MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Conditions for Number of Solutions of a Pair of Linear Equations

Updated 13 October 2026 · Fact-checked

For a pair of linear equations, compare the ratios a1/a2, b1/b2 and c1/c2. If the first two differ, there is one solution. If all three are equal, there are infinitely many. If only the first two are equal, there is none. The board asks you to classify pairs and to find k for a stated number of solutions.

Compare the ratios: different a and b ratios mean one solution; all three equal means infinite; only a and b equal means none.

Illustration drawn for QuestionBuddy.

Understand Conditions for Number of Solutions

Every linear equation in two variables draws a straight line. Two straight lines can only cross once, lie on top of each other, or run parallel. So a pair of linear equations has exactly one solution, infinitely many, or none. This topic lets you tell which one it is without drawing a graph, just by comparing coefficients.

Write both equations in the same form, $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ . Then form three ratios: $\frac{a _{1}}{a _{2}}$ , $\frac{b _{1}}{b _{2}}$ and $\frac{c _{1}}{c _{2}}$ .

$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ : the lines intersect, so there is a unique solution (consistent).
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ : the lines coincide, so there are infinitely many solutions (consistent, dependent).
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ : the lines are parallel, so there is no solution (inconsistent).

Why it works. The ratio of the $x$ and $y$ coefficients fixes the direction (slope) of a line. If $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ , both lines point the same way, so they either never meet or are the same line. If the constants are in the same ratio too, the first equation is just a multiple of the second, so it is the same line. If not, the same expression $a_{1} x + b_{1} y$ would need two different values at once, which is impossible. If the first two ratios differ, the directions differ, and two non-parallel lines always meet exactly once.

This connects the graphical method (what the lines look like) with the algebraic methods (what happens when you substitute or eliminate). A unique solution means elimination gives definite values. Infinitely many means you end with $0 = 0$ . No solution means you end with a false statement such as $0 = 4$ .

Conditions to remember. The constants must sit on the same side in both equations. Mixing $a x + b y = c$ with $a x + b y + c = 0$ flips the sign of one constant and gives the wrong answer. If a coefficient in the second equation is zero, compare by cross-multiplying, for example $a_{1} b_{2} = a_{2} b_{1}$ , rather than dividing by zero. In find- $k$ questions, the first two ratios usually give a value of $k$ , and the constant ratio decides whether that value gives infinitely many solutions or none.

Conditions for Number of Solutions: formulas to remember

Unique solution (intersecting lines): $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$; The constants play no part. Consistent pair.
Infinitely many solutions (coincident lines): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; All three ratios equal. Consistent and dependent pair.
No solution (parallel lines): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$; First two equal, third different. Inconsistent pair.
Cross-multiplied form of the parallel test: $a_{1} b_{2} = a_{2} b_{1}$; Same as the first two ratios being equal. Safer when a coefficient can be zero.
Consistent pair: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} or \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; Consistent means at least one solution: unique or infinitely many.

How to solve Conditions for Number of Solutions questions

This method works for classify-the-pair questions and for find-k questions.

1Rewrite both equations in the same form, either both as $a x + b y + c = 0$ or both as $a x + b y = c$ .
2Read off $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ with their signs.
3Write the condition the question asks for: $\neq =$ for unique, all equal for infinitely many, first two equal and third unequal for none.
4If there is an unknown, set $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ and solve. A quadratic may give two values.
5Test every value of the unknown in the constant ratio $\frac{c _{1}}{c _{2}}$ and reject any value that breaks the condition.
6For two unknowns, equate both pairs of ratios to the known ratio and solve the resulting pair.
7State the answer with its meaning: the value, the type of lines, and consistent or inconsistent.

Conditions for Number of Solutions 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.

Algebraic conditions for the number of solutions

Suppose you are comparing two mobile data top-up plans for a month. Plan A costs Rs 150 plus Rs 12 for each extra GB used. Plan B costs Rs 210 plus Rs $k$ for each extra GB.

