Class 10 Maths · Original EZT practice
Pair of Linear Equations (Class 10): original worked practice
Two conditions must hold at the same time. Sometimes that gives one point; sometimes every point on a line works, and sometimes no point can satisfy both conditions. Check the pair before choosing a method.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 17 October 2026:
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Test whether a second condition is new
Choose the method: Compare whole equations. Multiplying every term by the same nonzero number describes the same line.
EZT practice question
Classify 3x − 2y = 8 and 9x − 6y = 24 without finding a single pair.
Show the worked solution
- Multiply the first equation by 3. It becomes exactly the second.
- The second adds no new restriction. Every point on the first line satisfies both.
Answer: Infinitely many solutions; the pair is dependent and consistent.
Check: For instance, (4, 2) gives 12 − 4 = 8 and 36 − 12 = 24.
Watch for this: One working pair does not prove uniqueness.
Now try it yourself
Classify 5x + 4y = 13 and 10x + 8y = 29.
Check your answer
No solution; the pair is inconsistent.
Doubling the first left side gives the second left side but requires 26, not 29.
Handle zero coefficients without dividing by zero
Choose the method: A valid line ax + by = c has at least one of a or b nonzero. Read vertical and horizontal lines directly.
EZT practice question
Solve 0x + 3y = 12 and 2x + 0y = −6.
Show the worked solution
- The first equation fixes y = 4 and leaves x unrestricted.
- The second fixes x = −3. Both hold at their intersection.
Answer: (x, y) = (−3, 4)
Check: 0(−3) + 3(4) = 12; 2(−3) + 0(4) = −6.
Watch for this: Coefficient-ratio tests written as fractions cannot divide by a zero coefficient.
Now try it yourself
Solve 4x = 28 and −2y = 10.
Check your answer
(7, −5)
A vertical line x = 7 and a horizontal line y = −5 cross once.
Isolate a variable with a small coefficient
Choose the method: Substitute the entire expression, using brackets when multiplying or subtracting it.
EZT practice question
Solve x − 2y = 1 and 3x + y = 31 by substitution.
Show the worked solution
- The first equation gives x = 1 + 2y.
- Substitute into the second: 3(1 + 2y) + y = 31.
- Thus 7y = 28, y = 4 and x = 9.
Answer: (9, 4)
Check: 9 − 2(4) = 1; 3(9) + 4 = 31.
Watch for this: 3(1 + 2y) is 3 + 6y, not 3 + 2y.
Now try it yourself
Solve x + 3y = 19 and 2x − y = 3.
Check your answer
(4, 5)
x = 19 − 3y gives 38 − 7y = 3, so y = 5 and x = 4.
Scale every term, including the constant
Choose the method: Match opposite coefficients to add, or equal coefficients to subtract.
EZT practice question
Solve 4x + 3y = 29 and 3x − 2y = 9 by elimination.
Show the worked solution
- Multiply the first by 2: 8x + 6y = 58.
- Multiply the second by 3: 9x − 6y = 27.
- Add to get 17x = 85, so x = 5. Substitute to get y = 3.
Answer: (5, 3)
Check: 4(5) + 3(3) = 29; 3(5) − 2(3) = 9.
Watch for this: Multiplying an equation by 3 also multiplies its right side by 3.
Now try it yourself
Solve 5x + 2y = 37 and 2x − 3y = −8.
Check your answer
(5, 6)
Multiply the first by 3 and the second by 2, then add: 19x = 95. Substitution gives y = 6.
A zero determinant needs a second check
Choose the method: When coefficient rows are proportional, compare the constants too. A zero determinant alone does not distinguish no solution from infinitely many.
EZT practice question
For which k does 4x + ky = 10 and 6x + 9y = 15 fail to have a unique solution? Classify that case.
Show the worked solution
- The coefficient determinant is 4 × 9 − 6k = 36 − 6k. It is zero at k = 6.
- At k = 6, multiplying 4x + 6y = 10 by 3/2 gives the second equation exactly.
Answer: k = 6 gives infinitely many solutions. Every other real k gives one solution.
Check: At k = 6, (1, 1) fits both equations, but so do infinitely many other points on the same line.
Watch for this: Do not label every zero-determinant pair inconsistent.
Now try it yourself
If the second right side is changed from 15 to 16, what happens at k = 6?
Check your answer
There is no solution.
The proportional left sides would require 15 and 16 at once.
Extra revision: a determinant check
Choose the method: For ax + by = e and cx + dy = f, if D = ad − bc is nonzero, x = (ed − bf)/D and y = (af − ec)/D.
