MathsCBSE Class 10 · Pair of Linear Equations in Two Variables

Consistent and Inconsistent Pairs of Equations Class 10

Updated 13 October 2026 · Fact-checked

A pair of linear equations is consistent if it has at least one solution (intersecting or coincident lines) and inconsistent if it has none (parallel lines). A consistent pair with infinitely many solutions is called dependent. The board asks you to classify pairs from graphs or coefficient ratios, find k for a given type, and explain your answer.

Lines that cross or coincide share a point, so they are consistent; parallel lines never meet, so the pair is inconsistent.

Illustration drawn for QuestionBuddy.

Understand Consistent and Inconsistent Pairs of Equations

Every linear equation in two variables is a straight line. When you take two such equations together, you are really asking: which points lie on both lines? The answer depends only on how the two lines sit in the plane.

Two lines in a plane can do only three things. They can cross at one point, so the pair has exactly one solution. They can lie on top of each other (coincide), so every point of one is on the other and there are infinitely many solutions. Or they can be parallel, never meeting, so there is no solution at all.

The words describe this. A pair is consistent if it has at least one solution: intersecting lines and coincident lines are both consistent. A pair is inconsistent if it has no solution: parallel lines. A consistent pair with infinitely many solutions is also called dependent, because one equation is just a non-zero multiple of the other. So every dependent pair is consistent, but not every consistent pair is dependent.

Why can you tell without drawing? For $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ , the ratio $\frac{a _{1}}{a _{2}}$ against $\frac{b _{1}}{b _{2}}$ tells you whether the lines have the same slant. If these differ, the lines must cross: consistent. If they are equal, the lines are parallel or coincident, and the constants decide. If $\frac{c _{1}}{c _{2}}$ is also equal, the first equation is a multiple of the second (dependent). If not, the same left side would have to equal two different values, which is impossible (inconsistent).

This links straight to the solving methods. When you use substitution or elimination on a dependent pair, both variables cancel and you get a true statement like $0 = 0$ . On an inconsistent pair you get a false statement like $0 = 7$ . These conditions hold only when both equations are written in the same form, with all coefficients compared in the same order, and the denominators $a_{2}$ , $b_{2}$ , $c_{2}$ non-zero when you write ratios.

Consistent and Inconsistent Pairs of Equations: formulas to remember

Consistent, unique solution (intersecting lines): $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$; Constants do not matter. The pair is consistent but not dependent.
Consistent and dependent (coincident lines): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; Infinitely many solutions. One equation is a multiple of the other.
Inconsistent (parallel lines): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$; No solution. Lines never meet.
Consistent overall: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}} or \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; For 'find k so the pair is consistent', include both cases.
Algebraic signal while solving: $0 = 0 \Rightarrow dependent; 0 = k (k \neq = 0) \Rightarrow inconsistent$; Seen when both variables cancel in substitution or elimination.

How to solve Consistent and Inconsistent Pairs of Equations questions

Use this routine for any classify, find-k or graph-reading question on consistency.

1Write both equations in the same form, either $a x + b y + c = 0$ or $a x + b y = c$ .
2Read off $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ with their signs.
3Compute $\frac{a _{1}}{a _{2}}$ and $\frac{b _{1}}{b _{2}}$ and simplify. If they differ, the lines intersect: consistent, unique solution.
4If they are equal, compute $\frac{c _{1}}{c _{2}}$ . Equal gives coincident lines (consistent, dependent); different gives parallel lines (inconsistent).
5In find- $k$ questions, solve the equation from the first two ratios, then test every value of $k$ against the constant ratio and reject any that fails.
6If a graph is given, read the shape directly: crossing, overlapping or parallel, then name the type.
7State the answer with both the graph shape and the word: for example, 'parallel lines, so the pair is inconsistent'.

Consistent and Inconsistent Pairs of Equations 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.

Consistent, inconsistent and dependent pairs

Suppose a mobile recharge shop offers two plans. Plan P costs a fixed Rs 50 plus Rs 2 per GB of extra data. Plan Q costs a fixed Rs 80 plus Rs 2 per GB. Your cousin says that for some amount of data the two plans will cost the same.

Let $x$ be the extra GB used and $y$ the total cost in rupees. Write a pair of linear equations for the two plans, classify the pair as consistent or inconsistent using coefficient ratios, and decide whether your cousin is right. If Plan Q instead charged Rs 50 fixed plus Rs 3 per GB, find the data for which both plans cost the same.

