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) \(\dfrac{x}{2} + \dfrac{2y}{3} = -1\) and \(x - \dfrac{y}{3} = 3\)
(i) \(x + y = 5\), \(2x - 3y = 4\)
From \(x + y = 5\), express \(x\) as \(x = 5 - y\). Substitute in \(2x - 3y = 4\):
\[2(5 - y) - 3y = 4 \Rightarrow 10 - 2y - 3y = 4 \Rightarrow 10 - 5y = 4.\]
\[-5y = -6 \Rightarrow y = \frac{6}{5}.\]
Then \(x = 5 - y = 5 - \frac{6}{5} = \frac{25 - 6}{5} = \frac{19}{5}.\)
Solution: \(x = \dfrac{19}{5},\ y = \dfrac{6}{5}.\)
(ii) \(3x + 4y = 10\), \(2x - 2y = 2\)
From \(2x - 2y = 2\), divide by 2:
\[x - y = 1 \Rightarrow x = y + 1.\]
Substitute in \(3x + 4y = 10\):
\[3(y + 1) + 4y = 10 \Rightarrow 3y + 3 + 4y = 10 \Rightarrow 7y + 3 = 10.\]
\[7y = 7 \Rightarrow y = 1, \quad x = y + 1 = 2.\]
Solution: \(x = 2,\ y = 1.\)
(iii) \(3x - 5y - 4 = 0\) and \(9x = 2y + 7\)
Rewrite the equations:
\[3x - 5y = 4, \quad 9x - 2y = 7.\]
Multiply the first equation by 2:
\[6x - 10y = 8.\]
Multiply the second equation by 5:
\[45x - 10y = 35.\]
Subtract the first (multiplied) equation from the second:
\[(45x - 10y) - (6x - 10y) = 35 - 8 \Rightarrow 39x = 27.\]
\[x = \frac{27}{39} = \frac{9}{13}.\]
Substitute in \(3x - 5y = 4\):
\[3 \cdot \frac{9}{13} - 5y = 4 \Rightarrow \frac{27}{13} - 5y = 4.\]
\[-5y = 4 - \frac{27}{13} = \frac{52 - 27}{13} = \frac{25}{13} \Rightarrow y = -\frac{25}{65} = -\frac{5}{13}.\]
Solution: \(x = \dfrac{9}{13},\ y = -\dfrac{5}{13}.\)
(iv) \(\dfrac{x}{2} + \dfrac{2y}{3} = -1\), \(x - \dfrac{y}{3} = 3\)
Clear denominators by multiplying each equation by 6:
First equation:
\[6\left(\frac{x}{2} + \frac{2y}{3}\right) = 6(-1) \Rightarrow 3x + 4y = -6.\]
Second equation:
\[6\left(x - \frac{y}{3}\right) = 6 \cdot 3 \Rightarrow 6x - 2y = 18.\]
Divide the second equation by 2 to simplify:
\[3x - y = 9.\]
Now solve the pair \(3x + 4y = -6\) and \(3x - y = 9\) by elimination. Subtract the second from the first:
\[(3x + 4y) - (3x - y) = -6 - 9 \Rightarrow 5y = -15 \Rightarrow y = -3.\]
Substitute in \(3x - y = 9\):
\[3x - (-3) = 9 \Rightarrow 3x + 3 = 9 \Rightarrow 3x = 6 \Rightarrow x = 2.\]
Solution: \(x = 2,\ y = -3.\)
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 \(\dfrac{1}{2}\) 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.
(i) Fraction
Let the fraction be \(\dfrac{x}{y}\), where \(x\) is the numerator and \(y\) is the denominator.
