# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.4

Go back to  'Pair of Linear Equations in Two Variables'

## Question 1

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)

\begin{align}\frac{x}{2} + \frac{{2y}}{3} = - 1\end{align} and \begin{align}x - \frac{y}{3} = 3 \end{align}

### Solution

What is the Unknown?

Solution for the linear pair of equations.

Reasoning:

Substitution method:

Pick either of the equations and write one variable in terms of the other then substitute the value of the obtained variable in other equation to solve.

Elimination method:

First multiply one or both the equations by some suitable non-zero constants to make the coefficients of one variable numerically equal then add or subtract one equation from the other so that one variable gets eliminated

(i) What is the Known?

$$x + y = 5 \\2x - 3 y= 4$$

Steps:

Elimination method:

\begin{align}x + y = 5 \qquad \ldots \left( 1 \right)\\2x - 3 y= 4 \qquad \ldots \left( 2 \right)\end{align}

Multiplying equation $$(1)$$ by $$2$$

$\begin{array}{l} \left[ {x + {\rm{ }}y = 5} \right] \times 2\\ 2x + 2y = 10 \qquad \ldots \left( 3 \right) \end{array}$

By subtracting equation $$(2)$$ from equation $$(3)$$

\begin{align}\left( {2x + 2y} \right) - \left( {2x - 3y} \right)& = 10 - 4\\2x + 2y - 2x + 3y &= 6\\ 5y &= 6\\y &= \frac{6}{5}\end{align}

Substituteing \begin{align}y = \frac {{6}}{5}\end{align} in equation (1)

\begin{align}x + \frac{6}{5} &= 5\\x &= 5 - \frac{6}{5}\\x &= \frac{{25 - 6}}{5}\\x &= \frac{{19}}{5}\end{align}

Substitution method:

\begin{align}x + y &= 5 \qquad \ldots \left( 1 \right)\\2x - 3 y&= 4 \qquad \ldots \left( 2 \right)\end{align}

By solving equation $$(1)$$

\begin{align}x + {\rm{ }}y &= 5\\y &= 5 - x \qquad \ldots \left( 3 \right)\end{align}

Substitute \begin{align}y = 5 - x\end{align} in equation $$(2)$$

\begin{align}2x - 3(5 - x) &= 4\\2x - 15 + 3x &= 4\\5x &= 4 + 15\\x &= \frac{{19}}{5}\end{align}

Substituting \begin{align}x = \frac{{19}}{5}\end{align} in equation $$(3)$$

\begin{align}y &= 5 - \frac{{19}}{5}\\y &= \frac{{25 - 19}}{5}\\y &= \frac{6}{5}\end{align}

The answer is,\begin{align}x = \frac{{19}}{5} \quad \text{and} \quad y = \frac{6}{5}\end{align}

(ii) What is the Known?

\begin{align}3x+4y&=10\\2x-2y&=2\end{align}

Steps:

Elimination method:

\begin{align}3x+4y&=10…………(1)\\2x-2y&=2…………. (2)\end{align}

Multiplying equation $$(2)$$ by $$2$$

\begin{align}&\left[ {2x - 2y = 2} \right] \times 2\\&4x - 4y = 4 \qquad \ldots \left( 3 \right)\end{align}

By adding equation $$(1)$$ and equation $$(3)$$

\begin{align}\left( {3x + 4y} \right) + \left( {4x - 4y} \right) &= 10 + {4^{}}\\3x + 4y + 4x - 4y &= 14\\ 7x &= 14\\x &= \frac{{14}}{7}\\x &= 2\end{align}

Substituting $$x = 2$$ in equation $$(2)$$

\begin{align}2 x\, 2 - 2y &= 2\\4 - 2y &= 2\\2y &= 4 - 2\\y &= \frac{2}{2}\\y &= 1\end{align}

Substitution method:

\begin{align}3x+4y&=10 \quad \dots(1)\\2x-2y&=2\quad \;\;\dots(2)\end{align}

By solving equation $$(1)$$

\begin{align}3x + 4y & = 10\\4y &= 10 - 3x\\y &= \frac{{10 - 3x}}{4} \qquad \ldots \left( 3 \right) \end{align}

Substituteing \begin{align}y = \frac{{10 - 3x}}{4}\end{align} in equation $$(2)$$

\begin{align}2x - 2\left( {\frac{{10 - 3x}}{4}} \right) &= 2\\\frac{{4x - 10 + 3x}}{2} &= 2\\4x - 10 + 3x &= 4\\7x &= 4 + 10\\x &= \frac{{14}}{7}\\x &= 2\end{align}

Substituting \begin{align}x = 2\end{align} in equation (3)

\begin{align}y &= \frac{{10 - 3 \times 2}}{4}\\y &= \frac{{10 - 6}}{4}\\y &= \frac{4}{4}\\y &= 1\end{align}

The answer is,\begin{align}x = 2 \quad \text{and} \quad y = 1\end{align}

(iii) What is the Known?

\begin{align}3x - 5y - 4 &= 0\\9x &= 2y + 7\end{align}

Steps:

Elimination method:

\begin{align}3x - 5y - 4 &= 0 \qquad \qquad \ldots \left( 1 \right)\\9x &= 2y + 7 \qquad \ldots \left( 2 \right)\end{align}