Let $x$ be the extra GB and $y$ the total cost in rupees. (i) Write both plans as linear equations in the form $a x + b y + c = 0$ . (ii) For what value of $k$ will the two plans never cost the same? (iii) If $k = 8$ , for how many extra GB do the plans cost the same, and what is that cost?

Try it first, then show the solutionSolution

Plan A: $y = 150 + 12 x$ , so $12 x - y + 150 = 0$ .
Plan B: $y = 210 + k x$ , so $k x - y + 210 = 0$ .
Never the same cost means no solution: $\frac{12}{k} = \frac{- 1}{- 1} \neq = \frac{150}{210}$ .
$\frac{12}{k} = 1$ gives $k = 12$ .
Check constants: $\frac{150}{210} = \frac{5}{7} \neq = 1$ , so $k = 12$ gives parallel lines.
For $k = 8$ : $\frac{12}{8} \neq = 1$ , so there is a unique solution.
$150 + 12 x = 210 + 8 x$ gives $4 x = 60$ , so $x = 15$ .
Cost $= 150 + 12 \times 15 = 330$ ; check Plan B: $210 + 8 \times 15 = 330$ .

Answer: (ii) $k = 12$ (same rate, different fixed charge, so parallel lines). (iii) At 15 extra GB both plans cost Rs 330.

Memory hook: Same rate, different start: parallel lines that never meet.

Tricks to remember Conditions for Number of Solutions

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

I-C-P ladder
Ratios differ: Intersect. All three equal: Coincide. Only two equal: Parallel. Read it as 'I Can Picture it'.
Use it when: Classify any pair in an MCQ within seconds.
a and b decide direction, c decides distance
The a and b ratios decide whether the lines point the same way. Only then does c decide if they are the same line or two separate parallel lines.
Use it when: Find-k questions: solve with a and b first, then use c as the judge.
Consistent = at least one
Consistent means the lines touch somewhere (once or everywhere). Inconsistent means they never touch.
Use it when: Assertion-reason items that mix the words dependent, consistent and inconsistent.

Conditions for Number of Solutions: solved examples

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

Example 1: Value of k for no solution · 1 mark · MCQBoard level

For what value of $k$ does the pair $3 x + k y = 7$ and $6 x + 10 y = 11$ have no solution?
(a) $5$
(b) $10$
(c) $- 5$
(d) $2$

Try it first, then show the solutionSolution

No solution needs $\frac{3}{6} = \frac{k}{10} \neq = \frac{7}{11}$ .
$\frac{k}{10} = \frac{1}{2}$ gives $k = 5$ .
Check: $\frac{7}{11} \neq = \frac{1}{2}$ , so the condition holds.

Answer: (a) $k = 5$

Example 2: Classifying a pair given in mixed forms · 3 marks · Short answerBoard level

Without drawing, decide the nature of the pair $3 x = 2 y + 7$ and $y = \frac{6 x - 8}{4}$ . Also find the value of $c$ for which the pair $3 x - 2 y = 7$ and $6 x - 4 y = c$ has infinitely many solutions.

Try it first, then show the solutionSolution

First equation: $3 x - 2 y = 7$ .
Second: $4 y = 6 x - 8$ , so $6 x - 4 y = 8$ . (½ mark for both standard forms, ½ for the first)
Ratios: $\frac{3}{6} = \frac{1}{2}$ , $\frac{- 2}{- 4} = \frac{1}{2}$ , $\frac{7}{8}$ . (½ mark)
$\frac{1}{2} = \frac{1}{2} \neq = \frac{7}{8}$ , so the lines are parallel: no solution, inconsistent. (½ mark)
Infinitely many needs $\frac{7}{c} = \frac{1}{2}$ , so $c = 14$ . (1 mark)

Answer: The pair is inconsistent (parallel lines, no solution). $c = 14$ gives infinitely many solutions.

