Mathematics · Pair of Linear Equations in Two Variables · NCERT Exercises

Exercise 3.3

Complete, independently verified solutions for NCERT Exercise 3.3.

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Mathematics · Chapter 3 · NCERT Exercises

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Question 1

elimination method · substitution method

Solve each pair by elimination and substitution.

  1. x+y=5; 2x−3y=4x+y=5;\ 2x-3y=4
  2. 3x+4y=10; 2x−2y=23x+4y=10;\ 2x-2y=2
  3. 3x−5y−4=0; 9x=2y+73x-5y-4=0;\ 9x=2y+7
  4. x2+2y3=−1; x−y3=3\frac{x}{2}+\frac{2y}{3}=-1;\ x-\frac{y}{3}=3

Solution

  1. x=5−y⇒y=65, x=195x=5-y\Rightarrow y=\frac65,\ x=\frac{19}{5}
  2. x=y+1⇒y=1, x=2x=y+1\Rightarrow y=1,\ x=2
  3. 3x−5y=4, 9x−2y=7⇒y=−513, x=9133x-5y=4,\ 9x-2y=7\Rightarrow y=-\frac5{13},\ x=\frac9{13}
  4. 3x+4y=−6, 3x−y=9⇒y=−3, x=23x+4y=-6,\ 3x-y=9\Rightarrow y=-3,\ x=2

Substitution of each ordered pair verifies both original equations.

Question

Question 2

elimination method · word problems

Form equations and solve by elimination.

  1. Adding 1 to a fraction’s numerator and subtracting 1 from its denominator makes 1; adding only 1 to the denominator makes 1/2.

  2. Five years ago Nuri was three times Sonu; ten years later Nuri will be twice Sonu.

  3. A two-digit number has digit sum 9; nine times it is twice its reversal.

  4. ₹2000 is withdrawn in ₹50 and ₹100 notes, 25 notes total.

  5. A library charges a fixed amount for three days plus a daily amount thereafter; seven days cost ₹27 and five days ₹21.

Solution

  1. x+1y−1=1, xy+1=12⇒x−y=−2, 2x−y=1\frac{x+1}{y-1}=1,\ \frac{x}{y+1}=\frac12\Rightarrow x-y=-2,\ 2x-y=1
    (x,y)=(3,5)(x,y)=(3,5)
  2. N−3S=−10, N−2S=10⇒(N,S)=(50,20)N-3S=-10,\ N-2S=10\Rightarrow(N,S)=(50,20)
  3. x+y=9, 88x−11y=0⇒(x,y)=(1,8)x+y=9,\ 88x-11y=0\Rightarrow(x,y)=(1,8)
  4. x+y=25, x+2y=40⇒(x,y)=(10,15)x+y=25,\ x+2y=40\Rightarrow(x,y)=(10,15)
  5. f+4d=27, f+2d=21⇒(f,d)=(15,3)f+4d=27,\ f+2d=21\Rightarrow(f,d)=(15,3)

Final answer

Answers: 3/5; Nuri 50 and Sonu 20; 18; ten ₹50 notes and fifteen ₹100 notes; ₹15 fixed and ₹3 per extra day.