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A pair of linear equations is two straight lines, and solving them means finding where they meet. Class 10 asks you to do this three ways: draw the lines, use substitution, or use elimination, and to tell in advance whether there is one solution, none or infinitely many. This post covers what the current syllabus includes, solves one system both ways, works a word problem, and lists the slips to avoid.
What does the syllabus include this year?
CBSE's Class X Mathematics syllabus for 2026-27 lists the graphical method and "consistency/inconsistency", the "algebraic conditions for number of solutions", and solving "by substitution, by elimination", with "simple situational problems". The NCERT chapter in its rationalised 2026-27 reprint matches this: section 3.2 on the graphical method and section 3.3 on the two algebraic methods. Cross-multiplication is not in the current syllabus, so do not use it as your main method unless your teacher says otherwise.
How do you know how many solutions there are?
For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, compare the ratios of coefficients:
a₁/a₂ ≠ b₁/b₂: the lines intersect. One solution. The pair is consistent.
a₁/a₂ = b₁/b₂ ≠ c₁/c₂: the lines are parallel. No solution. The pair is inconsistent.
a₁/a₂ = b₁/b₂ = c₁/c₂: the lines coincide. Infinitely many solutions. The pair is dependent, and still consistent.
Example. 2x + 3y = 5 and 4x + 6y = 7. Write them as 2x + 3y − 5 = 0 and 4x + 6y − 7 = 0. Then a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, and c₁/c₂ = −5/−7 = 5/7. The first two match but the third does not, so the lines are parallel and there is no solution.
Solve one system two ways
Solve: 2x + 3y = 13 and 3x − y = 3.
By substitution.
From the second equation, y = 3x − 3.
Substitute into the first: 2x + 3(3x − 3) = 13, so 2x + 9x − 9 = 13.
11x = 22, so x = 2.
Then y = 3(2) − 3 = 3.
By elimination.
Multiply the second equation by 3: 9x − 3y = 9.
Add it to the first: (2x + 3y) + (9x − 3y) = 13 + 9, so 11x = 22, x = 2.
Substitute back: 2(2) + 3y = 13, so y = 3.
Check in both equations: 2(2) + 3(3) = 13, and 3(2) − 3 = 3. The solution is x = 2, y = 3. On a graph, the two lines cross at the point (2, 3).
Substitution is quicker when one variable already has a coefficient of 1 or −1, as y does here. Elimination is quicker when the coefficients are larger, as in the next example.
Why compare the two methods?
Seeing two methods side by side teaches you when to use each. Rittle-Johnson and Star (2007) randomly assigned 70 seventh-graders learning to solve equations either to compare alternative methods or to study the same methods one at a time. The comparison group made greater gains in procedural knowledge and flexibility, with similar gains in conceptual knowledge. Solving each system both ways and asking "which was shorter, and why?" is the same idea.
A word problem, worked in full
Problem. Two adult tickets and three child tickets cost ₹1300. Three adult tickets and two child tickets cost ₹1450. Find the price of each.
Let an adult ticket cost ₹x and a child ticket ₹y.
Equations: 2x + 3y = 1300 and 3x + 2y = 1450.
Multiply the first by 3 and the second by 2: 6x + 9y = 3900 and 6x + 4y = 2900.
Subtract: 5y = 1000, so y = 200.
Substitute: 2x + 600 = 1300, so x = 350.
Check: 3(350) + 2(200) = 1050 + 400 = 1450.
Answer: an adult ticket costs ₹350 and a child ticket ₹200.
Which mistakes lose marks?
Sign errors when subtracting. In step 4, subtract every term: 9y − 4y and 3900 − 2900. Write the subtraction in brackets if it helps.
Stopping after one variable. Find both, and write them as a pair.
No check. Substituting into the second equation takes ten seconds and catches most slips.
Not defining variables. In word problems, write "Let ₹x be..." before any equation.
Ending with x and y. Answer in the words of the question, with units.
Inaccurate graphs. Plot at least two points per line from a table, use a ruler and label the intersection.
If you can follow a solution but cannot start one yourself, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see why the first move was chosen.
A one-week practice routine
Day 1: Six pairs: decide one, none or infinitely many solutions by ratios alone.
Day 2: Four systems, each solved by substitution and then by elimination. Note which was shorter.
Day 3: Two systems solved on graph paper. Confirm the crossing point algebraically.
Day 4: Four word problems: ages, tickets, digits and speeds.
Day 5: A mixed set of six, timed, with ratios, solving and word problems shuffled.
The same comparison habit works in other chapters; our guide to the self-explanation technique shows how to talk through each step as you go.
Today, take one system from your textbook and solve it both ways, then write one line on which method was quicker.