Example 3: A quadratic in k with two cases · 3 marks · Short answerHarder board level

Find the values of $k$ for which the pair $k x + y = k^{2}$ and $x + k y = 1$ has (i) infinitely many solutions, (ii) no solution.

Try it first, then show the solutionSolution

$a_{1} = k, b_{1} = 1, c_{1} = k^{2}$ ; $a_{2} = 1, b_{2} = k, c_{2} = 1$ (all in form $a x + b y = c$ ).
Equal direction: $\frac{k}{1} = \frac{1}{k}$ , so $k^{2} = 1$ , $k = 1$ or $k = - 1$ . (1 mark)
$k = 1$ : ratios $1, 1, \frac{1}{1} = 1$ , all equal, so infinitely many. (1 mark)
$k = - 1$ : ratios $- 1, - 1$ , and $\frac{c _{1}}{c _{2}} = \frac{1}{1} = 1 \neq = - 1$ , so no solution. (1 mark)
Check $k = - 1$ : equations $- x + y = 1$ and $x - y = 1$ give $x - y = - 1$ and $x - y = 1$ , clearly parallel.

Answer: (i) $k = 1$ ; (ii) $k = - 1$ .

Example 4: Two unknowns, then reasoning about a fixed value · 5 marks · Long answerHarder board level

Consider the pair $(a - 1) x + 3 y = 2$ and $6 x + (1 - 2 b) y = 6$ .
(i) Find $a$ and $b$ for which the pair has infinitely many solutions.
(ii) Take $b = - 4$ . Show that no value of $a$ makes the pair inconsistent.
(iii) With $b = - 4$ , for which values of $a$ does the pair have a unique solution?

Try it first, then show the solutionSolution

Infinitely many: $\frac{a - 1}{6} = \frac{3}{1 - 2 b} = \frac{2}{6} = \frac{1}{3}$ . (1 mark)
$\frac{a - 1}{6} = \frac{1}{3}$ gives $a - 1 = 2$ , so $a = 3$ . (½ mark)
$\frac{3}{1 - 2 b} = \frac{1}{3}$ gives $1 - 2 b = 9$ , so $b = - 4$ . (½ mark)
(ii) With $b = - 4$ : $\frac{b _{1}}{b _{2}} = \frac{3}{9} = \frac{1}{3}$ and $\frac{c _{1}}{c _{2}} = \frac{1}{3}$ . (½ mark)
No solution needs $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{1}{3}$ and $\frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . (½ mark)
But $\frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ already, so the second part always fails; the pair can never be inconsistent. (1 mark)
(iii) Unique needs $\frac{a - 1}{6} \neq = \frac{1}{3}$ , so $a \neq = 3$ . (1 mark)

Answer: (i) $a = 3$ , $b = - 4$ . (ii) With $b = - 4$ the $b$ and $c$ ratios are equal, so the pair is either coincident or intersecting, never parallel. (iii) Every $a \neq = 3$ .

Example 5: Case study: border lines of a rangoli grid · 4 marks · Case-basedHOTS / competency

Imagine a student plans a rangoli on squared paper treated as a coordinate grid. One straight border is the line $L_{1} : 2 x + 3 y = 12$ . A second straight border is $L_{2} : 4 x + p y = q$ .
(i) If $p = 6$ and $q = 24$ , how are the two borders placed? (1 mark)
(ii) If $p = 6$ and $q = 18$ , is the pair consistent? Give a reason. (1 mark)
(iii) She wants $L_{2}$ to pass through $(3, 2)$ and $(0, 8)$ . Find $p$ and $q$ , and check that $L_{1}$ and $L_{2}$ then meet at exactly one point. (2 marks)

