Label equations and write 'substituting (3) in (2)' every time.
Usually costs: Usually all marks for the solution.
Mistake: Dropping brackets while substituting
The negative sign or multiplier only reaches the first term.
Wrong working
$4 x - 3 (2 x - 5) = 7 \Rightarrow 4 x - 6 x - 5 = 7$
Right working
$4 x - 3 (2 x - 5) = 7 \Rightarrow 4 x - 6 x + 15 = 7 \Rightarrow x = 4$
Do this: Write the full expression in brackets first, then expand every term.
Usually costs: 1 to 2 marks; the final answer is wrong.
Mistake: Stopping after finding one variable
Students forget to back-substitute or to answer the 'hence' part.
Wrong working
$x = 3$ . (End of answer.)
Right working
$x = 3$ , then from $y = 2 x - 1$ , $y = 5$ ; state $x = 3$ , $y = 5$ .
Do this: Underline everything the question asks for and tick each off at the end.
Usually costs: ½ to 1 mark.
Mistake: Misreading a vanished variable
Confusing a true statement with a false one, or calling 0 = 4 'zero solutions, so consistent'.
Wrong working
$0 = 4$ , so infinitely many solutions.
Right working
$0 = 4$ is false, so there is no solution; the lines are parallel and the pair is inconsistent.
Do this: Remember: true, many; false, none.
Usually costs: The full 1 to 3 marks for that part.

Exam tips

If the question says 'by substitution', you must show substitution; another method can lose method marks.
Write 'from (1)' and 'substituting in (2)' so the examiner can follow and award step marks.
Clear fractions or decimals by multiplying through by the LCM before choosing what to isolate.
Show a one-line check in the unused equation; it protects 'hence' parts that depend on your values.
In word problems, end with a sentence answer and units.