Add 1 to the numerator and subtract 1 from the denominator; the fraction becomes 1:
\[\frac{x + 1}{y - 1} = 1 \Rightarrow x + 1 = y - 1 \Rightarrow x - y + 2 = 0.\]
If only 1 is added to the denominator, the fraction becomes \(\dfrac{1}{2}\):
\[\frac{x}{y + 1} = \frac{1}{2} \Rightarrow 2x = y + 1 \Rightarrow 2x - y - 1 = 0.\]
Now solve the system
\[x - y + 2 = 0, \quad 2x - y - 1 = 0.\]
Subtract the first from the second:
\[(2x - y - 1) - (x - y + 2) = 0 \Rightarrow x - 3 = 0 \Rightarrow x = 3.\]
Then \(x - y + 2 = 0 \Rightarrow 3 - y + 2 = 0 \Rightarrow 5 - y = 0 \Rightarrow y = 5.\]
The fraction is \(\dfrac{3}{5}.\)
(ii) Ages of Nuri and Sonu
Let the present ages (in years) of Nuri and Sonu be \(x\) and \(y\) respectively.
Five years ago, their ages were \(x - 5\) and \(y - 5\). At that time, Nuri was thrice as old as Sonu:
\[x - 5 = 3(y - 5) \Rightarrow x - 5 = 3y - 15 \Rightarrow x - 3y + 10 = 0.\]
Ten years later, their ages will be \(x + 10\) and \(y + 10\). Then Nuri will be twice as old as Sonu:
\[x + 10 = 2(y + 10) \Rightarrow x + 10 = 2y + 20 \Rightarrow x - 2y - 10 = 0.\]
Solve
\[x - 3y + 10 = 0, \quad x - 2y - 10 = 0.\]
Subtract the second from the first:
\[(x - 3y + 10) - (x - 2y - 10) = 0 \Rightarrow -y + 20 = 0 \Rightarrow y = 20.\]
Then \(x - 2y - 10 = 0 \Rightarrow x - 40 - 10 = 0 \Rightarrow x = 50.\]
Nuri is 50 years old and Sonu is 20 years old.
(iii) Two-digit number
Let the tens digit be \(x\) and the units digit be \(y\). The number is \(10x + y\).
The sum of the digits is 9:
\[x + y = 9.\]
The number formed by reversing the digits is \(10y + x\). Nine times the original number is twice the reversed number:
\[9(10x + y) = 2(10y + x).\]
Simplify:
\[90x + 9y = 20y + 2x \Rightarrow 90x - 2x + 9y - 20y = 0 \Rightarrow 88x - 11y = 0.\]
Divide by 11:
\[8x - y = 0.\]
Now solve
\[x + y = 9, \quad 8x - y = 0.\]
Add the equations:
\[(x + y) + (8x - y) = 9 + 0 \Rightarrow 9x = 9 \Rightarrow x = 1.\]
Then \(y = 8x = 8.\)
The number is \(18\).
(iv) Notes of ₹ 50 and ₹ 100
Let \(x\) be the number of ₹ 50 notes and \(y\) the number of ₹ 100 notes.
Total money is ₹ 2000:
\[50x + 100y = 2000.\]
Total number of notes is 25:
\[x + y = 25.\]
Divide the money equation by 50:
\[x + 2y = 40.\]
Now solve
\[x + 2y = 40, \quad x + y = 25.\]
Subtract the second from the first:
\[(x + 2y) - (x + y) = 40 - 25 \Rightarrow y = 15.\]
Then \(x + y = 25 \Rightarrow x + 15 = 25 \Rightarrow x = 10.\]
Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
(v) Library charges
Let \(x\) be the fixed charge (in ₹) for the first three days and \(y\) be the additional charge (in ₹) per extra day.
Saritha kept the book for 7 days, that is 3 days fixed + 4 extra days:
\[x + 4y = 27.\]
Susy kept the book for 5 days, that is 3 days fixed + 2 extra days:
\[x + 2y = 21.\]
Subtract the second from the first:
\[(x + 4y) - (x + 2y) = 27 - 21 \Rightarrow 2y = 6 \Rightarrow y = 3.\]
Then \(x + 2y = 21 \Rightarrow x + 6 = 21 \Rightarrow x = 15.\]
The fixed charge is ₹ 15 and the charge for each extra day is ₹ 3.