Try it first, then show the solutionSolution

(i) Ratios $\frac{2}{4}, \frac{3}{6}, \frac{12}{24}$ are all $\frac{1}{2}$ , so the borders coincide (same line). (1 mark)
(ii) Ratios $\frac{1}{2}, \frac{1}{2}, \frac{12}{18} = \frac{2}{3}$ ; first two equal, third different, so parallel lines, inconsistent. (1 mark)
(iii) $(0, 8)$ on $L_{2}$ : $8 p = q$ . $(3, 2)$ on $L_{2}$ : $12 + 2 p = q$ . (½ mark)
$8 p = 12 + 2 p$ gives $p = 2$ , $q = 16$ . (½ mark)
Unique check: $\frac{2}{4} = \frac{1}{2}$ , $\frac{3}{2}$ ; these differ, so the lines intersect once. (½ mark)
$(3, 2)$ also lies on $L_{1}$ : $6 + 6 = 12$ , so they meet at $(3, 2)$ . (½ mark)

Answer: (i) Coincident. (ii) Inconsistent, parallel. (iii) $p = 2$ , $q = 16$ ; the lines meet only at $(3, 2)$ .

Example 6: Assertion-reason: coincident lines (with a trap version) · 1 mark · Assertion-reasonBoard level

Assertion (A): The pair $2 x + 3 y = 5$ and $4 x + 6 y = 10$ has infinitely many solutions.
Reason (R): A pair $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ has infinitely many solutions when $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .
(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

Ratios: $\frac{2}{4} = \frac{1}{2}$ , $\frac{3}{6} = \frac{1}{2}$ , $\frac{5}{10} = \frac{1}{2}$ , all equal, so A is true.
R is the standard condition for coincident lines, so R is true.
A follows directly from applying R, so R correctly explains A: option (a).
Trap version: if the second equation were $4 x + 6 y = 12$ , the constant ratio would be $\frac{5}{12} \neq = \frac{1}{2}$ . The lines would be parallel with no solution, so A would be false while R stays true: option (d).

Answer: (a) Both A and R are true and R is the correct explanation of A. (In the trap version with $4 x + 6 y = 12$ , the answer would be (d).)

Example 7: Find k for a unique solution · 2 marks · Short answerBoard level

For what values of $k$ does the pair $2 x + 3 y = 7$ and $(k - 1) x + (k + 2) y = 3 k$ have a unique solution?

Try it first, then show the solutionSolution

Unique solution needs $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , i.e. $a_{1} b_{2} \neq = a_{2} b_{1}$ (this form is safe even if $k - 1 = 0$ ). (½ mark)
$2 (k + 2) \neq = 3 (k - 1)$ . (½ mark)
$2 k + 4 \neq = 3 k - 3$ , so $k \neq = 7$ . (½ mark)
Check $k = 7$ : the second equation is $6 x + 9 y = 21$ , which is 3 times the first, so the lines coincide and the solution is not unique. Hence every other $k$ works. (½ mark)

Answer: The pair has a unique solution for every real value of $k$ except $k = 7$ .

Example 8: Writing a second equation for a given type of pair · 2 marks · Short answerBoard level

Given the linear equation $2 x + 3 y - 8 = 0$ , write another linear equation in two variables such that the pair formed represents (i) parallel lines, (ii) coincident lines. Justify each answer using ratios.

Try it first, then show the solutionSolution

(i) Keep $a$ and $b$ in proportion but change the constant: take $4 x + 6 y - 5 = 0$ . Ratios $\frac{2}{4} = \frac{1}{2}$ , $\frac{3}{6} = \frac{1}{2}$ , $\frac{- 8}{- 5} = \frac{8}{5} \neq = \frac{1}{2}$ , so the lines are parallel. (1 mark)
(ii) Multiply the whole equation by 2: $4 x + 6 y - 16 = 0$ . Ratios $\frac{1}{2}, \frac{1}{2}, \frac{- 8}{- 16} = \frac{1}{2}$ , all equal, so the lines coincide. (1 mark)

Answer: (i) For example $4 x + 6 y - 5 = 0$ (parallel). (ii) For example $4 x + 6 y - 16 = 0$ (coincident). Many other correct answers exist if the ratios are justified.