Practice questions on Substitution Method

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

Q1 · 3 marks · Short answerHarder board level
Solve by substitution: $2 x + 3 y = - 8$ and $6 x - y = 6$ . Hence find $λ$ if the line $y = λ x - 5$ passes through the solution.
Hint
Isolate $y$ from the second equation, where its coefficient is $- 1$ . Expect a fraction.
Answer
$y = 6 x - 6$ . Then $2 x + 3 (6 x - 6) = - 8$ , so $20 x = 10$ , $x = \frac{1}{2}$ and $y = 3 - 6 = - 3$ . Check: $1 - 9 = - 8$ . Then $- 3 = \frac{λ}{2} - 5$ , so $\frac{λ}{2} = 2$ and $λ = 4$ .
Q2 · 3 marks · Short answerHarder board level
Solve $4 x + 3 y = 5$ and $3 x - 2 y = 8$ by substitution. Hence find $p$ if $p x - 3 y = 11$ .
Hint
No coefficient is 1: isolate $x$ from the first equation and clear the fraction by multiplying by 4.
Answer
$x = \frac{5 - 3 y}{4}$ ; $\frac{3 ( 5 - 3 y )}{4} - 2 y = 8$ ; multiplying by 4, $15 - 9 y - 8 y = 32$ , so $y = - 1$ , $x = 2$ . Then $2 p + 3 = 11$ , so $p = 4$ .
Q3 · 3 marks · Short answerBoard level
Using substitution, decide whether the pair $2 x - 6 y = 8$ and $x - 3 y = 5$ has a solution. State the graphical meaning.
Hint
Make $x$ the subject of the second equation and substitute in the first.
Answer
$x = 5 + 3 y$ ; $2 (5 + 3 y) - 6 y = 8$ gives $10 = 8$ , a false statement. No solution; the lines are parallel and the pair is inconsistent.
Q4 · 3 marks · Short answerHarder board level
The sum of two numbers is 75. Four times the smaller number exceeds the larger by 15. Find the numbers by substitution, and check that their difference is 39.
Hint
Let the larger be $x$ and smaller $y$ ; write $x = 75 - y$ .
Answer
$x + y = 75$ , $4 y - x = 15$ . $4 y - (75 - y) = 15$ gives $5 y = 90$ , $y = 18$ , $x = 57$ . Difference $57 - 18 = 39$ .
Q5 · 3 marks · Short answerHOTS / competency
Aman solved $3 x + y = 11$ and $2 x + 3 y = 12$ . He wrote: $y = 11 - 3 x$ , so $2 x + 3 (11 - 3 x) = 12 \Rightarrow 2 x + 33 - 3 x = 12 \Rightarrow x = 21$ . Find his error and solve correctly.
Hint
Check how he expanded the bracket.
Answer
He wrote $3 (- 3 x) = - 3 x$ instead of $- 9 x$ . Correct: $2 x + 33 - 9 x = 12$ , so $- 7 x = - 21$ , $x = 3$ , $y = 11 - 9 = 2$ . Check: $6 + 6 = 12$ .
Q6 · 3 marks · Short answerHarder board level
Find $a$ and $b$ so that the pair $3 x + 2 y = 5$ and $(a - 1) x + (b + 2) y = 15$ has infinitely many solutions. Confirm your answer by substitution.
Hint
Set all three ratios equal to $\frac{5}{15}$ .
Answer
Need $\frac{3}{a - 1} = \frac{2}{b + 2} = \frac{5}{15} = \frac{1}{3}$ , so $a - 1 = 9$ , $a = 10$ and $b + 2 = 6$ , $b = 4$ . The second equation is $9 x + 6 y = 15$ . Substituting $y = \frac{5 - 3 x}{2}$ : $9 x + 3 (5 - 3 x) = 15$ gives $15 = 15$ , true, so infinitely many solutions.
Q7 · 5 marks · Long answerHarder board level
A two-digit number is 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Form a pair of linear equations, solve it by substitution and find the number. Also find the sum of the number and the number formed by reversing its digits.
Hint
Let the tens digit be $x$ and the units digit $y$ ; the number is $10 x + y$ .
Answer
$10 x + y = 4 (x + y)$ gives $6 x = 3 y$ , so $y = 2 x$ . $10 x + y + 18 = 10 y + x$ gives $x - y = - 2$ . Substituting $y = 2 x$ : $x - 2 x = - 2$ , so $x = 2$ , $y = 4$ . The number is 24 (check: $4 \times 6 = 24$ and $24 + 18 = 42$ ). Sum $= 24 + 42 = 66$ .
Q8 · 4 marks · Case-basedHOTS / competency
Suppose a motorboat on a river travels 30 km downstream in 2 hours and 18 km upstream in 2 hours. Let the speed of the boat in still water be $u$ km/h and the speed of the stream be $v$ km/h.
(i) Write the pair of linear equations. (1 mark)
(ii) Find the speed of the stream by substitution. (1 mark)
(iii) How long will the boat take to go 45 km downstream and come back the same distance upstream? (2 marks)
Hint
Downstream speed is $u + v$ and upstream speed is $u - v$ .
Answer
(i) $u + v = 15$ and $u - v = 9$ . (ii) $u = 15 - v$ , so $15 - 2 v = 9$ , $v = 3$ km/h (and $u = 12$ km/h). (iii) Downstream time $= \frac{45}{15} = 3$ hours, upstream time $= \frac{45}{9} = 5$ hours, total 8 hours.
Q9 · 3 marks · Short answerHarder board level
The perimeter of a rectangular plot is 64 m, and its length is 4 m less than twice its breadth. Find the length and breadth by substitution. If both the length and the breadth are increased by 2 m, by how much does the area increase?
Hint
Half the perimeter is $l + b$ . Substitute $l = 2 b - 4$ .
Answer
$l + b = 32$ and $l = 2 b - 4$ . So $3 b - 4 = 32$ , $b = 12$ m and $l = 20$ m (check: $2 (32) = 64$ ). Old area $= 240$ m²; new area $= 22 \times 14 = 308$ m²; increase $= 68$ m².
Q10 · 4 marks · Case-basedHOTS / competency
Case study: Suppose your school buys 50 saplings for a green drive: neem saplings at Rs 40 each and mango saplings at Rs 60 each, spending exactly Rs 2400.
(i) Write the pair of linear equations, taking $x$ neem and $y$ mango saplings. (1 mark)
(ii) Find $x$ and $y$ by substitution. (1 mark)
(iii) Next year each sapling costs Rs 5 more. With the same Rs 2400, the school wants twice as many mango saplings as neem saplings. Find the largest number of each it can buy, and the money left over. (2 marks)
Hint
In (iii) the cost is $45 n + 65 (2 n)$ for $n$ neem saplings, and it must not exceed 2400.
Answer
(i) $x + y = 50$ and $40 x + 60 y = 2400$ . (1)
(ii) $y = 50 - x$ ; $40 x + 3000 - 60 x = 2400$ , so $20 x = 600$ , $x = 30$ neem and $y = 20$ mango. Check: $1200 + 1200 = 2400$ . (1)
(iii) For $n$ neem and $2 n$ mango: cost $= 45 n + 130 n = 175 n \leq 2400$ , so $n \leq 13.7$ . The largest whole number is 13: 13 neem and 26 mango cost $175 \times 13 = 2275$ , leaving Rs 125. (2)

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.What is the first step in the substitution method?Show answer
Make one variable the subject of one equation, ideally one with coefficient 1 or -1.
2.Into which equation do you substitute?Show answer
The other equation, never the one you used to isolate the variable.
3.Substitution gives 0 = 0. What does it mean?Show answer
Infinitely many solutions; the lines coincide.
4.Substitution gives 0 = 7. What does it mean?Show answer
No solution; the lines are parallel and the pair is inconsistent.
5.Solve $y = 2 x$ and $x + y = 9$ .Show answer
$3 x = 9$ , so $x = 3$ , $y = 6$ .

Substitution Method: frequently asked questions

When should I use substitution instead of elimination?

Use substitution when one variable has coefficient 1 or minus 1, or is already the subject, such as y = 3x + 2. If no such variable exists, elimination is often quicker. Both give the same answer.

Do I lose marks if I use elimination when substitution is asked?

You can lose method marks, because the question tests that method. Always use the method named in the question.

Can substitution tell me if there is no solution?

Yes. If both variables cancel and you get a false statement like 0 = 5, the pair has no solution. A true statement like 0 = 0 means infinitely many solutions.

Is the cross-multiplication method still in the syllabus?

No. Cross-multiplication and equations reducible to linear form are not in the 2026-27 syllabus. Only the graphical, substitution and elimination methods are needed.