MathsCBSE Class 9 · Two Variables, One Line

Solving Pairs of Linear Equations: Substitution, Elimination and Graphs

Updated 17 October 2026 · Fact-checked

A pair of linear equations in two variables is solved by substitution (express one variable from one equation and put it in the other) or by elimination (add or subtract multiples to remove one variable). Graphically, the solution is the intersection of the two lines. Two lines meet at one point (a unique solution), are parallel (no solution), or coincide (infinitely many solutions), which can be read from the ratios of the coefficients.

The solution of the pair is the point where the two lines cross: (4, 2), which satisfies both equations.

Illustration drawn for QuestionBuddy.

Understand Solving Pairs of Linear Equations: Substitution, Elimination and Graphs

A pair of linear equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ asks for the ordered pair $(x, y)$ that satisfies both. Geometrically, each equation is a line and a common solution is a common point.

Substitution. (1) Solve one equation for one variable. (2) Substitute into the other, getting an equation in one variable. (3) Solve it. (4) Back-substitute for the other variable. (5) Check in both equations. For $y = 2 x - 1$ and $3 x + 2 y = 12$ : $3 x + 2 (2 x - 1) = 12$ gives $7 x = 14$ , $x = 2$ , $y = 3$ ; check $6 + 6 = 12$ .

Elimination. Make the coefficients of one variable equal (multiply the equations by suitable numbers), then add or subtract to cancel it. For $2 x + 3 y = 12$ and $5 x - 3 y = 9$ the $y$ -coefficients are already opposite: adding gives $7 x = 21$ , $x = 3$ , then $y = 2$ ; check $15 - 6 = 9$ . For $2 p + 3 e = 19$ and $3 p + 2 e = 16$ : multiply the first by $3$ and the second by $2$ to get $6 p + 9 e = 57$ and $6 p + 4 e = 32$ ; subtracting gives $5 e = 25$ , $e = 5$ and $p = 2$ .

Graphical method. Draw both lines on the same axes using intercepts or a table of values; the coordinates of the intersection point are the solution. For $x + y = 6$ and $x - y = 2$ the lines meet at $(4, 2)$ . They cut the y-axis at $6$ and $- 2$ and the triangle formed by the two lines and the y-axis has base $∣6 - (- 2) ∣ = 8$ and height $4$ (the x-coordinate of the intersection), so its area is $16$ .

Nature of the solutions. Compare the ratios $\frac{a _{1}}{a _{2}}$ , $\frac{b _{1}}{b _{2}}$ , $\frac{c _{1}}{c _{2}}$ .

$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ : the lines intersect at one point: a unique solution (consistent). Example: $x + 2 y = 4$ and $x - y = 1$ give $3 y = 3$ , $y = 1$ , $x = 2$ .
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ : parallel lines: no solution (inconsistent). Example: $x + 2 y = 4$ and $2 x + 4 y = 7$ .
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ : the same line: infinitely many solutions. Example: $x + 2 y = 4$ and $2 x + 4 y = 8$ .

Finding a parameter. For $2 x + 3 y = 7$ and $(k - 1) x + (k + 2) y = 3 k$ to have infinitely many solutions we need $\frac{2}{k - 1} = \frac{3}{k + 2} = \frac{7}{3 k}$ . From the first equality $2 k + 4 = 3 k - 3$ gives $k = 7$ ; check $\frac{3}{9} = \frac{1}{3}$ and $\frac{7}{21} = \frac{1}{3}$ , so $k = 7$ works.

Word problems. (1) Name the unknowns with units. (2) Translate each statement into an equation. (3) Solve by a suitable method. (4) Check in the original words. (5) Answer in a sentence. For a two-digit number with tens digit $x$ and units digit $y$ (value $10 x + y$ ): digit sum $9$ and "reversed number is $27$ more" give $x + y = 9$ and $(10 y + x) - (10 x + y) = 27$ , i.e. $y - x = 3$ ; adding: $2 y = 12$ , $y = 6$ , $x = 3$ , and the number is $36$ (check: $63 - 36 = 27$ ).