Example 9: Reading 0 = 0 and 0 = 4 after elimination · 3 marks · Short answerHarder board level

Consider the pair $x + 2 y = 4$ and $3 x + 6 y = k$ . Use the elimination method to find the value of $k$ for which the pair has infinitely many solutions. Then show, by the same method, what happens when $k = 16$ , and confirm both conclusions with the ratio test.

Try it first, then show the solutionSolution

Multiply the first equation by 3: $3 x + 6 y = 12$ . (½ mark)
Subtract this from the second equation: $(3 x + 6 y) - (3 x + 6 y) = k - 12$ , so $0 = k - 12$ . (½ mark)
Both variables have vanished. The statement is true for every $(x, y)$ only when $k - 12 = 0$ , i.e. $k = 12$ ; then it reads $0 = 0$ , so there are infinitely many solutions (coincident lines). (1 mark)
When $k = 16$ it reads $0 = 4$ , which is false, so no $(x, y)$ satisfies both: no solution (parallel lines). (½ mark)
Ratio check: $\frac{1}{3}$ , $\frac{2}{6} = \frac{1}{3}$ ; for $k = 12$ , $\frac{4}{12} = \frac{1}{3}$ (all equal), for $k = 16$ , $\frac{4}{16} = \frac{1}{4} \neq = \frac{1}{3}$ (parallel). Both agree. (½ mark)

Answer: $k = 12$ gives $0 = 0$ : infinitely many solutions. $k = 16$ gives $0 = 4$ : no solution (inconsistent).

Where students lose marks in Conditions for Number of Solutions

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

Mistake: Mixing the two standard forms
Moving the constant across changes its sign, so the constant ratio gets the wrong sign.
Wrong working
$x + 2 y = 5$ and $2 x + 4 y - 10 = 0$ : $\frac{c _{1}}{c _{2}} = \frac{5}{- 10} = - \frac{1}{2} \neq = \frac{1}{2}$ , so no solution.
Right working
Write $2 x + 4 y = 10$ : $\frac{5}{10} = \frac{1}{2}$ ; all ratios equal, so infinitely many solutions.
Do this: Rewrite both equations in one form before writing any ratio.
Usually costs: The whole 1-mark MCQ, or 1-2 marks in a longer answer.
Mistake: Not testing the constant ratio in a find-k question
Students stop after solving $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ and accept every root.
Wrong working
$k x + y = k^{2}$ , $x + k y = 1$ infinitely many: $k^{2} = 1$ , so $k = \pm 1$ .
Right working
$k = - 1$ gives $\frac{c _{1}}{c _{2}} = 1 \neq = - 1$ , so reject it; only $k = 1$ .
Do this: Substitute each root into all three ratios and write a line accepting or rejecting it.
Usually costs: 1 mark for the final answer in a 3-mark question.
Mistake: Using the constants for a unique-solution question
Students apply all three ratios when only two are needed.
Wrong working
Unique solution for $2 x + 3 y = 4$ , $4 x + k y = 8$ : need $\frac{2}{4} \neq = \frac{3}{k} \neq = \frac{4}{8}$ .
Right working
Only $\frac{2}{4} \neq = \frac{3}{k}$ matters, so $k \neq = 6$ .
Do this: Unique solution: compare $a$ and $b$ ratios only.
Usually costs: 1-2 marks.
Mistake: Calling a dependent pair inconsistent
The words dependent and inconsistent both sound 'bad'.
Wrong working
All three ratios equal, so the pair is inconsistent.
Right working
All three ratios equal: coincident lines, consistent and dependent.
Do this: Inconsistent only ever means no solution.
Usually costs: The full mark in an MCQ or assertion-reason item.

Exam tips

Write the three ratios on separate lines with their values; examiners give step marks for the ratio line.
In find-k answers, end with a sentence such as 'k = 3 is rejected because the constant ratio differs'.
Assertion-reason items often pair a correct rule with a wrongly classified pair; check the assertion by computing the ratios yourself.
For two unknowns, equate each of the first two ratios to the known constant ratio; that gives a simple pair to solve.
When a coefficient could be zero, use $a_{1} b_{2} = a_{2} b_{1}$ instead of dividing.