Try it first, then show the solutionSolution

Plan P: $y = 2 x + 50$ , i.e. $2 x - y + 50 = 0$ .
Plan Q: $y = 2 x + 80$ , i.e. $2 x - y + 80 = 0$ .
$\frac{a _{1}}{a _{2}} = \frac{2}{2} = 1$ , $\frac{b _{1}}{b _{2}} = \frac{- 1}{- 1} = 1$ , $\frac{c _{1}}{c _{2}} = \frac{50}{80} = \frac{5}{8}$ .
First two ratios equal, third different: parallel lines, pair inconsistent.
So the costs are never equal; Plan Q always costs Rs 30 more. Your cousin is wrong.
New Plan Q: $y = 3 x + 50$ . Ratios $\frac{2}{3} \neq = \frac{- 1}{- 1}$ , so consistent, unique solution.
Equate: $2 x + 50 = 3 x + 50$ , so $x = 0$ , $y = 50$ .

Answer: The original pair is inconsistent, so the plans never cost the same (cousin is wrong). With the new Plan Q, both cost Rs 50 only at 0 GB of extra data.

Memory hook: Same rate, different fixed charge: parallel lines, they never meet.

Dependent pair

Imagine you buy 2 samosas and 3 cups of tea for Rs 60 at a school fete. Your friend, at the same stall, says 4 samosas and 6 cups of tea cost Rs 120.

Show that these two statements cannot tell you the price of a samosa. Then, if a third friend says 1 samosa and 1 cup of tea cost Rs 25, find both prices.

Try it first, then show the solutionSolution

Equations: $2 s + 3 t = 60$ and $4 s + 6 t = 120$ .
Ratios: $\frac{2}{4} = \frac{3}{6} = \frac{60}{120} = \frac{1}{2}$ .
All equal: coincident lines, dependent pair, infinitely many solutions, so prices cannot be fixed.
Use $2 s + 3 t = 60$ with $s + t = 25$ , i.e. $s = 25 - t$ .
$2 (25 - t) + 3 t = 60 \Rightarrow 50 + t = 60 \Rightarrow t = 10$ .
$s = 15$ . Check: $2 (15) + 3 (10) = 60$ .

Answer: The first two statements are dependent (same information twice). With the third statement, a samosa costs Rs 15 and a cup of tea Rs 10.

Memory hook: Doubling the order doubles the bill: same line, no new information.

Tricks to remember Consistent and Inconsistent Pairs of Equations

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

Consistent = Can meet
Consistent pairs Cross or Coincide (at least one meeting point). Inconsistent pairs are like railway tracks: parallel, never meet.
Use it when: Naming the type from a graph or description.
Dependent is a subset
Dependent means 'depends on the other': one equation is a copy times a number. Copies are always consistent.
Use it when: Avoid saying a dependent pair is inconsistent.
Two-then-three check
Compare the first two ratios first. Different: done, one solution. Same: now look at the third ratio.
Use it when: Fast MCQ and find-k answers.

Consistent and Inconsistent Pairs of Equations: solved examples

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

Example 1: Classify three pairs · 3 marks · Short answerBoard level

Without drawing, state whether each pair is consistent or inconsistent, and say whether the lines intersect, coincide or are parallel.
(i) $3 x - 2 y = 5$ and $6 x - 4 y = 7$
(ii) $\frac{3}{2} x + \frac{5}{3} y = 7$ and $9 x - 10 y = 14$
(iii) $5 x - 3 y = 11$ and $- 10 x + 6 y = - 22$

Try it first, then show the solutionSolution

(i) $\frac{3}{6} = \frac{1}{2}$ , $\frac{- 2}{- 4} = \frac{1}{2}$ , $\frac{5}{7}$ . First two equal, third different: parallel, inconsistent. (1 mark)
(ii) $\frac{3/2}{9} = \frac{1}{6}$ , $\frac{5/3}{- 10} = - \frac{1}{6}$ . Ratios differ: intersecting, consistent. (1 mark)
(iii) $\frac{5}{- 10} = - \frac{1}{2}$ , $\frac{- 3}{6} = - \frac{1}{2}$ , $\frac{11}{- 22} = - \frac{1}{2}$ . All equal: coincident, consistent (dependent). (1 mark)

Answer: (i) Inconsistent, parallel. (ii) Consistent, intersecting. (iii) Consistent and dependent, coincident.

