EXERCISE 3.3
Pair Of Linear Equations In Two Variables • 2 Questions
Question 1
Hint available
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) 2 1 and 3 2 3 3 x y y x
Key Idea
For a pair of linear equations in two variables, the elimination method involves making the coefficients of one variable equal (or opposite) in the two equations and then adding or subtracting the equations to eliminate that variable. The substitution method requires expressing one variable in terms of the other from one equation and substituting this expression into the second equation. Both methods lead to a single‑variable equation whose solution gives one variable; the other variable is then obtained by back‑substitution.
Step-by-Step Solution
### (i) \(x + y = 5\) and \(2x - 3y = 4\)
Elimination
1. Multiply the first equation by 2: \(2x + 2y = 10\).
2. Subtract the second equation from this result:
\[(2x + 2y) - (2x - 3y) = 10 - 4\]
\[5y = 6 \Rightarrow y = \frac{6}{5}\].
3. Substitute \(y\) in \(x + y = 5\):
\[x = 5 - \frac{6}{5} = \frac{19}{5}\].
Substitution
1. From the first equation, \(x = 5 - y\).
2. Substitute in the second equation:
\[2(5 - y) - 3y = 4 \Rightarrow 10 - 5y = 4 \Rightarrow 5y = 6 \Rightarrow y = \frac{6}{5}\].
3. Hence \(x = 5 - \frac{6}{5} = \frac{19}{5}\).
### (ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
Elimination
1. Multiply the second equation by 2: \(4x - 4y = 4\).
2. Add to the first equation:
\[(3x + 4y) + (4x - 4y) = 10 + 4\]
\[7x = 14 \Rightarrow x = 2\].
3. Substitute \(x\) in \(2x - 2y = 2\):
\[4 - 2y = 2 \Rightarrow 2y = 2 \Rightarrow y = 1\].
Substitution
1. From the second equation, \(x - y = 1\) so \(x = y + 1\).
2. Substitute in the first equation:
\[3(y + 1) + 4y = 10 \Rightarrow 7y + 3 = 10 \Rightarrow 7y = 7 \Rightarrow y = 1\].
3. Hence \(x = 1 + 1 = 2\).
### (iii) \(3x - 5y = 4\) and \(9x = 2y + 7\)
Elimination
1. Rewrite the second equation as \(9x - 2y = 7\).
2. Multiply the first equation by 3: \(9x - 15y = 12\).
3. Subtract the second equation from this result:
\[(9x - 15y) - (9x - 2y) = 12 - 7\]
\[-13y = 5 \Rightarrow y = -\frac{5}{13}\].
4. Substitute \(y\) in \(3x - 5y = 4\):
\[3x + \frac{25}{13} = 4 \Rightarrow 3x = \frac{27}{13} \Rightarrow x = \frac{9}{13}\].
Substitution
1. From the second equation, \(x = \frac{2y + 7}{9}\).
2. Substitute in the first equation:
\[3\left(\frac{2y + 7}{9}\right) - 5y = 4 \Rightarrow \frac{2y + 7}{3} - 5y = 4\].
3. Multiply by 3: \[2y + 7 - 15y = 12 \Rightarrow -13y = 5 \Rightarrow y = -\frac{5}{13}\].
4. Hence \(x = \frac{2(-5/13) + 7}{9} = \frac{9}{13}\).
### (iv) \(2x + y = 1\) and \(3x - 2y = 3\)
Elimination
1. Multiply the first equation by 2: \(4x + 2y = 2\).
2. Add to the second equation:
\[(3x - 2y) + (4x + 2y) = 3 + 2\]
\[7x = 5 \Rightarrow x = \frac{5}{7}\].
3. Substitute \(x\) in \(2x + y = 1\):
\[\frac{10}{7} + y = 1 \Rightarrow y = 1 - \frac{10}{7} = -\frac{3}{7}\].
Substitution
1. From the first equation, \(y = 1 - 2x\).
2. Substitute in the second equation:
\[3x - 2(1 - 2x) = 3 \Rightarrow 3x - 2 + 4x = 3 \Rightarrow 7x = 5 \Rightarrow x = \frac{5}{7}\].
3. Hence \(y = 1 - 2\left(\frac{5}{7}\right) = -\frac{3}{7}\).
Elimination
1. Multiply the first equation by 2: \(2x + 2y = 10\).
2. Subtract the second equation from this result:
\[(2x + 2y) - (2x - 3y) = 10 - 4\]
\[5y = 6 \Rightarrow y = \frac{6}{5}\].
3. Substitute \(y\) in \(x + y = 5\):
\[x = 5 - \frac{6}{5} = \frac{19}{5}\].
Substitution
1. From the first equation, \(x = 5 - y\).
2. Substitute in the second equation:
\[2(5 - y) - 3y = 4 \Rightarrow 10 - 5y = 4 \Rightarrow 5y = 6 \Rightarrow y = \frac{6}{5}\].
