# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3

Pair of Linear Equations in Two Variables

## Question 1

Solve the following pair of linear equations by the substitution method.

(i)

\(\begin{align}x + y &= 14\\x-y &= 4\end{align}\)

(ii)

\(\begin{align}s-t &= 3\\\frac{s}{3} + \frac{t}{2} &= 6\end{align}\)

(iii)

\(\begin{align}x-y = 3\\9x-3y = 9\end{align}\)

(iv)

\(\begin{align}0.2x + 0.3y& = 1.3\\0.4x + 0.5y &= 2.3\end{align}\)

(v)

\(\begin{align}\sqrt 2 x + \sqrt 3 y = 0\\\sqrt 3 x - \sqrt {8y} = 0\end{align}\)

(vi)

\(\begin{align}\frac{{3x}}{2} - \frac{{5y}}{3} &= - 2\\\frac{x}{3} + \frac{y}{2} &= \frac{{13}}{6} \end{align}\)

### Solution

**Video Solution**

**What is Known?**

Pair of linear equations.

**What is Unknown?**

Solution for the given pair of linear equations.

**Reasoning:**

Pick any one of two equations, write one variable in terms of other. Now substituting this in other equation will result in one variable equation and easy to solve.

** (i) Steps:**

\[\begin{align}x + y &= 14 \qquad \dots(1)\\x - y &= 4 \qquad \;\;\dots(2)\end{align}\]

By solving the equation (1)

\[y = 14 - x \qquad \dots(3)\]

Substitute \(y = 14 - x\) in equation (2), we get

\[\begin{align}x - \left( {14 - x} \right) &= 4\\2x - 14 &= 4\\2x &= 4 + 14\\2x &= 18\\x &= 9\end{align}\]

Substituting \(x = 9\) in equation (3), we get

\[\begin{align}y &= 14 - 9\\y &= 5\end{align}\]

The answer is

\[\begin{align}x &= 9\\y &= 5\end{align}\]

**(ii) Steps:**

\[\begin{align}s - t &= 3 \qquad \dots(1)\\\frac{s}{3} + \frac{t}{2} &= 6 \qquad \dots(2)\end{align}\]

By solving the equation (1)

\[\begin{align}s - t &= 3\\s &= 3 + t \qquad \dots(3)\end{align}\]

Substitute \(s = 3 + t\) in equation (2), we get

\[\begin{align}\frac{{3 + t}}{3} + \frac{t}{2} &= 6\\\frac{{6 + 2t + 3t}}{6} &= 6\\6 + 5t &= 6 \times 6\\5t &= 36 - 6\\t &= \frac{{30}}{5}\\t &= 6\end{align}\]

Substituting \(t = 6\) in equation (3), we get

\[\begin{align}s &= 3 + 6\\s &= 9\end{align}\]

The answer is

\[\begin{align}s = 9\\ t= 6\end{align}\]

**(iii) Steps:**

\[\begin{align}3x - y &= 3 \qquad \dots(1)\\9x - 3y &= 9 \qquad \dots(2)\end{align}\]

By solving the equation (1)

\[\begin{align}3x - y &= 3\\y &= 3x - 3 \qquad \dots(1)\end{align}\]

Substitute \(y = 3x - 3\) in equation (2), we get

\[\begin{align}9x - 3(3x - 3) &= 9\\9x - 9x + 9 &= 9\\9& = 9\end{align}\]

Shows that the lines are coincident and having infinitely many solutions.

The answer is \(y = 3x - 3\)

Where *\(x\)* can take any value. i.e. Infinitely many Solutions.

**(iv) Steps:**

\[\begin{align}0.2x + 0.3y &= 1.3 \qquad \dots(1)\\0.4x + 0.5y &= 2.3 \qquad \dots(2)\end{align}\]

Multiply both the equations (1) and (2) by \(10,\) to remove the decimal number and making it easier for calculation.

\[\begin{align}&[ 0.2x + 0.3y = 1.3 ] \times 10\\& \Rightarrow 2x + 3y = 13 \;\;\;\;\qquad \dots(3)\\\\&\left[ {0.4x + 0.5y = 23} \right] \times (10)\\& \Rightarrow 4x + 5y = 23\;\;\;\;\qquad \dots(4)\end{align}\]