Example 2: Find k for an inconsistent pair · 3 marks · Short answerHarder board level

Find the value of $k$ for which the pair $k x + 3 y = k - 3$ and $12 x + k y = k$ is inconsistent.

Try it first, then show the solutionSolution

Inconsistent needs $\frac{k}{12} = \frac{3}{k} \neq = \frac{k - 3}{k}$ . (½ mark)
$\frac{k}{12} = \frac{3}{k} \Rightarrow k^{2} = 36 \Rightarrow k = 6$ or $k = - 6$ . (1 mark)
$k = 6$ : $\frac{3}{k} = \frac{1}{2}$ , $\frac{k - 3}{k} = \frac{3}{6} = \frac{1}{2}$ . Equal, so coincident, not inconsistent. Reject. (½ mark)
$k = - 6$ : $\frac{3}{k} = - \frac{1}{2}$ , $\frac{k - 3}{k} = \frac{- 9}{- 6} = \frac{3}{2}$ . Different, so parallel. Accept. (1 mark)

Answer: $k = - 6$ (for $k = 6$ the pair is dependent instead).

Example 3: Graph check and classification · 3 marks · Short answerHarder board level

Draw the graphs of $x - 2 y = - 2$ and $2 x - 4 y = 8$ on the same axes. Use them to classify the pair, confirm with ratios, and find the distance between the points where the two lines meet the $x$ -axis.

Try it first, then show the solutionSolution

Line 1: points $(0, 1)$ , $(- 2, 0)$ , $(2, 2)$ . Line 2: points $(0, - 2)$ , $(4, 0)$ , $(2, - 1)$ . (1 mark)
Plotted, the lines have the same slant and do not meet: parallel, so inconsistent. (½ mark)
Ratios: $\frac{1}{2} = \frac{- 2}{- 4} = \frac{1}{2}$ , $\frac{- 2}{8} = - \frac{1}{4}$ (form $a x + b y = c$ ). Third differs, confirmed. (½ mark)
$x$ -intercepts are $(- 2, 0)$ and $(4, 0)$ , so distance $= 4 - (- 2) = 6$ units. (1 mark)

Answer: Parallel lines, pair inconsistent; the $x$ -intercepts are 6 units apart.

Example 4: Two unknowns for a dependent pair, then a twist · 5 marks · Long answerHarder board level

(i) Find $a$ and $b$ so that the pair $2 x + 3 y = 7$ and $(a + b) x + (2 a - b) y = 21$ is dependent.
(ii) For the pair $2 x + 3 y = 7$ and $(a + b) x + (2 a - b) y = 20$ , using the same $a$ and $b$ , state the type of pair and explain.
(iii) Find a value of $c$ such that $2 x + 3 y = 7$ and $4 x + cy = 10$ is consistent with a unique solution, and solve the pair for $c = 2$ .

Try it first, then show the solutionSolution

(i) Dependent: $\frac{2}{a + b} = \frac{3}{2 a - b} = \frac{7}{21} = \frac{1}{3}$ . (1 mark)
$a + b = 6$ and $2 a - b = 9$ . Adding: $3 a = 15$ , $a = 5$ , $b = 1$ . (1 mark)
(ii) Second equation becomes $6 x + 9 y = 20$ . Ratios $\frac{1}{3} = \frac{1}{3}$ but $\frac{7}{20} \neq = \frac{1}{3}$ : parallel, inconsistent. (1 mark)
(iii) Unique needs $\frac{2}{4} \neq = \frac{3}{c}$ , i.e. $c \neq = 6$ ; $c = 2$ works. (1 mark)
Solve $2 x + 3 y = 7$ , $4 x + 2 y = 10$ : double first, $4 x + 6 y = 14$ ; subtract: $4 y = 4$ , $y = 1$ , $x = 2$ . Check: $4 (2) + 2 (1) = 10$ . (1 mark)

Answer: (i) $a = 5$ , $b = 1$ . (ii) Inconsistent (parallel lines). (iii) Any $c \neq = 6$ ; for $c = 2$ , $x = 2$ , $y = 1$ .