Choosing the method. Substitution is quick when one coefficient is $1$ ; elimination suits equations in standard form; the graph is best for seeing the nature of solutions.

Solving Pairs of Linear Equations: Substitution, Elimination and Graphs: formulas to remember

Pair of linear equations: $a_{1} x + b_{1} y = c_{1}, a_{2} x + b_{2} y = c_{2}$; Solution: a common point.
Unique solution: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$; Lines intersect.
No solution: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$; Parallel lines.
Infinitely many solutions: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; Coincident lines.
Two-digit number: $10 x + y, reversed 10 y + x$; Difference $= 9 (y - x)$ .
Check: $substitute in both equations$; Always verify.

How to solve Solving Pairs of Linear Equations: Substitution, Elimination and Graphs questions

Choose the method; solve; check in both equations.

1Write both equations in standard form $a x + b y = c$ and note the coefficients.
2For substitution, solve the simpler equation for one variable and substitute into the other.
3For elimination, multiply to equalise the coefficient of one variable, then add or subtract.
4Find the second variable by back-substitution.
5Check the pair in both equations.
6For the nature of solutions, compare the ratios of the coefficients; for word problems define variables with units and answer in a sentence.

Solving Pairs of Linear Equations: Substitution, Elimination and Graphs 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 in a pricing problem

Suppose 2 pencils and 3 erasers cost 19 rupees in total, while 3 pencils and 2 erasers cost 16 rupees.

Find the price of a pencil and of an eraser.

Try it first, then show the solutionSolution

Let a pencil cost $p$ and an eraser $e$ : $2 p + 3 e = 19$ and $3 p + 2 e = 16$ .
Multiply the first by 3 and the second by 2: $6 p + 9 e = 57$ and $6 p + 4 e = 32$ .
Subtract: $5 e = 25$ , so $e = 5$ ; then $2 p + 15 = 19$ gives $p = 2$ . Check: $3 (2) + 2 (5) = 16$ .

Answer: A pencil costs 2 rupees and an eraser 5 rupees.

Memory hook: Equalise, subtract, back-substitute, check.

Tricks to remember Solving Pairs of Linear Equations: Substitution, Elimination and Graphs

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

Equalise, then add or subtract
Make one pair of coefficients equal in size; same signs: subtract; opposite signs: add.
Use it when: Elimination.
Ratios tell the story
Compare $a_{1} / a_{2}$ , $b_{1} / b_{2}$ and $c_{1} / c_{2}$ : unequal, one solution; first two equal but not the third, none; all three equal, infinitely many.
Use it when: Nature of solutions.
Always check both
Substitute the answer in both original equations.
Use it when: Catching slips.

Solving Pairs of Linear Equations: Substitution, Elimination and Graphs: solved examples

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

Example 1: Substitution · 3 marks · Short answerBoard level

Solve by substitution: $y = 2 x - 1$ and $3 x + 2 y = 12$ . Verify your solution.

Try it first, then show the solutionSolution

Substitute $y = 2 x - 1$ in the second equation: $3 x + 2 (2 x - 1) = 12$ . (1)
$7 x - 2 = 12$ , so $x = 2$ and $y = 2 (2) - 1 = 3$ . (1)
Check: $3 (2) + 2 (3) = 12$ ✓ and $3 = 2 (2) - 1$ ✓. (1)

Answer: $x = 2$ , $y = 3$ .

Example 2: Elimination · 3 marks · Short answerBoard level

Solve by elimination: $2 x + 3 y = 12$ and $5 x - 3 y = 9$ .

Try it first, then show the solutionSolution

The coefficients of $y$ are $+ 3$ and $- 3$ : add the equations. (½)
$7 x = 21$ , so $x = 3$ . (1)
$2 (3) + 3 y = 12$ gives $y = 2$ . Check: $5 (3) - 3 (2) = 9$ ✓. (1½)

Answer: $x = 3$ , $y = 2$ .

Example 3: Digits problem · 5 marks · Long answerHarder board level

The sum of the digits of a two-digit number is 9. If the digits are interchanged, the new number is 27 more than the original. Find the number.