Multiplying equation $$(1)$$ by $$3$$

\begin{align}&\left[ {3x - 5y - 4 = 0} \right] \times 3\\&9x - 15y - 12 = 0 \qquad \ldots \left( 3 \right) \end{align}

By solving equation $$(2)$$

$9x - 2y - 7 = 0 \qquad \ldots \left( 4 \right)$

By subtracting equation $$(4)$$ from equation $$(3)$$

\begin{align}\left( {9x \! - \! 15y \! - \! 12} \right) \! - \left( {9x \! - \! 2y \! - \! 7} \right) & \! = \! 0\\9x \! - \! 15y \! - \! 12 \! - \! 9x \! + \! 2y \! + \! 7 & \! = \! 0\\ - 13y \! - \! 5 & \! = \! 0\\ - 13y & \! = \! 5\\y & \! = \! - \! \frac{5}{{13}}\end{align}

Substituteting \begin{align} y = {-\frac {{5}}{13}}\end{align} in equation $$(2)$$

\begin{align}9x &= 2\left( { - \frac{5}{{13}}} \right) + 7\\9x &= \frac{{ - 10 + 91}}{{13}}\\ x &= \frac{{81}}{{13}} \times \frac{1}{9}\\x& = \frac{9}{{13}}\end{align}

Substitution method:

\begin{align}&3x - 5y - 4 = 0 \qquad \ldots \left( 1 \right)\\&9x = 2y + 7 \qquad \qquad \ldots \left( 2 \right)\end{align}

By solving equation $$(1)$$

\begin{align}3x - 5y - 4 &= 0\\5y &= 3x - 4\\y &= \frac{{3x - 4}}{5} \qquad \ldots \left( 3 \right)\end{align}

Substituteing \begin{align}y = \frac {{3x - 4}}{5} \end{align} in equation $$(2)$$

\begin{align}9x &= 2\left( {\frac{{3x - 4}}{5}} \right) + 7\\9x &= \frac{{6x - 8 + 35}}{5}\\ 45x &= 6x + 27\\45x - 6x &= 27\\39x &= 27\\x &= \frac{{27}}{{39}}\\x& = \frac{9}{{13}}\end{align}

Substitute \begin{align} x = \frac {{9}}{13}\end{align} in equation$$(3)$$

\begin{align}y &=\frac{{3\left( {\frac{9}{{13}}} \right) - 4}}{5}\\y& = \left( {\frac{{27 - 52}}{{13}}} \right) \times \frac{1}{5}\\y &= - \frac{{25}}{{13}} \times \frac{1}{5}\\y &= - \frac{5}{{13}}\end{align}

The answer is,\begin{align}y = \frac{{ - 5}}{{13}} \quad \text{and} \quad x = \frac{9}{{13}}\end{align}

(iv) What is the Known?

\begin{align}\frac{x}{2} + \frac{{2y}}{3} &= - 1\\x - \frac{y}{3} &= 3\end{align}

Steps:

Elimination method:

\begin{align}\frac{x}{2} + \frac{{2y}}{3} &= - 1 \qquad \dots(1)\\x - \frac{y}{3}& = 3 \qquad\;\;\; \dots(2)\end{align}

Multiplying equation $$(1)$$ by $$6$$ and equation $$(2)$$ by $$3$$

\begin{align}\left[ {\frac{x}{2} + \frac{{2y}}{3} = - 1} \right] \times 6\\ 3x + 4y = - 6& \qquad \dots (3)\\\\\left[ {x - \frac{y}{3} = 3} \right] \times 3\\3x - y = 9 &\qquad \dots(4) \end{align}

By subtracting equation $$(4)$$ from equation $$(3)$$

\begin{align}\left( {3x + 4y} \right) - \left( {3x - y} \right) &= - 6 - 9\\3x + 4y - 3x + y &= - 15\\5y &= - 15\\y &= - \frac{{15}}{5}\\y &= - 3\end{align}

Substitute $$y = - 3$$ in equation (2)

\begin{align}x - \frac{{ - 3}}{3} &= 3\\x + 1 &= 3\\x &= 3 - 1\\x &= 2\end{align}

Substitution method:

\begin{align}\frac{x}{2} + \frac{{2y}}{3} &= - 1 \qquad \dots(1)\\x - \frac{y}{3} &= 3 \qquad \dots(2) \end{align}

By solving equation $$(2)$$

\begin{align}x - \frac{y}{3} &= 3\\x &= \frac{y}{3} + 3\\x &= \frac{{y + 9}}{3} \qquad \dots(3)\end{align}

Substituteing \begin{align}x = \frac {{y+9}}{3} \end{align} in equation (1)

\begin{align}\frac{1}{2}\left( {\frac{{y + 9}}{3}} \right) + \frac{{2y}}{3} &= - 1\\\frac{{y + 9 + 4y}}{6} &= - 1\\5y + 9 &= - 6\\5y &= - 6 - 9\\y &= \frac{{ - 15}}{5}\\y &= - 3\end{align}

Substituiting $$y = -3$$ in equation (3)

\begin{align}x &= \frac{{ - 3 + 9}}{3}\\x &= \frac{6}{3}\\x &= 2\end{align}

The answer is,\begin{align}x = 2 \quad \text{and} \quad y = - 3\end{align}

## Question 2

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.

### Solution

(i)

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 $$₹\, 3$$