Example 5: Case study: two coaching batches · 4 marks · Case-basedHOTS / competency

Suppose a sports academy records fees. In one week it collects Rs 2600 from 4 cricket and 6 football trainees, each sport having a fixed fee per trainee. An assistant records another week as 6 cricket and 9 football trainees paying Rs 3900, and a third week as 2 cricket and 3 football trainees paying Rs 1400.
(i) Write the equations for weeks 1 and 2 and classify the pair. (1 mark)
(ii) Classify the pair formed by weeks 1 and 3. (1 mark)
(iii) The manager says week 3 must be a recording error. Explain why, and say what the week 3 total should have been if weeks 1 and 2 are correct. (2 marks)

Try it first, then show the solutionSolution

(i) $4 c + 6 f = 2600$ , $6 c + 9 f = 3900$ . Ratios $\frac{4}{6} = \frac{6}{9} = \frac{2600}{3900} = \frac{2}{3}$ : dependent, consistent. (1 mark)
(ii) $4 c + 6 f = 2600$ , $2 c + 3 f = 1400$ . Ratios $2 = 2$ , $\frac{2600}{1400} = \frac{13}{7} \neq = 2$ : inconsistent. (1 mark)
(iii) Week 3 has exactly half the trainees of week 1, so its fee must be exactly half; an inconsistent pair means no fee values fit both records. (1 mark)
Correct total $= \frac{1}{2} \times 2600 =$ Rs 1300. (1 mark)

Answer: (i) Dependent (consistent). (ii) Inconsistent. (iii) No fees can satisfy both weeks 1 and 3; week 3 should have been Rs 1300.

Where students lose marks in Consistent and Inconsistent Pairs of Equations

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

Mistake: Calling a coincident pair inconsistent
Students think 'infinitely many' is somehow not a proper solution.
Wrong working
$x + y = 2$ , $2 x + 2 y = 4$ : all ratios equal, so inconsistent.
Right working
All ratios equal $\frac{1}{2}$ : coincident lines, consistent and dependent.
Do this: Consistent means at least one solution. Only parallel lines are inconsistent.
Usually costs: The full 1 mark in MCQ or assertion-reason.
Mistake: Not testing the constant ratio in find-k
Both roots from the first two ratios are accepted.
Wrong working
$k x + 3 y = k - 3$ , $12 x + k y = k$ inconsistent: $k^{2} = 36$ , so $k = \pm 6$ .
Right working
$k = 6$ makes all ratios $\frac{1}{2}$ (dependent), so only $k = - 6$ .
Do this: Always put each value back into the third ratio and reject what fails.
Usually costs: 1 to 1½ marks of a 3-mark question.
Mistake: Mixing forms when reading constants
One equation is in $a x + b y = c$ form and the other in $a x + b y + c = 0$ form.
Wrong working
$x + y = 3$ and $2 x + 2 y - 6 = 0$ : $\frac{c _{1}}{c _{2}} = \frac{3}{- 6}$ , so parallel.
Right working
Rewrite as $2 x + 2 y = 6$ : $\frac{3}{6} = \frac{1}{2}$ , all equal, coincident.
Do this: Put both equations in the same form before taking ratios.
Usually costs: Wrong classification, so the whole part.
Mistake: Writing a solution after getting a false statement
Students feel an answer must be a pair of numbers.
Wrong working
Elimination gives $0 = 5$ , so $x = 0$ , $y = 0$ .
Right working
$0 = 5$ is false, so no solution: parallel lines, inconsistent.
Do this: If both variables vanish, read the statement: true means dependent, false means inconsistent.
Usually costs: The conclusion mark, often 1 mark.

Exam tips

Always write the final line with both the graph shape and the word, for example 'coincident lines, so consistent (dependent)'.
In assertion-reason items, watch for a Reason that says consistent means 'exactly one solution'; that statement is false.
Show the three ratios explicitly, even for MCQs worked in rough; it prevents sign slips with negative coefficients.
For 'consistent' find-k questions, remember both possibilities: first two ratios unequal, or all three equal.
In graph questions, a third plotted point per line proves your lines are drawn correctly before you call them parallel.

