This page holds the complete exercise 3.2 class 10 maths ncert solutions — every question of Exercise 3.2 reproduced word for word and solved line by line with the substitution method. Work through each step as you check your homework; do not just read the final values.
Exercise 3.2 is the substitution exercise of Chapter 3, Pair of Linear Equations in Two Variables (NCERT Class 10 Mathematics). Question 1 gives six equation pairs to solve, Question 2 adds a line-verification twist to a solved pair, and Question 3 contains six word problems where forming the equations earns the marks.
If you follow the method here, you can verify your own working all the way through.
Exercise 3.2 Solutions
Exercise 3.2 (NCERT, p. 34) is built around one technique: the substitution method. You apply it to pure equation pairs in Question 1, carry the result into a line equation in Question 2, and translate words into equations before solving in Question 3.
Here is the shape of the exercise at a glance:
- Question 1 — six pairs of equations in different coefficient styles: whole numbers, one decimal pair, one surd pair and two fraction pairs.
- Question 2 — solve a pair, then use the solution point to find the slope \(m\) of the line \(y = mx + 3\).
- Question 3 — six word problems (numbers, angles, cricket prices, taxi fares, a fraction, ages), where the marks go to forming the equations first.
In Question 3, the examiner rewards the equations you set up, not just the arithmetic that follows. Name the unknowns, translate each sentence, then solve mechanically by substitution. Every equation used here matches the printed source — cross-check the surds, fractions and decimals of Exercise 3.2 against the official NCERT Class 10 Mathematics textbook PDF (Chapter 3) page by page before you start.
Substitution Method: Steps and the Two Special Cases
The textbook builds the substitution method in three steps (NCERT, p. 31). Once you can run these without thinking, the whole exercise becomes a decision about which variable to isolate and how to handle the coefficients.
- Isolate a variable. Choose either equation and write one variable in terms of the other. Pick the variable with coefficient 1 when you can.
- Substitute. Put that expression into the other equation. It becomes a one-variable equation; solve it.
- Back-substitute. Put the value found in Step 2 into the expression from Step 1 to get the second variable.
A quick worked example with fresh numbers — solve \(2x + y = 7\) and \(x – y = 2\). From the first, \(y = 7 – 2x\). Substituting into the second gives \(x – (7 – 2x) = 2\), so \(3x = 9\) and \(x = 3\). Then \(y = 7 – 6 = 1\). The solution is \((3, 1)\), and it checks in both equations.
Two outcomes need special handling. If Step 2 collapses to a true statement with no variable — like \(18 = 18\) — the pair has infinitely many solutions (NCERT, p. 32). If it collapses to a false statement — like \(-4 = 0\) — the lines are parallel and there is no solution (NCERT, p. 33).
You can predict either outcome before working, using the coefficient ratios from Table 3.1 (NCERT, p. 26). For a pair written as \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
| Compare the ratios | The lines | Number of solutions |
|---|---|---|
| \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) | Intersecting | Exactly one (unique) solution |
| \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) | Coincident | Infinitely many solutions |
| \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) | Parallel | No solution |
Compare the three ratios first; they tell you whether to expect a single point, a whole line, or nothing. This same “where do the graphs meet” idea appears with polynomial graphs in Chapter 2’s graph study, which is a useful refresher.
Question-wise Solutions
Question 1: Solve the following pair of linear equations by the substitution method.
- Part (i): \(x + y = 14\), \(x – y = 4\)
- Part (ii): \(s – t = 3\), \(\frac{s}{3} + \frac{t}{2} = 6\)
- Part (iii): \(3x – y = 3\), \(9x – 3y = 9\)
- Part (iv): \(0.2x + 0.3y = 1.3\), \(0.4x + 0.5y = 2.3\)
- Part (v): \(\sqrt{2}x + \sqrt{3}y = 0\), \(\sqrt{3}x – \sqrt{8}y = 0\)
- Part (vi): \(\frac{3x}{2} – \frac{5y}{3} = -2\), \(\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)
Part (i): In \(x + y = 14\) the variable \(y\) has coefficient 1, so isolate it: \(y = 14 – x\).
Substitute this into the other equation, \(x – y = 4\):
\[ x – (14 – x) = 4 \]
This gives \(2x – 14 = 4\), so \(2x = 18\) and \(x = 9\).
Back-substitute: \(y = 14 – 9 = 5\).
Final answer: \(x = 9,\ y = 5\). Check: \(9 + 5 = 14\) and \(9 – 5 = 4\).