EZT practice question
Use the formula to check 3x + 5y = 2 and 2x − 3y = 14.
Show the worked solution
- D = 3(−3) − 5(2) = −19.
- x = [2(−3) − 5(14)]/(−19) = (−76)/(−19) = 4.
- y = [3(14) − 2(2)]/(−19) = 38/(−19) = −2.
Answer: (4, −2)
Check: 3(4) + 5(−2) = 2 and 2(4) − 3(−2) = 14.
Watch for this: This optional formula needs D ≠ 0; it cannot solve a zero-determinant case by division.
Now try it yourself
Check 2x + 3y = 7 and 4x − y = 7 with the formula.
Check your answer
(2, 1)
D = −14; the x numerator is −28 and the y numerator is −14.
Translate the quantities before solving
Choose the method: Define a unit price, then multiply it by each item count. Treat these as fictional practice prices.
EZT practice question
A practice stall charges ₹122 for 4 folders and 3 pens, and ₹124 for 2 folders and 5 pens. Find the unit prices.
Show the worked solution
- Let f be rupees per folder and p be rupees per pen. Then 4f + 3p = 122 and 2f + 5p = 124.
- Double the second equation: 4f + 10p = 248. Subtract the first: 7p = 126.
- Thus p = 18 and 2f + 90 = 124, so f = 17.
Answer: Folder ₹17; pen ₹18.
Check: 4(17) + 3(18) = 122; 2(17) + 5(18) = 124.
Watch for this: The letters represent unit prices, not the total money for that type of item.
Now try it yourself
A fictional kiosk charges ₹98 for 3 notebooks and 2 cards, and ₹102 for 2 notebooks and 3 cards. Find the unit prices.
Check your answer
Notebook ₹18; card ₹22.
Adding gives 5n + 5c = 200, so n + c = 40. Subtracting the first equation from the second gives c − n = 4; hence n = 18 and c = 22.
Distinguish a plotted estimate from an exact solution
Choose the method: A graph suggests the intersection; substitution into both equations checks whether the coordinates are exact.
EZT practice question
A rough drawing suggests (3, 2) for x + 2y = 7 and 2x − y = 4. Is the point exact?
Show the worked solution
- Substitute x = 3 and y = 2 into both equations.
- 3 + 2(2) = 7 and 2(3) − 2 = 4. Both are exact equalities.
Answer: Yes. The point is an exact solution, not just a visual estimate.
Check: The coefficients have determinant −5, which is nonzero, so the intersection is unique.
Watch for this: A thick pencil line is not proof of exact coordinates.
Now try it yourself
A sketch suggests (2, 3) for x + y = 5 and 3x − y = 4. Check and correct it.
Check your answer
The guess fails; the exact point is (9/4, 11/4).
The guess gives 6 − 3 = 3, not 4. Adding the equations gives 4x = 9, then y = 5 − 9/4 = 11/4.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Two numbers total 43 and the first exceeds the second by 9. Find them.
Show answer and reason
26 and 17.
Adding x + y = 43 and x − y = 9 gives 2x = 52.
2. Does a pair of linear equations always have a unique solution?
Show answer and reason
No.
Dependent pairs have infinitely many solutions; inconsistent pairs have none. A unique solution needs independent coefficient rows.
3. For 7x − 4y = 9, a learner proposes 14x − 8y = 19 as an equivalent equation. Identify the error.
Show answer and reason
The constant must be 18, not 19.
An equivalent equation requires multiplying every term by the same nonzero factor.
4. Can 0x + 0y = 5 be treated as a straight line?
Show answer and reason
No; it is an impossible statement.
There is no coordinate pair satisfying 0 = 5; a line equation needs at least one nonzero variable coefficient.
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
For cost context, see home tuition fees in Delhi NCR. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- Which method should I choose?
- Use a method that keeps the arithmetic clear. A coefficient of 1 often helps substitution; easily matched coefficients often help elimination. Check the answer in both original equations.
- Can coefficient ratios include zero denominators?
- No. Compare proportional equations directly or use a determinant test, then inspect the constants. Do not divide by zero.
- Is cross-multiplication required here?
- The linked chapter summary lists substitution and elimination. Our determinant formula is optional extra revision, not a statement about your assessment requirements.
- Are graphs only rough estimates?
- Not always. A graph can show an exact intersection at clear coordinates, but a numerical claim should still be checked in both equations.
- Are these textbook exercise answers?
- No. They are original selected practice with worked checks, not a full NCERT answer key or a guarantee of examination coverage.