Exercise 3.3 is the first point in Class 10 Maths where the textbook asks you to solve the same pair of linear equations by two algebraic routes — elimination and substitution — and check that both give the same ordered pair.
These exercise 3.3 class 10 maths ncert solutions walk you through every sub-part of both questions with full working, so you can match each step against the textbook page. The chapter introduces substitution just before this exercise (Section 3.3.1) and elimination immediately after (Section 3.3.2), and the exercise trains you in both before moving on.
The two methods feel different but always land on the same answer. Substitution isolates one variable and plugs it into the other equation; elimination multiplies one or both equations so a variable cancels on addition or subtraction.
Knowing which route is shorter for a given pair is itself an exam skill, so this page ends with a method-selection note and a verification habit you can reuse across the remaining exercises of the chapter.
Exercise 3.3 Solutions
This exercise has two numbered questions. Question 1 has four algebraic pairs you must solve by both elimination and substitution; Question 2 has five word problems where you first form the equations yourself, then solve by elimination.
The chapter text on substitution and elimination tells you when each method is convenient: elimination suits pairs whose coefficients align easily after a small multiplier, while substitution is short when one variable already has coefficient 1 (or is already isolated) — for example, in Question 2(ii) one equation gives an age relation you can isolate quickly.
Key Concepts and Formulas for Solving Pairs of Linear Equations
Write every pair in the standard form \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) (NCERT, p. 26) before you compare coefficients or pick a method. The two algebraic routes Exercise 3.3 needs are:
- Substitution (Section 3.3.1): isolate one variable in one equation, substitute that expression into the other equation, solve the resulting one-variable equation, then back-substitute to get the second variable.
- Elimination (Section 3.3.2): multiply one or both equations by suitable non-zero constants so the coefficients of one variable become numerically equal, then add or subtract to cancel that variable. Solve for the other, then substitute back.
Before forming any word-problem pair, use the ratio test from the chapter summary to anticipate whether a solution exists (NCERT, p. 34):
| Ratio comparison | Graphical meaning | Solution type |
|---|---|---|
| \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) | Intersecting lines | Unique solution (consistent) |
| \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) | Coincident lines | Infinitely many solutions (dependent, consistent) |
| \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) | Parallel lines | No solution (inconsistent) |
Question 1: Solve the following pair of linear equations by the elimination method and the substitution method :
- (i) \( x + y = 5 \) and \( 2x – 3y = 4 \)
- (ii) \( 3x + 4y = 10 \) and \( 2x – 2y = 2 \)
- (iii) \( 3x – 5y – 4 = 0 \) and \( 9x = 2y + 7 \)
- (iv) \( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x – \frac{y}{3} = 3 \)
Elimination multiplies one or both equations so a chosen variable cancels on addition or subtraction; substitution isolates one variable in one equation and plugs it into the other — both must give the same ordered pair.
Working each part by both routes is the surest way to catch a sign or arithmetic slip, which is exactly why the textbook asks for both.
Part (i): Equations are \( x + y = 5 \) and \( 2x – 3y = 4 \).
Step 1 (Eliminate): Multiply Eq. (1) by 3: \( 3x + 3y = 15 \).
Add to Eq. (2):
\[ (3x + 3y) + (2x – 3y) = 15 + 4 \Rightarrow 5x = 19 \Rightarrow x = \frac{19}{5} \]
Step 2: Substitute into \( x + y = 5 \): \( y = 5 – \frac{19}{5} = \frac{6}{5} \).
Step 3 (Substitute method): From Eq. (1), \( x = 5 – y \).
Put in Eq. (2): \( 2(5 – y) – 3y = 4 \Rightarrow 10 – 5y = 4 \Rightarrow y = \frac{6}{5} \), then \( x = \frac{19}{5} \).
Final answer: \( x = \frac{19}{5},\ y = \frac{6}{5} \) — same by both methods.
Part (ii): Equations are \( 3x + 4y = 10 \) and \( 2x – 2y = 2 \).
Step 1 (Eliminate): Divide Eq. (2) by 2 to simplify first: \( x – y = 1 \).
Multiply this by 4: \( 4x – 4y = 4 \).
Add to Eq. (1):
\[ (3x + 4y) + (4x – 4y) = 10 + 4 \Rightarrow 7x = 14 \Rightarrow x = 2 \]
Step 2: \( 2 – y = 1 \Rightarrow y = 1 \).
Step 3 (Substitute method): From \( x – y = 1 \), \( x = y + 1 \).
Put in \( 3x + 4y = 10 \): \( 3(y + 1) + 4y = 10 \Rightarrow 7y = 7 \Rightarrow y = 1 \), so \( x = 2 \).
Final answer: \( x = 2,\ y = 1 \) — same by both methods.
Part (iii): Rewrite as \( 3x – 5y = 4 \) and \( 9x – 2y = 7 \).
Step 1 (Eliminate): Multiply Eq. (1) by 3: \( 9x – 15y = 12 \).
