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NCERT Solutions · Class 10 · Chapter 3: Pair of Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2

Exercise 3.2: Solving by substitution. Make one variable the subject of one equation, substitute into the other, solve, then substitute back. If the variables vanish, the lines are coincident (a true statement such as \(3 = 3\)) or parallel (a false one).

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Exercise 3.2 questions and solutions

Exercise 3.2, Question 1

Solve by substitution.
(i) \(x + y = 14,\ x - y = 4\)
Show solution
  1. \(x = 14 - y\), so \(14 - 2y = 4 \Rightarrow y = 5\).
  2. \(x = 9\).
Answer: \(x = 9,\ y = 5\)
(ii) \(s - t = 3,\ \tfrac s3 + \tfrac t2 = 6\)
Show solution
  1. \(s = t + 3\); then \[\begin{aligned}&\tfrac{t + 3}{3} + \tfrac t2 = 6 \\ \Rightarrow\ &2t + 6 + 3t = 36\end{aligned}\]
  2. \(t = 6,\ s = 9\).
Answer: \(s = 9,\ t = 6\)
(iii) \(3x - y = 3,\ 9x - 3y = 9\)
Show solution
  1. \(y = 3x - 3\); then \(9x - 9x + 9 = 9\), always true.
  2. The equations describe the same line.
Answer: Infinitely many solutions: \(y = 3x - 3\) for any \(x\)
(iv) \(0.2x + 0.3y = 1.3,\ 0.4x + 0.5y = 2.3\)
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  1. Multiply by 10: \(2x + 3y = 13\), \(4x + 5y = 23\).
  2. \(x = \dfrac{13 - 3y}{2}\); then \(26 - 6y + 5y = 23 \Rightarrow y = 3\), \(x = 2\).
Answer: \(x = 2,\ y = 3\)
(v) \(\sqrt2\,x + \sqrt3\,y = 0,\ \sqrt3\,x - \sqrt8\,y = 0\)
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  1. \(x = -\dfrac{\sqrt3}{\sqrt2}y\); then \[\begin{aligned}&-\dfrac{3}{\sqrt2}y - 2\sqrt2\,y = 0 \\ \Rightarrow\ &y\left(-\dfrac{7}{\sqrt2}\right) = 0\end{aligned}\]
  2. \(y = 0\), so \(x = 0\).
Answer: \(x = 0,\ y = 0\)
(vi) \(\tfrac{3x}{2} - \tfrac{5y}{3} = -2,\ \tfrac x3 + \tfrac y2 = \tfrac{13}{6}\)
Show solution
  1. Clear fractions: \(9x - 10y = -12\) and \(2x + 3y = 13\).
  2. \(x = \dfrac{13 - 3y}{2}\); then \[\begin{aligned}&\dfrac{9(13 - 3y)}{2} - 10y = -12 \\ \Rightarrow\ &117 - 27y - 20y = -24\end{aligned}\]
  3. \(y = 3,\ x = 2\).
Answer: \(x = 2,\ y = 3\)

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Exercise 3.2, Question 2

Solve \(2x + 3y = 11\) and \(2x - 4y = -24\), and hence find \(m\) for which \(y = mx + 3\).
Show solution
  1. Subtracting, \(7y = 35 \Rightarrow y = 5\); then \(2x = 11 - 15 \Rightarrow x = -2\).
  2. \(5 = -2m + 3 \Rightarrow m = -1\).
Answer: \(x = -2,\ y = 5,\ m = -1\)

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Exercise 3.2, Question 3

Form the pair of equations and solve by substitution.
(i) Two numbers differ by 26, and one is three times the other.
Show solution
  1. \(x - y = 26\), \(x = 3y\).
  2. \(2y = 26 \Rightarrow y = 13,\ x = 39\).
Answer: \(39\) and \(13\)
(ii) Of two supplementary angles, the larger exceeds the smaller by \(18^\circ\).
Show solution
  1. \(x + y = 180\), \(x - y = 18\).
  2. \(x = 99,\ y = 81\).
Answer: \(99^\circ\) and \(81^\circ\)
(iii) 7 bats and 6 balls cost ₹3800; 3 bats and 5 balls cost ₹1750. Find the price of each.
Show solution
  1. \(7x + 6y = 3800\), \(3x + 5y = 1750\).
  2. \(x = \dfrac{1750 - 5y}{3}\); then \[\begin{aligned}&7(1750 - 5y) + 18y = 11400 \\ \Rightarrow\ &17y = 850\end{aligned}\]
  3. \(y = 50,\ x = 500\).
Answer: Bat ₹500, ball ₹50
(iv) A taxi charges a fixed amount plus a rate per km: ₹105 for 10 km and ₹155 for 15 km. Find the fixed charge and the rate. How much for 25 km?
Show solution
  1. \(x + 10y = 105\), \(x + 15y = 155\).
  2. \(5y = 50 \Rightarrow y = 10\), \(x = 5\).
  3. 25 km: \(5 + 250 = 255\).
Answer: Fixed ₹5, ₹10 per km; ₹255 for 25 km
(v) A fraction becomes \(\tfrac{9}{11}\) if 2 is added to both numerator and denominator, and \(\tfrac56\) if 3 is added to both.
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  1. \(11(x + 2) = 9(y + 2)\) and \(6(x + 3) = 5(y + 3)\): \(11x - 9y = -4\), \(6x - 5y = -3\).
  2. \(x = 7,\ y = 9\).
Answer: \(\tfrac79\)
(vi) In five years a father will be three times as old as his son; five years ago he was seven times as old.
Show solution
  1. \(x + 5 = 3(y + 5)\) and \(x - 5 = 7(y - 5)\): \(x - 3y = 10\), \(x - 7y = -30\).
  2. \(4y = 40 \Rightarrow y = 10\), \(x = 40\).
Answer: Father 40 years, son 10 years

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