Try it first, then show the solutionSolution

Let the tens digit be $x$ and the units digit $y$ ; the number is $10 x + y$ . (½)
Digit sum: $x + y = 9$ . (1)
Reversed number $10 y + x$ is 27 more: $(10 y + x) - (10 x + y) = 27$ , so $9 y - 9 x = 27$ and $y - x = 3$ . (1½)
Adding $x + y = 9$ and $y - x = 3$ : $2 y = 12$ , $y = 6$ , $x = 3$ . (1)
The number is $36$ . Check: reversed is $63$ and $63 - 36 = 27$ ✓. (1)

Answer: The number is 36.

Example 4: Nature of solutions · 3 marks · Short answerHarder board level

For each pair, decide whether there is a unique solution, no solution or infinitely many, and solve if unique: (i) $x + 2 y = 4$ , $2 x + 4 y = 8$ ; (ii) $x + 2 y = 4$ , $2 x + 4 y = 7$ ; (iii) $x + 2 y = 4$ , $x - y = 1$ .

Try it first, then show the solutionSolution

(i) $\frac{1}{2} = \frac{2}{4} = \frac{4}{8}$ : the same line, infinitely many solutions. (1)
(ii) $\frac{1}{2} = \frac{2}{4} \neq = \frac{4}{7}$ : parallel lines, no solution. (1)
(iii) $\frac{1}{1} \neq = \frac{2}{- 1}$ : unique solution. Subtract: $3 y = 3$ , $y = 1$ , $x = 2$ . (1)

Answer: (i) infinitely many; (ii) none; (iii) $x = 2$ , $y = 1$ .

Example 5: Notebooks and pens · 4 marks · Case-basedHOTS / competency

Case: 3 notebooks and 2 pens cost 130 rupees; 2 notebooks and 5 pens cost 160 rupees. (i) Write the equations. (ii) Find the price of each. (iii) Find the cost of 5 notebooks and 4 pens. (iv) At which point do the two lines meet when plotted, and what does it mean?

Try it first, then show the solutionSolution

(i) With a notebook costing $x$ and a pen $y$ rupees: $3 x + 2 y = 130$ and $2 x + 5 y = 160$ . (½)
(ii) Multiply the first by $5$ and the second by $2$ : $15 x + 10 y = 650$ and $4 x + 10 y = 320$ . Subtract: $11 x = 330$ , $x = 30$ ; $y = \frac{130 - 90}{2} = 20$ . (1½)
(iii) $5 (30) + 4 (20) = 230$ rupees. (1)
(iv) The lines meet at $(30, 20)$ : the only prices that satisfy both purchases. (1)

Answer: (i) as above; (ii) notebook 30, pen 20 rupees; (iii) 230 rupees; (iv) $(30, 20)$ .

Where students lose marks in Solving Pairs of Linear Equations: Substitution, Elimination and Graphs

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

Mistake: Not checking in both equations
Verifying only one.
Wrong working
$x = 3$ , $y = 2$ satisfies the first equation, so done.
Right working
Check both: $5 (3) - 3 (2) = 9$ also holds.
Do this: Substitute in both original equations.
Usually costs: ½ to 1 mark.
Mistake: Sign error when subtracting equations
Subtracting only the first term.
Wrong working
$(15 x + 10 y) - (4 x + 10 y) = 11 x + 20 y$ .
Right working
$(15 x + 10 y) - (4 x + 10 y) = 11 x$ .
Do this: Write the second equation in brackets with a minus sign.
Usually costs: The answer.
Mistake: Wrong conclusion about parallel lines
Calling the pair solvable.
Wrong working
$x + 2 y = 4$ and $2 x + 4 y = 7$ gives $x =$ some value.
Right working
Elimination gives $0 = - 1$ , so there is no solution.
Do this: A false statement like $0 = - 1$ means no solution; $0 = 0$ means infinitely many.
Usually costs: 2 marks.

Exam tips

Label the equations (1) and (2) and state which operation you do.
Show back-substitution and a check line.
Define variables with units in word problems.
Use ratios to decide the nature of the solutions before solving.