Part (ii): From \(s – t = 3\), write \(s = t + 3\).
Substitute into \(\frac{s}{3} + \frac{t}{2} = 6\):
\[ \frac{t+3}{3} + \frac{t}{2} = 6 \]
Clear the fractions by multiplying the whole equation by 6: \(2(t+3) + 3t = 36\), so \(5t = 30\) and \(t = 6\).
Then \(s = 6 + 3 = 9\).
Final answer: \(s = 9,\ t = 6\). Check: \(9 – 6 = 3\) and \(\frac{9}{3} + \frac{6}{2} = 3 + 3 = 6\).
Part (iii): From \(3x – y = 3\), write \(y = 3x – 3\).
Substitute into \(9x – 3y = 9\):
\[ 9x – 3(3x – 3) = 9 \]
This simplifies to \(9x – 9x + 9 = 9\), that is \(9 = 9\), a true statement with no variable left.
The second equation is exactly three times the first, so both describe the same line.
Final answer: infinitely many solutions — every point \((x,\ 3x – 3)\) satisfies both equations.
Part (iv): Decimals make substitution messy, so clear them first.
Multiply each equation by 10: \(2x + 3y = 13\) and \(4x + 5y = 23\).
From \(2x + 3y = 13\), write \(x = \frac{13 – 3y}{2}\).
Substitute:
\[ 4\left(\frac{13 – 3y}{2}\right) + 5y = 23 \]
This gives \(2(13 – 3y) + 5y = 23\), so \(26 – 6y + 5y = 23\) and \(y = 3\).
Then \(x = \frac{13 – 9}{2} = 2\).
Final answer: \(x = 2,\ y = 3\). Check: \(0.2(2) + 0.3(3) = 1.3\) and \(0.4(2) + 0.5(3) = 2.3\).
Part (v): Both equations equal 0, so the unknowns are tied proportionally.
From \(\sqrt{2}x + \sqrt{3}y = 0\), write \(x = -\frac{\sqrt{3}}{\sqrt{2}}y\).
Substitute into \(\sqrt{3}x – \sqrt{8}y = 0\):
\[ \sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) – \sqrt{8}y = 0 \]
Collect the \(y\) terms: \(\left(-\frac{3}{\sqrt{2}} – \sqrt{8}\right)y = 0\).
The bracket is not zero, so \(y = 0\) and then \(x = 0\).
Final answer: \(x = 0,\ y = 0\).
Part (vi): Clear both sets of fractions by multiplying each equation by 6.
The first becomes \(9x – 10y = -12\) and the second \(2x + 3y = 13\).
From \(2x + 3y = 13\), write \(x = \frac{13 – 3y}{2}\).
Substitute:
\[ 9\left(\frac{13 – 3y}{2}\right) – 10y = -12 \]
Multiply through by 2: \(117 – 27y – 20y = -24\), so \(47y = 141\) and \(y = 3\).
Then \(x = \frac{13 – 9}{2} = 2\).
Final answer: \(x = 2,\ y = 3\). Check: \(\frac{3(2)}{2} – \frac{5(3)}{3} = 3 – 5 = -2\) and \(\frac{2}{3} + \frac{3}{2} = \frac{13}{6}\).
Common errors: In Part (iii), the moment a true statement like \(9 = 9\) appears, stop — it means infinitely many solutions, not a single guessed pair. In Part (v), never divide both sides by \(y\); that cancels away the only solution, \(y = 0\). In Parts (ii) and (vi), clear every fraction term by the same multiplier, never just one term.
Question 2: Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Step 1: Isolate \(x\) from \(2x + 3y = 11\):
\[ x = \frac{11 – 3y}{2} \]
Step 2: Substitute into \(2x – 4y = -24\):
\[ 2\left(\frac{11 – 3y}{2}\right) – 4y = -24 \]
The fraction cancels: \(11 – 3y – 4y = -24\), so \(11 – 7y = -24\), giving \(7y = 35\) and \(y = 5\).
Step 3: Back-substitute: \(x = \frac{11 – 15}{2} = -2\).
The solution point is \((-2,\ 5)\).
Find m: The word “hence” tells you to use this point.
The line \(y = mx + 3\) passes through \((-2,\ 5)\):
\[ 5 = m(-2) + 3 \]
So \(5 – 3 = -2m\), giving \(2 = -2m\) and \(m = -1\).
