# Ex.3.4 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

## Question

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 \(\begin{align}\frac{1}{2}\end{align}\) 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)**

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** Video Solution**

**Pair Of Linear Equations In Two Variables**

**Ex 3.4 | Question 2**

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## Text Solution

**What is the Known?**

Fraction becomes \(1\) , if \(1\) is added to the numerator and \(1\) is subtracted from the denominator, and fraction becomes \(\begin{align}\frac{1}{2}\end{align}\) if \(1\) is added to the denominator.

**What is the Unknown?**

The pair of linear equations and fractions

**Reasoning:**

Fractions has two parts numerator and denominator so assuming the numerator as \(x,\) and denominator as \( y, \)two linear equations can be formed for the known situation.

**Steps:**

Let the numerator \(= x\)

And the denominator \(= y\)

Then the fraction \(\begin{align} = \frac {{x}}{y} \end{align}\)

When \(1\) is added to the numerator and \(1\) is subtracted from the denominator;

\[\begin{align}\frac{{x + 1}}{{y - 1}} &= 1\\x + 1 &= y - 1\\x + 1 &= y - 1\\x - y + 1 + 1 &= 0\\x - y + 2 &= 0 \qquad \dots(1)\end{align}\]

When \(1\) is added to the denominator;

\[\begin{align}\frac{x}{{y + 1}} &= \frac{1}{2}\\2x &= y + 1\\2x - y - 1 &= 0 \qquad \dots(2)\end{align}\]

By subtracting equation (2) from equation (1)

\[\begin{align}\left( {x - y + 2} \right) - \left( {2x - y - 1} \right) &= 0\\x - y + 2 - 2x + y + 1 &= 0\\

- x + 3 &= 0\\x &= 3\end{align}\]

Substitute \(x = 3\) in equation (1)

\[\begin{align}3 - y + 2 &= 0\\y &= 5\end{align}\]

The equations are \(2x - y - 1 = 0\) and \(x - y + 2 = 0\) where the numerator of the fraction is \(x\), and denominator is \(y\) .

Fraction is \(\begin{align}\frac{3}{5}\end{align}\)

**(ii) **

**What is the Known?**

\(5\) years ago, Nuri was thrice as old as Sonu and \(10\) years later, Nuri will be twice as old as Sonu

**What is the Unknown?**

The pair of linear equations and ages of Nuri and Sonu.

**Reasoning:**

Assuming the present age of Nuri as *\(x\)* years and Sonu as \(y\) years,two linear equations can be formed for the Known Solutions.

**Steps:**

Let the present age of Nuri \(=x\) years

And the present age of Sonu \(=y\) years

\(5\) years ago,

Nuri's age \(= (x-5)\)years

Sonu's age \(= (y-5)\) years

\[\begin{align}x - 5 &= 3(y - 5)\\x - 5 &= 3y - 15\\x - 3y - 5 + 15 &= 0\\x - 3y + 10 &= 0 \qquad \dots(1)

\end{align}\]

\(10\) years later,

Nuri's age \(= (x +10 )\) years

Sonu's age \(= (y+10)\) years

\[\begin{align}x + 10 &= 2(y + 10)\\x + 10 &= 2y + 20\\x - 2y + 10 - 20 &= 0\\x - 2y - 10 &= 0 \qquad \dots(2)\end{align}\]

By subtracting equation \((2)\) from equation \((1)\)

\[\begin{align}\left( {x - 3y + 10} \right) - \left( {x - 2y - 10} \right) &= 0\\x - 3y + 10 - x + 2y + 10 &= 0\\

- y + 20 &= 0\\y &= 20\end{align}\]

Figure Substitute \(y = 20\) in equation \((1)\)

\[\begin{align}x - 3 \times 20 + 10 &= 0\\x - 60 + 10 &= 0\\x - 50 &= 0\\x &= 50\end{align}\]

Liner equations are \(x - 3y + 10 = 0\) and \(x - 2y - 10 = 0\) where the present age of Nuri is \(x\) and Sonu is \(y\)

Age of Nuri is \(50\) years.

Age of Sonu is \(20\) years.

**(iii) **

**What is the Known?**

Sum of digits of a two-digit number is \(9\) and nine times this number twice the number obtained by reversing the order of the digits.

**What is the Unknown?**

The pair of linear equations and two digit number.

**Reasoning:**

A two-digit number's form is \(10y+ x\) where \(y\) and \(x\) are ten's and one's digit respectively.