Subtract Eq. (2):
\[ (9x – 15y) – (9x – 2y) = 12 – 7 \Rightarrow -13y = 5 \Rightarrow y = -\frac{5}{13} \]
Step 2: Put in \( 3x – 5y = 4 \): \( 3x + \frac{25}{13} = 4 \Rightarrow 3x = \frac{27}{13} \Rightarrow x = \frac{9}{13} \).
Step 3 (Substitute method): From Eq. (1), \( 3x = 5y + 4 \Rightarrow x = \frac{5y + 4}{3} \).
Substitute in \( 9x – 2y = 7 \): \( 3(5y + 4) – 2y = 7 \Rightarrow 13y = -5 \Rightarrow y = -\frac{5}{13} \), hence \( x = \frac{9}{13} \).
Final answer: \( x = \frac{9}{13},\ y = -\frac{5}{13} \) — same by both methods.
Part (iv): Clear denominators first. Multiply the first equation by 6 (the LCM of 2 and 3) and the second by 3:
\[ 3x + 4y = -6 \quad (1),\qquad 3x – y = 9 \quad (2) \]
Step 1 (Eliminate): Subtract Eq. (2) from Eq. (1):
\[ (3x + 4y) – (3x – y) = -6 – 9 \Rightarrow 5y = -15 \Rightarrow y = -3 \]
Step 2: Put in \( 3x – y = 9 \): \( 3x + 3 = 9 \Rightarrow x = 2 \).
Step 3 (Substitute method): From \( 3x – y = 9 \), \( y = 3x – 9 \).
Substitute in \( 3x + 4y = -6 \): \( 3x + 4(3x – 9) = -6 \Rightarrow 15x = 30 \Rightarrow x = 2 \), then \( y = -3 \).
Final answer: \( x = 2,\ y = -3 \) — same by both methods.
After equalising coefficients, students often add when they should subtract (or vice versa). Before combining, check the sign of the term you want to eliminate: if the coefficients have the same sign, subtract the equations; if they have opposite signs, add them.
Question 2: Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
- (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?
- (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
- (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
- (iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
- (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
A word problem becomes a pair of linear equations once you name the two unknowns and write two independent facts about them; elimination then removes one unknown so the other falls out directly. The key skill here is translating each sentence — the solve itself is the easy part.
Part (i): Let the fraction be \( \frac{x}{y} \) (numerator \( x \), denominator \( y \)).
Step 1 (Form): “Add 1 to numerator, subtract 1 from denominator” → \( \frac{x + 1}{y – 1} = 1 \).
“Only add 1 to denominator” → \( \frac{x}{y + 1} = \frac{1}{2} \).
Clear fractions:
\[ x + 1 = y – 1 \Rightarrow x – y = -2 \quad (1); \qquad 2x = y + 1 \Rightarrow 2x – y = 1 \quad (2) \]
Step 2 (Eliminate): Subtract Eq. (1) from Eq. (2): \( (2x – y) – (x – y) = 1 – (-2) \Rightarrow x = 3 \).
Step 3: From \( x – y = -2 \): \( 3 – y = -2 \Rightarrow y = 5 \).
Final answer: The fraction is \( \frac{3}{5} \).
Part (ii): Let Nuri’s present age \( = n \), Sonu’s present age \( = s \).
Step 1 (Form): Five years ago: \( n – 5 = 3(s – 5) \Rightarrow n – 3s = -10 \) (Eq. 1).
Ten years later: \( n + 10 = 2(s + 10) \Rightarrow n – 2s = 10 \) (Eq. 2).
Step 2 (Eliminate): Subtract Eq. (1) from Eq. (2): \( (n – 2s) – (n – 3s) = 10 – (-10) \Rightarrow s = 20 \).
Step 3: \( n – 2(20) = 10 \Rightarrow n = 50 \).
Final answer: Nuri is 50 years old; Sonu is 20 years old.
Part (iii): Let the tens digit \( = x \) and the units digit \( = y \), so the number is \( 10x + y \) and the reversed number is \( 10y + x \).
Step 1 (Form): Digit sum: \( x + y = 9 \) (Eq. 1).
Nine times the number equals twice the reversed number:
\[ 9(10x + y) = 2(10y + x) \Rightarrow 90x + 9y = 20y + 2x \Rightarrow 88x – 11y = 0 \Rightarrow 8x – y = 0 \quad \text{(Eq. 2)} \]
Step 2 (Eliminate): From Eq. (2), \( y = 8x \).
Substitute in Eq. (1): \( x + 8x = 9 \Rightarrow x = 1 \), so \( y = 8 \).
Final answer: The number is \( 10(1) + 8 = 18 \).
Part (iv): Let the number of ₹ 50 notes \( = x \) and the number of ₹ 100 notes \( = y \).
Step 1 (Form): Total notes: \( x + y = 25 \) (Eq. 1).
Total value: \( 50x + 100y = 2000 \).
Divide by 50: \( x + 2y = 40 \) (Eq. 2).
Step 2 (Eliminate): Subtract Eq. (1) from Eq. (2): \( (x + 2y) – (x + y) = 40 – 25 \Rightarrow y = 15 \).
Step 3: \( x + 15 = 25 \Rightarrow x = 10 \).