Final answer: \(x = -2,\ y = 5\) and \(m = -1\). Check: \(2(-2) + 3(5) = 11\), \(2(-2) – 4(5) = -24\), and \(5 = -1(-2) + 3\).
Common error: Solving the pair correctly and stopping there loses the second half of the marks. Also, substituting the point with coordinates swapped uses \(x = 5, y = -2\) and gives a wrong \(m\) — always place \(x\) in the \(x\)-position and \(y\) in the \(y\)-position.
Question 3: Form the pair of linear equations for the following problems and find their solution by substitution method.
- Part (i): The difference between two numbers is 26 and one number is three times the other. Find them.
- Part (ii): The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
- Part (iii): The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
- Part (iv): The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
- Part (v): A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
- Part (vi): Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Part (i): Let the numbers be \(x\) and \(y\), with \(x \gt y\).
“Difference is 26” gives \(x – y = 26\); “one number is three times the other” gives \(x = 3y\).
Substitute \(x = 3y\) into the first:
\[ 3y – y = 26 \]
So \(2y = 26\), giving \(y = 13\), and then \(x = 39\).
Final answer: the numbers are 39 and 13.
Part (ii): Let the larger angle be \(x\) and the smaller be \(y\).
Supplementary angles sum to \(180^\circ\), so \(x + y = 180\); “larger exceeds smaller by 18 degrees” gives \(x – y = 18\).
From the second, \(x = y + 18\).
Substitute:
\[ (y + 18) + y = 180 \]
So \(2y = 162\), giving \(y = 81\), and \(x = 99\).
Final answer: the angles are \(99^\circ\) and \(81^\circ\).
Part (iii): Let a bat cost \(b\) rupees and a ball cost \(l\) rupees.
The two purchases give \(7b + 6l = 3800\) and \(3b + 5l = 1750\).
From the second, \(b = \frac{1750 – 5l}{3}\).
Substitute into the first:
\[ 7\left(\frac{1750 – 5l}{3}\right) + 6l = 3800 \]
Multiply through by 3: \(12250 – 35l + 18l = 11400\), so \(17l = 850\) and \(l = 50\).
Then \(b = \frac{1750 – 250}{3} = 500\).
Final answer: a bat costs ₹500 and a ball costs ₹50.
Part (iv): Let the fixed charge be \(f\) rupees and the per-kilometre charge be \(k\) rupees.
Ten kilometres cost ₹105 and fifteen kilometres cost ₹155:
\[ f + 10k = 105,\qquad f + 15k = 155 \]
From the first, \(f = 105 – 10k\).
Substitute into the second: \(105 – 10k + 15k = 155\), so \(5k = 50\) and \(k = 10\).
Then \(f = 105 – 100 = 5\).
For 25 km, the cost is \(5 + 10(25) = 255\).
Final answer: fixed charge ₹5, charge per km ₹10, and ₹255 for a 25 km journey.
Part (v): Let the fraction be \(\frac{x}{y}\) with numerator \(x\) and denominator \(y\).
Adding 2 to both parts gives \(\frac{x+2}{y+2} = \frac{9}{11}\); adding 3 gives \(\frac{x+3}{y+3} = \frac{5}{6}\).
Cross-multiply each:
\[ 11(x+2) = 9(y+2) \Rightarrow 9y – 11x = 4 \]
\[ 6(x+3) = 5(y+3) \Rightarrow 5y – 6x = 3 \]
From \(9y – 11x = 4\), write \(y = \frac{11x + 4}{9}\) and substitute into the second:
\[ 5\left(\frac{11x + 4}{9}\right) – 6x = 3 \]
Multiply through by 9: \(55x + 20 – 54x = 27\), so \(x = 7\).
Then \(y = \frac{77 + 4}{9} = 9\).
Final answer: the fraction is \(\frac{7}{9}\). Check: \(\frac{7+2}{9+2} = \frac{9}{11}\) and \(\frac{7+3}{9+3} = \frac{10}{12} = \frac{5}{6}\).
Part (vi): Let Jacob’s present age be \(j\) years and his son’s be \(s\) years.
“Five years hence” gives \(j + 5 = 3(s + 5)\).
“Five years ago” gives \(j – 5 = 7(s – 5)\).
Expand both:
\[ j = 3s + 10,\qquad j = 7s – 30 \]
Both equal \(j\), so equate them: \(3s + 10 = 7s – 30\), giving \(4s = 40\) and \(s = 10\).
Then \(j = 3(10) + 10 = 40\).