By solving the equation (3)

\[\begin{align}2x + 3y &= 13\\3y &= 13 - 2x\\y &= \frac{{13 - 2x}}{3} \qquad \dots(5)\end{align}\]

Substitute \(\begin{align} y = \frac{{13 - 2x}}{3} \end{align}\) in equation (4), we get

\[\begin{align}4x + 5\left( {\frac{{13 - 2x}}{3}} \right) &= 23\\\frac{{12x + 65 - 10x}}{3} &= 23\\

2x + 65 &= 23 \times 3\\2x &= 69 - 65\\x &= \frac{4}{2}\\x &= 2\end{align}\]

Substituting \(x = 2\) in equation (5), we get

\[\begin{align}y &= \frac{{13 - 2 \times 2}}{3}\\y &= \frac{9}{3}\\y &= 3\end{align}\]

The answer is

\[\begin{align}x = 2\\y = 3\end{align}\]

**(v) Steps:**

\[\begin{align}\sqrt 2 x + \sqrt 3 y &= 0\qquad \dots(1)\\\sqrt 3 x - \sqrt 8 y &= 0 \qquad \dots(2)\end{align}\]

By solving the equation (1)

\[\begin{align}\sqrt 2 x + \sqrt 3 y &= 0\\\sqrt 3 y& = - \sqrt 2 x\\y &= - \frac{{\sqrt 2 x}}{3} \qquad \dots(3)\end{align}\]

Substitute \(\begin{align}y = - \frac{{\sqrt 2 x}}{3} \end{align}\) in equation (2), we get

\[\begin{align}\sqrt 3 x - \sqrt 8 \left( {\frac{{ - \sqrt 2 x}}{3}} \right) &= 0\\

\sqrt 3 x + \frac{{\sqrt {16} x}}{3}& = 0\\\frac{{3\sqrt 3 x + 4x}}{3} &= 0\\x\left( {3\sqrt 3 x + 4} \right)& = 0\\x &= 0\end{align}\]

Substituting \(x = 0\) in equation (3), we get

\[\begin{align}y &= \frac{{\sqrt 2 \times 0}}{3}\\y &= 0\end{align}\]

The answer is

\[\begin{align}x &= 0\\y &= 0\end{align}\]

**(vi) Steps:**

\[\begin{align}\frac{{3x}}{2} - \frac{{5y}}{3} &= - 2 \qquad \dots(1)\\\frac{x}{3} + \frac{y}{2} &= \frac{{13}}{6} \qquad \dots\left( 2 \right)\end{align}\]

Multiply both the equations (1) and (2) by \(6,\) to remove the decimal number and making it easier for calculation.

\[\begin{align}&\left[ {\frac{{3x}}{2} - \frac{{5y}}{3} = - 2} \right] \times 6\\&9x - 10y = - 12 \qquad \qquad \dots(3)\\

\\&\left[ {\frac{x}{3} + \frac{y}{2} = \frac{{13}}{6}} \right] \times 6\\&2x + 3y = 13 \qquad \qquad\dots(4)\end{align}\]

By solving the equation (3)

\[\begin{align}9x - 10y &= - 12\\10y &= 9x + 12\\y &= \frac{{9x + 12}}{{10}} \qquad \dots(5)\end{align}\]

Substituting \(\begin{align}y = \frac{{9x + 12}}{{10}} \end{align}\) in equation (4), we get

\[\begin{align}2x + 3\left( {\frac{{9x + 12}}{{10}}} \right) &= 13\\

\frac{{20x + 27x + 36}}{{10}} &= 13\\47x &= 130 - 36\\x& = \frac{{94}}{{47}}\\x &= 2

\end{align}\]

Substituting \(x = 2\) in equation (5), we get

\[\begin{align}y &= \frac{{9 \times 2 + 12}}{{10}}\\y &= \frac{{30}}{{10}}\\y &= 3\end{align}\]

The answer is

\[\begin{align}x &= 2\\y &= 3\end{align}\]

## Question 2

Solve \(2x + 3y = 11\) and \(2x-4y = -24,\) hence find the value of \(m\) for which \(y = mx + 3.\)

### Solution

**Video Solution**

**Reasoning:**

Solve the linear equations (1) and (2) by substitution method and substitute the values of \(x\) and \(y\) in \(y = mx + 3\) to get the value of \(m.\)