Final answer: Meena got 10 notes of ₹ 50 and 15 notes of ₹ 100.
Part (v): Let the fixed charge \( = x \) and the per-day extra charge \( = y \). (The first three days cost only the fixed charge; each day after that adds \( y \).)
Step 1 (Form): Saritha kept the book 7 days → 4 extra days: \( x + 4y = 27 \) (Eq. 1).
Susy kept it 5 days → 2 extra days: \( x + 2y = 21 \) (Eq. 2).
Step 2 (Eliminate): Subtract Eq. (2) from Eq. (1): \( (x + 4y) – (x + 2y) = 27 – 21 \Rightarrow 2y = 6 \Rightarrow y = 3 \).
Step 3: \( x + 2(3) = 21 \Rightarrow x = 15 \).
Final answer: Fixed charge is ₹ 15; the per-day extra charge is ₹ 3.
In part (iii), students often write the reversed number as the “concatenation” \( yx \) instead of the place-value form \( 10y + x \). That notation has no arithmetic meaning and produces the wrong second equation. Always expand a two-digit number using place value: a number with tens digit \( a \) and units digit \( b \) is \( 10a + b \), and reversing the digits gives \( 10b + a \).
Method Recap: Choosing Between Elimination and Substitution
Picking the shorter method saves time in the exam. The thumb rule below compares the two routes:
| When elimination is faster | When substitution is faster |
|---|---|
| Coefficients of one variable align easily after a small multiplier (a 1× or 2× step). | One variable already has coefficient 1, or one equation is already of the form \( y = \dots \) or \( x = \dots \). |
| You suspect the pair may be inconsistent or dependent, because a true/false no-variable statement falls out directly. | The equations have fractional coefficients only in one variable that is easy to isolate. |
| Word problems where both equations are of similar complexity and you want a symmetric solve. | Word problems where one unknown is expressed in terms of the other in the problem statement itself. |
Common Mistakes and How to Check Your Answer
Three slips account for most of the marks lost on this exercise:
- Sign slip after equalising coefficients. Once the coefficients match, you add or subtract depending on whether you want the terms to cancel. Check the sign before combining — adding when you should subtract (or the reverse) flips the answer.
- Forgetting to clear denominators in Question 1 part (iv). Working with \( \frac{x}{2} \) and \( \frac{2y}{3} \) directly invites fraction errors. Multiply each equation by the LCM of its denominators (here, 6 for the first and 3 for the second) before eliminating.
- Writing the reversed two-digit number as concatenation. In Question 2 part (iii), writing the reversed number as \( yx \) gives a meaningless equation. Use the place-value expansion \( 10y + x \).
The verification habit is the single best checking move: after solving, substitute your values back into both original equations. The textbook models this in its worked examples — Example 4 (NCERT, p. 31) ends with a verification step after solving by substitution, and Example 10 (NCERT, p. 36) verifies both candidate numbers in the two-digit problem.
If both left-hand sides equal both right-hand sides, the solution is correct.
You can also read the full chapter — including the worked examples that precede this exercise — on the official NCERT textbook page, where the section on algebraic methods is printed with all its examples.
Frequently Asked Questions
What is the difference between the elimination method and the substitution method?
Substitution isolates one variable from one equation and plugs that expression into the other, reducing the pair to a single-variable equation (Section 3.3.1). Elimination multiplies one or both equations by suitable constants so the coefficients of one variable become numerically equal, then adds or subtracts to cancel that variable (Section 3.3.2).
Both methods give the same ordered pair; the choice is a matter of convenience.
How do I check if a pair of linear equations has no solution or infinitely many solutions before solving?
Use the ratio test. For the standard pair \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), compare \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), \( \frac{c_1}{c_2} \).
If all three ratios are equal, the pair has infinitely many solutions; if the first two are equal but the third differs, there is no solution; if the first two differ, there is a unique solution (NCERT, p. 34).
In the two-digit number problem, why is the reversed number written as 10y + x instead of yx?
A two-digit number with tens digit \( x \) and units digit \( y \) equals \( 10x + y \) by place value. Reversing the digits puts \( y \) in the tens place and \( x \) in the units place, so the reversed number is \( 10y + x \).
The symbol \( yx \) is just concatenation — it has no arithmetic value, so the textbook uses the expanded form \( 10y + x \) (Example 10, NCERT, p. 36).
When should I clear denominators before applying the elimination method?
Whenever an equation has fractional coefficients, clear the denominators first by multiplying through by the LCM of the denominators. In Question 1 part (iv), \( \frac{x}{2} + \frac{2y}{3} = -1 \) becomes \( 3x + 4y = -6 \) after multiplying by 6, and \( x – \frac{y}{3} = 3 \) becomes \( 3x – y = 9 \) after multiplying by 3. This converts the pair into integer-coefficient equations and sharply reduces the chance of fraction slips during elimination.
Attribution and Reference
Reference: NCERT Class 10 Mathematics textbook, chapter Pair of Linear Equations in Two Variables. Every question on this page is reproduced exactly from the official textbook exercise, so you can verify each one page by page against the source.
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