Practice questions on Consistent and Inconsistent Pairs of Equations

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
Which pair of equations is inconsistent?
(a) $x + y = 4$ and $x - y = 2$
(b) $2 x + 4 y = 6$ and $x + 2 y = 3$
(c) $3 x - y = 5$ and $6 x - 2 y = 9$
(d) $x + 2 y = 5$ and $2 x + y = 5$
Hint
Look for equal first two ratios with a different constant ratio.
Answer
(c): $\frac{3}{6} = \frac{- 1}{- 2} = \frac{1}{2}$ but $\frac{5}{9} \neq = \frac{1}{2}$ , so parallel lines.
Q2 · 3 marks · Short answerHarder board level
Find the value of $p$ for which the pair $2 x + p y = 5$ and $(p - 1) x + 3 y = 5$ has no solution.
Hint
Solve $\frac{2}{p - 1} = \frac{p}{3}$ , then test each root against the constant ratio.
Answer
$p (p - 1) = 6 \Rightarrow p^{2} - p - 6 = 0 \Rightarrow p = 3$ or $- 2$ . For $p = 3$ : ratios $1, 1, 1$ , dependent, reject. For $p = - 2$ : $\frac{2}{- 3} = \frac{- 2}{3}$ and $\frac{5}{5} = 1$ , different. So $p = - 2$ .
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The pair $4 x - 6 y = 10$ and $6 x - 9 y = 15$ is consistent.
Reason (R): A pair of linear equations is consistent only if the lines intersect at exactly one point.
(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
Check all three ratios, and remember coincident lines too.
Answer
(c). Ratios $\frac{4}{6} = \frac{- 6}{- 9} = \frac{10}{15} = \frac{2}{3}$ , so the pair is dependent, hence consistent: A true. R is false because coincident lines are also consistent.
Q4 · 5 marks · Long answerHarder board level
For what values of $a$ and $b$ does the pair $x + 2 y = 3$ and $(a - b) x + (a + b) y = a + b - 2$ have infinitely many solutions? Hence write the second equation and state the type of pair if $a + b - 2$ is replaced by $10$ .
Hint
Set the second-equation coefficients as $t$ , $2 t$ , $3 t$ and use $a + b = 2 t$ with $a + b - 2 = 3 t$ .
Answer
$\frac{1}{a - b} = \frac{2}{a + b} = \frac{3}{a + b - 2}$ . From the last two: $3 (a + b) = 2 (a + b) - 4 \Rightarrow a + b = - 4$ . From the first two: $a + b = 2 (a - b) \Rightarrow a - b = - 2$ . So $a = - 3$ , $b = - 1$ . Second equation: $- 2 x - 4 y = - 6$ . With constant $10$ : $- 2 x - 4 y = 10$ , ratios $- \frac{1}{2} = - \frac{1}{2} \neq = \frac{3}{10}$ , inconsistent.
Q5 · 3 marks · Short answerHOTS / competency
Arjun solved $3 x - 5 y = 4$ and $9 x - 15 y = 14$ by elimination. He multiplied the first by 3 and subtracted, got $0 = - 2$ , and wrote: 'Answer: $x = 0$ , $y = 0$ .' Identify his error, give the correct conclusion with its graphical meaning, and change only the constant 14 so that the pair becomes dependent.
Hint
What does a false statement like $0 = - 2$ tell you?
Answer
$9 x - 15 y = 12$ minus $9 x - 15 y = 14$ gives $0 = - 2$ , a false statement, so there is no solution; $(0, 0)$ does not even satisfy $3 x - 5 y = 4$ . The lines are parallel and the pair is inconsistent. Replacing 14 with 12 gives ratios all equal to $\frac{1}{3}$ , a dependent pair.

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 a consistent pair?Show answer
A pair with at least one solution: intersecting or coincident lines.
2.What is an inconsistent pair?Show answer
A pair with no solution: parallel lines.
3.What is a dependent pair?Show answer
A consistent pair with infinitely many solutions; one equation is a multiple of the other.
4.Condition for parallel lines?Show answer
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$
5.Elimination gives $0 = 0$ . Type?Show answer
Consistent and dependent (coincident lines).
6.Is $x + y = 1$ , $x - y = 1$ consistent?Show answer
Yes; $\frac{1}{1} \neq = \frac{1}{- 1}$ , so the lines intersect once.

Consistent and Inconsistent Pairs of Equations: frequently asked questions

Is a dependent pair consistent or inconsistent?

A dependent pair is always consistent. Its lines coincide, so it has infinitely many solutions, and consistent only needs at least one solution.

Can a pair of linear equations have exactly two solutions?

No. Two straight lines either meet once, never, or everywhere along their length. So the number of solutions is one, zero or infinitely many.

Do I need to draw a graph to decide if a pair is consistent?

Only if the question asks for it. Otherwise comparing the ratios of the coefficients is enough and faster, and you should state the graphical meaning in words.

Is the cross-multiplication method needed for this topic?

No. The cross-multiplication method is not in the 2026-27 syllabus. Use the ratio conditions, graphs, substitution and elimination.