3. Hence \(x = 5 - \frac{6}{5} = \frac{19}{5}\).
### (ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)
Elimination
1. Multiply the second equation by 2: \(4x - 4y = 4\).
2. Add to the first equation:
\[(3x + 4y) + (4x - 4y) = 10 + 4\]
\[7x = 14 \Rightarrow x = 2\].
3. Substitute \(x\) in \(2x - 2y = 2\):
\[4 - 2y = 2 \Rightarrow 2y = 2 \Rightarrow y = 1\].
Substitution
1. From the second equation, \(x - y = 1\) so \(x = y + 1\).
2. Substitute in the first equation:
\[3(y + 1) + 4y = 10 \Rightarrow 7y + 3 = 10 \Rightarrow 7y = 7 \Rightarrow y = 1\].
3. Hence \(x = 1 + 1 = 2\).
### (iii) \(3x - 5y = 4\) and \(9x = 2y + 7\)
Elimination
1. Rewrite the second equation as \(9x - 2y = 7\).
2. Multiply the first equation by 3: \(9x - 15y = 12\).
3. Subtract the second equation from this result:
\[(9x - 15y) - (9x - 2y) = 12 - 7\]
\[-13y = 5 \Rightarrow y = -\frac{5}{13}\].
4. Substitute \(y\) in \(3x - 5y = 4\):
\[3x + \frac{25}{13} = 4 \Rightarrow 3x = \frac{27}{13} \Rightarrow x = \frac{9}{13}\].
Substitution
1. From the second equation, \(x = \frac{2y + 7}{9}\).
2. Substitute in the first equation:
\[3\left(\frac{2y + 7}{9}\right) - 5y = 4 \Rightarrow \frac{2y + 7}{3} - 5y = 4\].
3. Multiply by 3: \[2y + 7 - 15y = 12 \Rightarrow -13y = 5 \Rightarrow y = -\frac{5}{13}\].
4. Hence \(x = \frac{2(-5/13) + 7}{9} = \frac{9}{13}\).
### (iv) \(2x + y = 1\) and \(3x - 2y = 3\)
Elimination
1. Multiply the first equation by 2: \(4x + 2y = 2\).
2. Add to the second equation:
\[(3x - 2y) + (4x + 2y) = 3 + 2\]
\[7x = 5 \Rightarrow x = \frac{5}{7}\].
3. Substitute \(x\) in \(2x + y = 1\):
\[\frac{10}{7} + y = 1 \Rightarrow y = 1 - \frac{10}{7} = -\frac{3}{7}\].
Substitution
1. From the first equation, \(y = 1 - 2x\).
2. Substitute in the second equation:
\[3x - 2(1 - 2x) = 3 \Rightarrow 3x - 2 + 4x = 3 \Rightarrow 7x = 5 \Rightarrow x = \frac{5}{7}\].
3. Hence \(y = 1 - 2\left(\frac{5}{7}\right) = -\frac{3}{7}\).
Question 2
Hint available
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 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. 37 (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.
Key Idea
Translate the word‑problem statements into a pair of linear equations in two unknowns, then solve the system using the elimination method as taught in NCERT.
Step-by-Step Solution
### (i) Fraction \(\dfrac{x}{y}\)
1. Adding 1 to numerator and subtracting 1 from denominator gives 1:
$$\frac{x+1}{y-1}=1 \;\Rightarrow\; x+1=y-1 \;\Rightarrow\; y=x+2 \tag{1}$$
2. Adding 1 only to denominator gives \(\frac12\):
$$\frac{x}{y+1}=\frac12 \;\Rightarrow\; 2x=y+1 \;\Rightarrow\; y=2x-1 \tag{2}$$
3. Eliminate \(y\) by equating (1) and (2):
$$x+2=2x-1 \;\Rightarrow\; x=3,\; y=5.$$ Hence the fraction is \(\frac{3}{5}\).
### (ii) Ages of Nuri (\(N\)) and Sonu (\(S\))
1. Five years ago: \(N-5 = 3(S-5)\)
$$N-5 = 3S-15 \;\Rightarrow\; N-3S = -10 \tag{1}$$
2. Ten years later: \(N+10 = 2(S+10)\)
$$N+10 = 2S+20 \;\Rightarrow\; N-2S = 10 \tag{2}$$
3. Eliminate \(N\): subtract (1) from (2):
$$(N-2S)-(N-3S)=10-(-10) \;\Rightarrow\; S = 20.$$ Substitute in (2):
$$N-2(20)=10 \;\Rightarrow\; N=50.$$ So Nuri is 50 years old and Sonu is 20 years old.
