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Pair of Linear Equations in Two Variables — Class 10 Maths Notes & Practice

July 31, 2026
Pair of Linear Equations in Two Variables — Class 10 Maths Notes & Practice

A pair of linear equations is just two straight-line equations considered together, and “solving” them means finding the point where the two lines meet. Once you see it that way, the whole chapter falls into place. This page teaches the three methods in plain language, with a plan and an original practice set (with answers you can reveal) at the end.

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Good to Know
The cross-multiplication method is no longer part of the current CBSE syllabus, so you can safely skip it — the focus is the graphical, substitution and elimination methods. Always cross-check your official syllabus.

What This Chapter Covers

Your Game Plan for This Chapter

  1. First — Understand the graphical method and what “consistent” and “inconsistent” mean.
  2. Next — Master substitution and elimination — these solve almost every question.
  3. Last — Learn the ratio conditions for the number of solutions, then attempt the practice set.

Study Notes

1. The Graphical Method

Plot both equations as straight lines on the same graph. Where they cross is the solution. If the lines intersect at one point, there is a unique solution (the pair is consistent). If they are parallel, there is no solution (inconsistent). If they are the same line (coincident), there are infinitely many solutions.

2. The Substitution Method

Make one variable the subject of one equation, then substitute that into the other equation to get a single-variable equation. For example, from x + y = 7 we get x = 7 − y; putting this into 2x − y = 8 gives 2(7 − y) − y = 8, which solves to y = 2 and then x = 5.

3. The Elimination Method

Multiply the equations so that one variable has the same coefficient in both, then add or subtract to eliminate it. For example, for 3x + 2y = 12 and 5x − 2y = 4, adding the two removes y (2y and −2y cancel) to give 8x = 16, so x = 2 and then y = 3.

Exam Tip
Board papers almost always ask one word problem and one direct algebraic solution from this chapter. Pick either substitution or elimination — whichever is quicker for the numbers — and show clean steps.

4. Conditions for the Number of Solutions

For a pair $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, you can tell the number of solutions just by comparing the ratios of the coefficients — without solving:

Three coordinate graphs side by side: intersecting lines labelled One Solution, coincident lines labelled Infinite Solutions, and parallel lines labelled No Solution.
Figure: The three possible cases for a pair of linear equations · चित्र: दो रैखिक समीकरणों की तीन संभावनाएँ
Compare the ratiosThe lines areNumber of solutions
$\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$IntersectingExactly one (unique) — consistent
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$ParallelNo solution — inconsistent
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$CoincidentInfinitely many — consistent
Common Mistake
In elimination, watch the signs when you add or subtract — a wrong sign flips the whole answer. And remember: parallel lines ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$) mean no solution, not infinite.

Practice Worksheet

Try each question fully on your own first, then click Show Answer to check yourself.

Q1. Solve by substitution: x + y = 7 and 2x − y = 8.

Show Answer
From the first equation, x = 7 − y. Substitute into the second: 2(7 − y) − y = 8 → 14 − 2y − y = 8 → 14 − 3y = 8 → 3y = 6 → y = 2. Then x = 7 − 2 = 5. So x = 5, y = 2.

Q2. Solve by elimination: 3x + 2y = 12 and 5x − 2y = 4.

Show Answer
Add the two equations (the 2y and −2y cancel): 8x = 16 → x = 2. Put x = 2 in the first: 3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3. So x = 2, y = 3.

Q3. Without solving, how many solutions does 2x + 3y = 7, 4x + 6y = 14 have, and what kind of lines are they?

Show Answer
Here $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, and $\frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$ — all equal. So the pair has infinitely many solutions and the lines are coincident (the same line).

Q4. Do x + 2y = 4 and 2x + 4y = 12 have a solution?

Show Answer
$\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$, but $\frac{c_1}{c_2} = \frac{4}{12} = \frac{1}{3}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, there is no solution — the lines are parallel.

Q5. The sum of two numbers is 20 and their difference is 4. Find the numbers.

Show Answer
Let the numbers be x and y. Then x + y = 20 and x − y = 4. Adding the two: 2x = 24 → x = 12. Then y = 20 − 12 = 8. So the numbers are 12 and 8.

Once these feel easy, you have genuinely finished this chapter. Do not aim for perfect on the first try — aim for one more correct question than yesterday.

Written & reviewed by Team Principal Saab — Meet the team →