Practice questions on Conditions for Number of Solutions

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
For which value of $c$ is the pair $2 x - 5 y = 3$ and $4 x - 10 y = c$ consistent?
(a) $3$
(b) $6$
(c) $9$
(d) $12$
Hint
The first two ratios are equal, so consistent here means coincident.
Answer
(b) $6$ . Ratios $\frac{1}{2}, \frac{1}{2}$ , so need $\frac{3}{c} = \frac{1}{2}$ ; any other $c$ gives parallel lines.
Q2 · 3 marks · Short answerHarder board level
Show that there is no value of $k$ for which the pair $k x - 4 y = 3$ and $6 x - 12 y = k$ has infinitely many solutions. Find the value of $k$ for which it has no solution.
Hint
Get $k$ from the first two ratios, then test it in the constant ratio.
Answer
$\frac{k}{6} = \frac{- 4}{- 12} = \frac{1}{3}$ gives $k = 2$ . Then $\frac{c _{1}}{c _{2}} = \frac{3}{2} \neq = \frac{1}{3}$ , so the only candidate fails: never infinitely many. For $k = 2$ the pair has no solution.
Q3 · 3 marks · Short answerHarder board level
Find the value(s) of $k$ for which the pair $k x + 9 y = k + 6$ and $x + k y = 3$ has infinitely many solutions.
Hint
$k^{2} = 9$ gives two candidates; only one passes the constant test.
Answer
$\frac{k}{1} = \frac{9}{k}$ gives $k = \pm 3$ . $k = 3$ : ratios $3, 3, \frac{9}{3} = 3$ , valid. $k = - 3$ : ratios $- 3, - 3, \frac{3}{3} = 1$ , so no solution there. Answer: $k = 3$ .
Q4 · 5 marks · Long answerHarder board level
Consider the pair $(2 k - 1) x + (k - 1) y = 2 k + 1$ and $3 x + y = 5$ .
(i) Find the value of $k$ for which the pair has infinitely many solutions, show that no value of $k$ gives no solution, and state when the solution is unique.
(ii) Take $k = 3$ . Solve the pair by elimination.
(iii) Find $p$ if the line $p x + 2 y = 1$ passes through the solution found in (ii).
Hint
Use $a_{1} b_{2} = a_{2} b_{1}$ for the direction test, then check the constant ratio at that value of $k$ .
Answer
(i) Equal directions: $\frac{2 k - 1}{3} = \frac{k - 1}{1}$ gives $2 k - 1 = 3 k - 3$ , so $k = 2$ . At $k = 2$ the first equation is $3 x + y = 5$ : ratios $1, 1, \frac{5}{5} = 1$ , so infinitely many solutions. Since $k = 2$ is the only value with equal direction ratios and it gives coincident lines, no value of $k$ gives no solution. Unique solution for every $k \neq = 2$ . (ii) $k = 3$ : $5 x + 2 y = 7$ and $3 x + y = 5$ . Double the second: $6 x + 2 y = 10$ . Subtract the first: $x = 3$ . Then $y = 5 - 9 = - 4$ . Check: $15 - 8 = 7$ . Solution $(3, - 4)$ . (iii) $3 p + 2 (- 4) = 1$ gives $3 p = 9$ , so $p = 3$ .
Q5 · 2 marks · Short answerHOTS / competency
A student says the pair $4 x - 6 y = 10$ and $6 x - 9 y - 15 = 0$ has no solution. His working: $\frac{4}{6} = \frac{- 6}{- 9} = \frac{2}{3}$ but $\frac{10}{- 15} = - \frac{2}{3}$ , so the ratios differ. Find his error and give the correct conclusion.
Hint
Are the constants on the same side in both equations?
Answer
He used the first equation as $a x + b y = c$ and the second as $a x + b y + c = 0$ . Writing both as $a x + b y = c$ : $6 x - 9 y = 15$ , so $\frac{10}{15} = \frac{2}{3}$ . All ratios equal, so the lines coincide: infinitely many solutions (consistent, dependent).
Q6 · 1 mark · Assertion-reasonBoard level
Assertion (A): The pair $x + 2 y = 3$ and $2 x + 4 y = 7$ has infinitely many solutions.