**Steps: **

Let the one's place \(=x\)

And the ten's place \(=y\)

Then the number \(=10y + x\)

Sum of the digits of the number;

\[x + y = 9 \qquad ...\,(1)\]

By reversing the order of the digits,the number \(=10y + x\)

Hence,

\[\begin{align}9\left( {10y + x} \right) &= 2\left( {10x + y} \right)\\90y + 9x &= 20x + 2y\\20x + 2y - 90y - 9x &= 0\\11x - 88y &= 0\\11\left( {x - 8y} \right) &= 0\\x - 8y &= 0 \qquad \dots (2)\end{align}\]

By subtracting equation \((2)\) from equation \((1)\)

\[\begin{align}\left( {x + y} \right) - \left( {x - 8y} \right) &= 9 - 0\\x + y - x + 8y &= 9\\

9y &= 9\\y &= 1\end{align}\]

Substitute \(y = 1\) in equation (1)

\[\begin{align}x + 1 &= 9\\x &= 9 - 1\\x& = 8\end{align}\]

Equations are \(x + y = 9\) and \(8y - y = 10\) where \(y\) and \(x\) are ten's and one's digit respectively.

The two-digit number is \(18.\)

**(iv) **

**What is the Known?**

Meena withdrew \(₹\,2000\) , got \(₹\, 50\) and \(₹\, 100\) notes only and \(25\) notes in all.

**What is the Unknown?**

The pair of linear equations and number of notes of \(₹\, 50\) and \(₹\, 100.\) each.

**Reasoning:**

Assuming the number of notes of \(₹\, 50\) as *\(x\)* and \(₹\, 100\) as \(y, \) two liner equations can be formed for the known Solutions.

**Steps:**

Let number of notes of \(₹\, 50 = x\)

and number of notes of \(₹\,100 = y\)

Meena got \(25\) notes in all;

\[x + y = 25 \qquad ...\,\left( 1 \right)\]

Meena withdrew \(₹\, 2000;\)

\[\begin{align}50x + 100y &= 2000\\50\left( {x + 2y} \right) &= 2000\\x + 2y &= \frac{{2000}}{{50}}\\

x + 2y &= 40 \qquad \dots(2)\end{align}\]

By subreacting equation (1) fromequation (2)

\[\begin{align}\left( {x + 2y} \right) - \left( {x + y} \right) &= 40 - 25\\x + 2y - x - y &= 15\\y &= 15\end{align}\]

Substitute \(y = 15\) in equation (1)

\[\begin{align}x + 15 &= 25\\x &= 10\end{align}\]

Equations are \(x + y = 25\) and \(x + 2y = 40\) where number of \(₹\,50\) and \(₹\,100\) notes are \(x\) and \(y\) respectively.

Number of \(₹\, 50\) notes is \(10\)

Number of \(₹\,100\) notes is \(15\)

**(v)**

**What is the Known?**

Saritha paid \(₹\,27\) for a book kept for \(7\) days while Sasy paid \(₹\,21\) for a book kept for \(5\) days, where fixed charges for first \(3\) days and an additional charge for each day thereafter.

**What is the Unknown?**

The pair of linear equations,fixed charge and charge for each extra day.

**Reasoning:**

Assuming fixed charges as \(₹\, x\) and additional charge for each extra day as \(₹\,y, \) two linear equations can be formed for the known situation

**Steps:**

Let the fixed charge \(= x\)

And charge per extra day \(= y\)

Saritha paid \(\rm Rs.27\) for a book kept for \(7\) days;

\[\begin{align}x + \left( {7 - 3} \right)y &= 27\\x + 4y &= 27 \qquad \dots\left( 1 \right)\end{align}\]

Susy paid \(₹\,21\) for a book kept for \(5\) days;

\[\begin{align}x + \left( {5 - 3} \right)y &= 21\\x + 2y &= 21 \qquad \dots \left( 2 \right)\end{align}\]

By subtracting equation \((2)\) from equation \((1)\)

\[\begin{align}\left( {x + 4y} \right) - \left( {x + 2y} \right) &= 27 - 21\\x + 4y - x - 2y &= 6\\2y &= 6\\y &= \frac{6}{2}\\y &= 3\end{align}\]

Substituting \(y = 3\) in equation \((3)\)

\[\begin{align}x + 4 \times 3 &= 27\\x + 12 &= 27\\x &= 27 - 12\\x &= 15\end{align}\]

Equations are \(x+ 2y = 21\) and \(x + 4y = 27\) where fixed charge is \(₹\,x\) and charge for each extra day is \(₹\, y.\)

Fixed charge is \(₹\,15\)

Charge for each extra day is \(₹\, 3\)