**What is Known?**

\[\begin{align}2x + 3y &= 11\\2x - 4y &= - 24\\y &= mx + 3\end{align}\]

**What is Unknown?**

Value of \(m\)

**Steps:**

\[\begin{align}2x + 3y &= 11 \qquad\;\;\; \dots(1)\\\\2x - 4y &= - 24 \qquad \dots(2)\end{align}\]

By solving the equation (1)

\[\begin{align}2x + 3y &= 11\\3y &= 11 - 2x\\y &= \frac{{11 - 2x}}{3} \qquad \dots(3)\end{align}\]

Substituting \(\begin{align} y = \frac{{11 - 2x}}{3}\end{align}\) in equation (2), we get

\[\begin{align}2x - 4\left( {\frac{{11 - 2x}}{3}} \right) &= - 24\\\frac{{6x - 44 + 8x}}{3} &= - 24\\14x - 44 &= - 72\\14x &= 44 - 72\\x &= - \frac{{28}}{{14}}\\x &= - 2\end{align}\]

Substituting \(x = - 2\) in equation (3)

\[\begin{align}y &= \frac{{11 - 2 \times \left( { - 2} \right)}}{3}\\y& = \frac{{11 + 4}}{3}\\y &= \frac{{15}}{3}\\y &= 5\end{align}\]

Now, Substituting \(x = - 2\) and \(y = 5\) in \(y = mx + 3\)

\[\begin{align}y &= mx + 3\\5 &= m( - 2) + 3\\5 - 3 &= - 2m\\2 &= - 2m\\m &= \frac{2}{{ - 2}}\\m &= - 1\end{align}\]

The answer is

\[\begin{align}x &= - 2\\y &= 5\\m &= - 1\end{align}\]

## Question 3

Form the pair of linear equations for the following problems and find their Solution by substitution method.

(i) The difference between two numbers is \(26\) and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by \(18\) degrees. Find them.

(iii) The coach of a cricket team buys \(7\) bats and 6 balls for ₹\(3800.\) Later, she buys \(3\) bats and \(5\) balls for ₹\(1750.\) Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of \(10\,\rm{ km,}\) the charge paid is ₹ \(105\) and for a journey of \(15\,\rm{ km,}\) the charge paid is ₹\(155.\) What are the fixed charges and the charge per \(\rm{km}?\) How much does a person have to pay for travelling a distance of \(25\,\rm{ km}\)?

(v) A fraction becomes \(\frac{9}{{11}},\) if \(2\) is added to both the numerator and the denominator. If, \(3\) is added to both the numerator and the denominator it becomes \(\frac{5}{6}.\) Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

### Solution

**Video Solution**

**What is Unknown?**

Formation of the pair of linear equations and their solution.

**(i) What is Known?**

The difference between two numbers is \(26\) and one number is three times the other.

**Reasoning:**

Assuming the numbers as *\(x\)* and \(y,\) two linear equations can be formed for the known situation.

**Steps:**

Let the first (larger) number \( = x\)

And the second number \( = y\)

The difference between two numbers is \(26.\)

\[x - y = 26 \qquad \ldots \left( 1 \right)\]

One number is three times the other

\[x = 3y \qquad \ldots \left( 2 \right)\]

Substituting \(x = 3y\) in equation (1), we get

\[\begin{align}3y - y &= 26\\2y &= 26\\y &= 13\end{align}\]

Substituting \(y = 13\) in equation (2)

\[\begin{align}x &= 3 \times 13\\x &= 39\end{align}\]

The two numbers are \(39\) and \(13.\)

**(ii) What is Known?**

Larger of two supplementary angles, exceeds the smaller by \(18\) degrees.

**Reasoning:**

Supplementary angles are two angles with a sum of \({180^ \circ }\) and assuming the angles as \({x^ \circ }\) and \({y^ \circ }\), two linear equations can be formed for the known situation.