Reason (R): For this pair, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ .
(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.
Hint
Equal $a$ and $b$ ratios alone do not decide the case; check the constants.
Answer
(d) A is false but R is true. $\frac{1}{2} = \frac{2}{4}$ , so R is true. But $\frac{3}{7} \neq = \frac{1}{2}$ , so the lines are parallel and the pair has no solution: A is false.
Q7 · 3 marks · Short answerHarder board level
Find $a$ and $b$ for which the pair $2 x - 3 y = 5$ and $(a + b) x - (a + 2 b) y = 15$ has infinitely many solutions. For these values, verify that $(4, 1)$ satisfies both equations.
Hint
Equate each of the first two ratios to the constant ratio $\frac{1}{3}$ .
Answer
Need $\frac{2}{a + b} = \frac{- 3}{- ( a + 2 b )} = \frac{5}{15} = \frac{1}{3}$ . So $a + b = 6$ and $a + 2 b = 9$ . Subtracting, $b = 3$ , then $a = 3$ . The second equation becomes $6 x - 9 y = 15$ . Check $(4, 1)$ : $8 - 3 = 5$ and $24 - 9 = 15$ .
Q8 · 4 marks · Case-basedHOTS / competency
Suppose an auto-rickshaw in your town charges a fixed amount of Rs $F$ plus Rs $r$ for every km. A 5 km ride costs Rs 70 and an 8 km ride costs Rs 100.
(i) Write the pair of linear equations in $F$ and $r$ . (1 mark)
(ii) Using the ratio test, show that the pair has a unique solution. (1 mark)
(iii) Find $F$ and $r$ by elimination, and hence the fare for a 12 km ride. (2 marks)
Hint
Subtracting the two equations removes $F$ .
Answer
(i) $F + 5 r = 70$ and $F + 8 r = 100$ . (ii) $\frac{1}{1} = 1$ and $\frac{5}{8}$ differ, so the lines intersect: unique solution, consistent pair. (iii) Subtract: $3 r = 30$ , so $r = 10$ ; then $F = 70 - 50 = 20$ . Check: $20 + 80 = 100$ . Fare for 12 km: $20 + 12 \times 10 = 140$ , so Rs 140 (fixed charge Rs 20, Rs 10 per km).

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1.Condition for a unique solution?Show answer
$\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ ; intersecting lines.
2.Condition for infinitely many solutions?Show answer
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ ; coincident lines.
3.Condition for no solution?Show answer
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ ; parallel lines.
4.Is a dependent pair consistent?Show answer
Yes. It has infinitely many solutions.
5.Nature of $x + y = 2$ and $2 x + 2 y = 5$ ?Show answer
Ratios $\frac{1}{2}, \frac{1}{2}, \frac{2}{5}$ : parallel, no solution.

Conditions for Number of Solutions: frequently asked questions

Do I need the c ratio for a unique solution?

No. If the x and y ratios are different, the lines cross once whatever the constants are. The constant ratio only matters when the first two ratios are equal.

Why do I get two values of k but the answer has only one?

Setting the first two ratios equal often gives a quadratic with two roots. One root may make the constant ratio different, which gives parallel lines instead of coincident ones, so it must be rejected.

Does it matter whether I use ax + by + c = 0 or ax + by = c?

Either form works, but both equations must be in the same form. Mixing them changes the sign of one constant and can turn infinitely many solutions into no solution.

Is the cross-multiplication method in the 2026-27 syllabus?

No. The cross-multiplication method is not in the 2026-27 syllabus.