### (iii) Two‑digit number \(10t+u\)
1. Sum of digits: \(t+u=9\) \(\tag{1}\)
2. Nine times the number equals twice the reversed number:
$$9(10t+u)=2(10u+t)$$
$$90t+9u=20u+2t \;\Rightarrow\; 88t=11u \;\Rightarrow\; 8t=u \tag{2}$$
3. Using (1) and (2): \(t+8t=9 \Rightarrow 9t=9 \Rightarrow t=1\), then \(u=8\).
The required number is \(10\times1+8=18\).
### (iv) Notes of \(\Rs 50\) (\(a\)) and \(\Rs 100\) (\(b\))
1. Total amount: \(50a+100b=2000\) \(\tag{1}\)
2. Total notes: \(a+b=25\) \(\tag{2}\)
3. From (2), \(a=25-b\). Substitute in (1):
$$50(25-b)+100b=2000$$
$$1250-50b+100b=2000 \;\Rightarrow\; 50b=750 \;\Rightarrow\; b=15,$$
$$a=25-15=10.$$ Hence 10 notes of \(\Rs 50\) and 15 notes of \(\Rs 100\).
### (v) Library charge: fixed charge \(F\) for first three days, extra‑day charge \(d\).
1. For 7 days (extra days = 4): \(F+4d=27\) \(\tag{1}\)
2. For 5 days (extra days = 2): \(F+2d=21\) \(\tag{2}\)
3. Eliminate \(F\): subtract (2) from (1):
$$(F+4d)-(F+2d)=27-21 \;\Rightarrow\; 2d=6 \;\Rightarrow\; d=3\) rupees.
Substitute back in (2): \(F+2(3)=21 \;\Rightarrow\; F=15\) rupees.
All solutions have been obtained by the elimination method as required.
1. Adding 1 to numerator and subtracting 1 from denominator gives 1:
$$\frac{x+1}{y-1}=1 \;\Rightarrow\; x+1=y-1 \;\Rightarrow\; y=x+2 \tag{1}$$
2. Adding 1 only to denominator gives \(\frac12\):
$$\frac{x}{y+1}=\frac12 \;\Rightarrow\; 2x=y+1 \;\Rightarrow\; y=2x-1 \tag{2}$$
3. Eliminate \(y\) by equating (1) and (2):
$$x+2=2x-1 \;\Rightarrow\; x=3,\; y=5.$$ Hence the fraction is \(\frac{3}{5}\).
### (ii) Ages of Nuri (\(N\)) and Sonu (\(S\))
1. Five years ago: \(N-5 = 3(S-5)\)
$$N-5 = 3S-15 \;\Rightarrow\; N-3S = -10 \tag{1}$$
2. Ten years later: \(N+10 = 2(S+10)\)
$$N+10 = 2S+20 \;\Rightarrow\; N-2S = 10 \tag{2}$$
3. Eliminate \(N\): subtract (1) from (2):
$$(N-2S)-(N-3S)=10-(-10) \;\Rightarrow\; S = 20.$$ Substitute in (2):
$$N-2(20)=10 \;\Rightarrow\; N=50.$$ So Nuri is 50 years old and Sonu is 20 years old.
### (iii) Two‑digit number \(10t+u\)
1. Sum of digits: \(t+u=9\) \(\tag{1}\)
2. Nine times the number equals twice the reversed number:
$$9(10t+u)=2(10u+t)$$
$$90t+9u=20u+2t \;\Rightarrow\; 88t=11u \;\Rightarrow\; 8t=u \tag{2}$$
3. Using (1) and (2): \(t+8t=9 \Rightarrow 9t=9 \Rightarrow t=1\), then \(u=8\).
The required number is \(10\times1+8=18\).
### (iv) Notes of \(\Rs 50\) (\(a\)) and \(\Rs 100\) (\(b\))
1. Total amount: \(50a+100b=2000\) \(\tag{1}\)
2. Total notes: \(a+b=25\) \(\tag{2}\)
3. From (2), \(a=25-b\). Substitute in (1):
$$50(25-b)+100b=2000$$
$$1250-50b+100b=2000 \;\Rightarrow\; 50b=750 \;\Rightarrow\; b=15,$$
$$a=25-15=10.$$ Hence 10 notes of \(\Rs 50\) and 15 notes of \(\Rs 100\).
### (v) Library charge: fixed charge \(F\) for first three days, extra‑day charge \(d\).
1. For 7 days (extra days = 4): \(F+4d=27\) \(\tag{1}\)
2. For 5 days (extra days = 2): \(F+2d=21\) \(\tag{2}\)
3. Eliminate \(F\): subtract (2) from (1):
$$(F+4d)-(F+2d)=27-21 \;\Rightarrow\; 2d=6 \;\Rightarrow\; d=3\) rupees.
Substitute back in (2): \(F+2(3)=21 \;\Rightarrow\; F=15\) rupees.
All solutions have been obtained by the elimination method as required.