**Steps:**

Let the larger angle \( = {x^ \circ }\)

and smaller angle \( = {y^ \circ }\)

Since the angles are supplementary

\[{x^ \circ } + {y^ \circ } = {180^ \circ } \qquad \ldots \left( 1 \right)\]

Larger angle exceeds the smaller by \({18^ \circ }\)

\[{x^ \circ } = {y^ \circ } + {18^ \circ } \qquad \ldots \left( 2 \right)\]

Substituting \({x^ \circ } = {y^ \circ } + {18^ \circ }\) in equation (1), we get

\[\begin{align}{y^ \circ } + {18^ \circ } + {y^ \circ } &= {180^ \circ }\\2{y^ \circ } &= {180^ \circ } - {18^ \circ }\\{y^ \circ } &= \frac{{{{162}^ \circ }}}{{{2^ \circ }}}\\{y^ \circ } &= {81^ \circ }\end{align}\]

Substituting \({y^ \circ } = {81^ \circ }\) in equation (2), we get

\[\begin{align}{x^ \circ } &= {81^ \circ } + {18^ \circ }\\{x^ \circ }& = {99^ \circ }\end{align}\]

The angles are \({99^ \circ }\) and \({81^ \circ }.\)

**(iii) What is Known?**

The cost of \(7\) bats and \(6\) balls is ₹ \(3800\) and the cost of \(3\) bats and \(5\) ball is ₹ \(1750\)

**Reasoning:**

Assuming the cost of \(1\) bat as ₹ *\(x\)* and cost of 1 ball as ₹ \(y,\) two linear equations can be formed for the Known situation.

**Steps:**

Let the cost of \(1\) bat \(=\) ₹ *\(x\)*

And the cost of \(1\) ball \(=\) ₹ *\(y\)*

Then,

\[\begin{align}7x + 6y &= 3800 \qquad \ldots \left( 1 \right)\\3x + 5y &= 1750 \qquad \ldots \left( 2 \right)\end{align}\]

By solving the equation (1)

\[\begin{align}7x + 6y &= 3800\\6y &= 3800 - 7x\\y &= \frac{{3800 - 7x}}{6} \qquad \ldots \left( 3 \right)

\end{align}\]

Substituting \(\begin{align} y = \frac{{3800 - 7x}}{6} \end{align}\) in equation (2), we get

\[\begin{align}3x \! + \! 5\left[ {\frac{{3800 \! - \! 7x}}{6}} \right] & \! = \! 1750\\\frac{{18x + 19000 \! - \! 35x}}{6} & \! = \! 1750\\ - 17x + 19000 & \! = \! 1750 \! \times \! 6\\17x & \! = \! 19000 \! - \! 10500\\x & \! = \! \frac{{8500}}{{17}}\\x & \! = \! 500\end{align}\]

Substituting \(x = 500\) in equation (3), we get

\[\begin{align}y &= \frac{{3800 - 7 \times 500}}{6}\\y &= \frac{{300}}{6}\\y &= 50\end{align}\]

Cost of 1 bat is ₹ \(500\)

Cost of 1 ball is ₹ \(50\)

**(iv) What is Known?**

Taxi Charge for a distance of \(10\,\rm{ km}\) is ₹\(105\) and for \(15\,\rm{ km}\) is ₹\(\rm{155.}\)

**Reasoning:**

Assuming fixed charge as ₹ *\(x\)* and charge for each kilometer as ₹ \(y,\) two linear equations can be formed.

**Steps:**

Let the fixed charge \(=\) ₹ *\(x\)*

And charge per km \(=\) ₹ *\(y\)*

Charge for a distance of \(10 \,\rm{km}\)

\[x + 10y = 105 \qquad \ldots \left( 1 \right)\]

Charge for a distance of \(15\,\rm{ km}\)

\[x + 15y = 155 \qquad \ldots \left( 2 \right)\]

By solving the equation (1)

\[\begin{align}x + 10y &= 105\\x &= 105 - 10y \qquad \ldots \left( 3 \right)\end{align}\]

Substituting \(x = 105 - 10y\) in equation (2), we get

\[\begin{align}105 - 10y + 15y &= 155\\5y &= 155 - 105\\y& = \frac{{50}}{5}\\y &= 10\end{align}\]

Substituting \(x = 5\) in equation (3)

\[\begin{align}x &= 105 - 10 \times 10\\x &= 105 - 100\\x &= 5\end{align}\]

Now, charge for a distance of \(25\,\rm{ km}\) \( = x + 25y\)

\[\begin{align} &= 5 + 25 \times 10\\ &= 5 + 250\\ &= 255\end{align}\]

Fixed charge \(=\) ₹ \(5\)

Charge per km \(=\) ₹ \(10\)

Charge for \(25 \,\rm{km}\) \(=\) ₹ \(255\)

**(v) What is Known?**

Fraction becomes \(\frac{9}{{11}}\), if \(2\) is added to both numerator and denominator and becomes \(\frac{5}{6},\) if \(3\) is added to both numerator and denominator.