Practice questions on Solving Pairs of Linear Equations: Substitution, Elimination and Graphs

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

Q1 · 1 mark · MCQBoard level
The solution of $x + y = 7$ and $x - y = 1$ is
(a) $(3, 4)$
(b) $(5, 2)$
(c) $(4, 3)$
(d) $(6, 1)$
Hint
Add the two equations.
Answer
(c) $(4, 3)$ . Adding: $2 x = 8$ , $x = 4$ , $y = 3$ .
Q2 · 3 marks · Short answerBoard level
Solve $3 x - y = 5$ and $x + 2 y = 4$ by substitution.
Hint
From the first equation $y = 3 x - 5$ .
Answer
$x + 2 (3 x - 5) = 4$ gives $7 x = 14$ , $x = 2$ , $y = 1$ . Check: $3 (2) - 1 = 5$ ✓ and $2 + 2 = 4$ ✓.
Q3 · 3 marks · Short answerHarder board level
Find the value of $k$ for which the pair $2 x + 3 y = 7$ and $(k - 1) x + (k + 2) y = 3 k$ has infinitely many solutions.
Hint
Equate $\frac{2}{k - 1}$ , $\frac{3}{k + 2}$ and $\frac{7}{3 k}$ .
Answer
$\frac{2}{k - 1} = \frac{3}{k + 2}$ gives $2 k + 4 = 3 k - 3$ , so $k = 7$ . Check: $\frac{3}{9} = \frac{1}{3}$ and $\frac{7}{21} = \frac{1}{3}$ . So $k = 7$ .
Q4 · 5 marks · Long answerHarder board level
Draw (or reason out) the lines $x + y = 6$ and $x - y = 2$ . Find their point of intersection, the y-intercepts, and the area of the triangle formed by the two lines and the y-axis.
Hint
The base lies on the y-axis between the two y-intercepts.
Answer
Adding: $2 x = 8$ , $x = 4$ , $y = 2$ : the lines meet at $(4, 2)$ . y-intercepts: $x = 0$ gives $y = 6$ for the first line and $y = - 2$ for the second. Base on the y-axis $= 6 - (- 2) = 8$ ; the height is the distance of $(4, 2)$ from the y-axis, $4$ . Area $= \frac{1}{2} \times 8 \times 4 = 16$ square units.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a taxi charges a fixed amount $a$ rupees plus $b$ rupees per kilometre. A 10 km trip costs 185 rupees and a 15 km trip costs 260 rupees. (i) Write the pair of equations and solve it. (ii) Find the fare for 22 km. (iii) How far can a passenger travel for 470 rupees? (iv) Why does this pair have a unique solution?
Hint
Subtract the two equations to find $b$ .
Answer
(i) $a + 10 b = 185$ and $a + 15 b = 260$ . Subtracting: $5 b = 75$ , $b = 15$ , $a = 35$ . (ii) $35 + 15 \times 22 = 365$ rupees. (iii) $35 + 15 d = 470$ gives $d = 29$ km. (iv) The ratios $\frac{1}{1}$ and $\frac{10}{15}$ are not equal, so the lines have different slopes and meet at exactly one point.

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.Intersecting lines give how many solutions?Show answer
Exactly one.
2.Parallel lines give?Show answer
No solution.
3.Coincident lines give?Show answer
Infinitely many solutions.
4.In elimination, when do we add?Show answer
When the coefficients of the variable to cancel have opposite signs.
5.What does $0 = - 1$ after elimination mean?Show answer
There is no solution.

Solving Pairs of Linear Equations: Substitution, Elimination and Graphs: frequently asked questions

Which method is better, substitution or elimination?

Both give the same answer; use substitution when a variable has coefficient 1 and elimination when equations are in standard form.

Can a pair of linear equations have exactly two solutions?

No. Two distinct lines meet in at most one point, so the number of solutions is 0, 1 or infinitely many.

How do I check a word problem answer?

Substitute the values in the words of the problem, not only in your equations.