Final answer: Jacob is 40 years old and his son is 10 years old. Check: in 5 years \(45 = 3 \times 15\); 5 years ago \(35 = 7 \times 5\).
Common errors: In Part (iv), students swap the two amounts — state “fixed ₹5, per km ₹10” with both units. In Part (v), careless cross-multiplication mis-signs the equation. In Part (vi), “five years hence” means add 5 and “five years ago” means subtract 5; inverting the signs gives impossible negative ages. End every part by naming the quantity asked for, with its unit.
Exercise 3.2 Class 10 Maths NCERT Solutions Recap: Substitution in Three Steps
Revision in six lines:
- Write one variable in terms of the other from either equation.
- Substitute that expression into the other equation and solve it.
- Back-substitute to get the second variable.
- Result is a true statement: infinitely many solutions.
- Result is a false statement: no solution.
- Always verify by substituting back into both original equations.
The slips below are the ones that actually cost marks on this exercise:
| Common mistake | Correct rule | How to check |
|---|---|---|
| Substituting back into the same equation you isolated from, getting \(0 = 0\) | Substitute the expression into the other equation only | The substituted equation should still contain one unknown; if both vanish, you used the same equation |
| Calling Part (iii) of Question 1 unsolvable instead of spotting the true statement | \(9x – 3y = 9\) is \(3 \times (3x – y = 3)\); predict with \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) | Every point \((x,\ 3x – 3)\) satisfies both equations |
| Clearing decimals on one equation but not the other | Multiply every term of both sides by the same power of 10 | \(0.4x + 0.5y = 2.3\) must become \(4x + 5y = 23\), never half-cleared |
| Dividing both sides by \(y\) in a product like \(y \cdot (\text{bracket}) = 0\) | A product is zero only when a factor is zero; never cancel the unknown | Keep the bracket factor: it forces \(y = 0\), then \(x = 0\) |
| Forgetting the “hence” part, or swapping coordinates in \(y = mx + 3\) | Place the solved point \((x,\ y)\) exactly in the line’s equation | \(5 = m(-2) + 3\) gives \(m = -1\); verify in both original equations |
Substitution is one of two algebraic routes; the next exercise, Exercise 3.3, introduces the elimination method, which is quicker when coefficients line up. Substitution also returns as a solving tool in quadratic equations later. For the full year, browse the complete set of Class 10 Maths solutions and the broader Class 10 revision hub.
Frequently Asked Questions
Why does the substitution method sometimes end in a true statement with no variable, like 18 = 18?
That happens when both equations describe the same line, so the statement holds for every value of the variable. This is exactly Part (iii) of Question 1: \(9x – 3y = 9\) is three times \(3x – y = 3\). The conclusion is infinitely many solutions, and the pair is called dependent (NCERT, p. 32).
Which variable should I make the subject when I start substituting?
Choose the variable that is cheapest to isolate — coefficient 1 or −1, or the equation already nearly solved. In Part (i), \(x + y = 14\) gives \(y = 14 – x\) in one line. In Part (ii), \(s – t = 3\) gives \(s = t + 3\). If every coefficient is awkward, clear fractions or decimals first, then pick the simplest expression.
How do I solve pairs with decimals such as 0.2x + 0.3y = 1.3?
Multiply the whole equation by a power of 10 before substituting. For \(0.2x + 0.3y = 1.3\) and \(0.4x + 0.5y = 2.3\), multiplying every term by 10 gives \(2x + 3y = 13\) and \(4x + 5y = 23\), which are the clean equations solved in Part (iv). For fractions, use the lowest common multiple of the denominators instead.
What does it mean if substitution gives a false statement like −4 = 0?
The two lines are parallel, so they never meet; the pair has no solution and is called inconsistent (NCERT, p. 33). The chapter’s rail example does this: substituting from \(x + 2y – 4 = 0\) into \(2x + 4y – 12 = 0\) ends at \(8 – 12 = 0\), i.e. \(-4 = 0\), a false statement, so the rails never cross.
How do I turn the Question 3 word problems into a pair of equations?
Give each unknown a letter first, then translate each sentence into an equation using the fact order. “Difference is 26” becomes \(x – y = 26\); “one number is three times the other” becomes \(x = 3y\). “Supplementary” means the sum is \(180^\circ\), so \(x + y = 180\). Write every equation in the same unit as the question, then solve by substitution.
Reference: NCERT Class 10 Mathematics textbook, chapter Pair of Linear Equations in Two Variables.
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