**Reasoning:**

Assuming the numerator as *\(x\)* and denominator as \(y, \)two linear equations can be formed.

**Steps:**

Let the numerator \( = x\)

And denominator \( = y\)

Then fraction\( = \frac{x}{y}\)

When \(2\) is added to both numerator and denominator

\[\begin{align}\frac{{x + 2}}{{y + 2}} &= \frac{9}{{11}}\\11(x + 2) &= 9(y + 2)\\11x + 22 &= 9y + 18\\11x - 9y + 22 - 18 &= 0\\11x - 9y + 4 &= 0 \qquad \dots(1)\end{align}\]

When \(3\) is added to both numerator and denominator

\[\begin{align}\frac{{x + 3}}{{y + 3}} &= \frac{5}{6}\\6(x + 3) &= 5(y + 3)\\6x + 18 &= 5y + 15\\

6x - 5y + 18 - 15 &= 0\\6x - 5y + 3 &= 0 \qquad \qquad\;\ldots \left( 2 \right)\\5y& = 6x + 3\\y &= \frac{{6x + 3}}{5} \qquad ...(3)\end{align}\]

Substituting \(\begin{align} y = \frac{{6x + 3}}{5} \end{align}\) in equation (1)

\[\begin{align}11x - 9\left( {\frac{{6x + 3}}{5}} \right) + 4 &= 0\\\frac{{55x - 9\left( {6x + 3} \right) + 20}}{5} &= 0\\55x - 54x - 27 + 20 &= 0\\x - 7 &= 0\\x &= 7\end{align}\]

Substituting \(x = 7\) in equation (1)

\[\begin{align}y &= \frac{{6 \times 7 + 3}}{5}\\y& = \frac{{42 + 3}}{5}\\y &= \frac{{45}}{5}\\y& = 9\end{align}\]

The fraction is \(\begin{align} \frac{7}{9} \end{align}\)

**(vi) What is Known?**

Five years hence, the age of Jacob will be three times that of his son and five years ago, Jacob was seven times that of his son.

**Reasoning:**

Assume their present age as *\(x\)* and \(y,\) then find their age \(5\) years from now and \(5\) years ago in terms of *\(x\)* and \(y;\) two linear equations can be formed.

**Steps:**

Let the present age of Jacob \(= x\) years and his son \(= y\) years

\(5\) years from now,

Jacob’s age \( = \left( {x + 5} \right)\) years

Son’s age \( = \left( {y + 5} \right)\) years

\[\begin{align}\left( {x + 5} \right) &= 3\left( {y + 5} \right)\\x + 5 &= 3y + 15\\x - 3y + 5 - 15 &= 0\\

x - 3y - 10 &= 0 \qquad \dots(1)\end{align}\]

\(5\) years ago,

Jacob’s age \( = \left( {x - 5} \right)\) years

Son’s age \( = \left( {y - 5} \right)\) years

\[\begin{align}\left( {x - 5} \right) &= 7(y - 5)\\x - 5 &= 7y - 35\\x - 7y - 5 + 35 &= 0\\x - 7y + 30 &= 0 \qquad \quad\;\;\; \dots\,(2)\\7y &= x + 30\\y &= \frac{{x + 30}}{7} \quad \;\dots\,(3)\end{align}\]

Substituting \(\begin{align} y = \frac{{x + 30}}{7} \end{align}\) in equation (1)

\[\begin{align}x - 3\left( {\frac{{x + 30}}{7}} \right) - 10 &= 0\\\frac{{7x - 3\left( {x + 30} \right) - 70}}{7} &= 0\\7x - 3x - 90 - 70 &= 0\\4x - 160 &= 0\\x &= \frac{{160}}{4}\\x &= 40\end{align}\]

Substituting \(x = 40\) in equation (3)

\[\begin{align}y &= \frac{{40 + 30}}{7}\\y &= \frac{{70}}{7}\\y &= 10\end{align}\]

Present age of Jacob is \(40\) years and his son is \(10\) years.

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