# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5

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

## Question 1

Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i)

\begin{align}\,&x-3y-3 = 0 \\&3x-9y-2 = 0\end{align}

(ii)

\begin{align}&2x + y = 5\\&3x + 2y = 8\end{align}

(iii)

\begin{align}3x-5y &= 20\\6x-10y &= 40\end{align}

(iv)

\begin{align}x-3y-7 = 0\\3x-3y-15 = 0\end{align}

### Solution

Steps:

\begin{align}\rm{(i)} \quad x-3y-3 &= 0\\3x-9y-2 &= 0\end{align}

\begin{align}& \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3},\quad \frac{{{b}_{1}}~}{{{b}_{2}}}=\frac{-3}{-9}~=~\frac{1}{3},\\&\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-3}{-2}=~\frac{3}{2} \\ \\& \frac{{{a}_{1}}}{{{a}_{2}}~}=\frac{{{b}_{1}}}{{{b}_{2}}~}\ne \frac{{{c}_{1}}}{~{{c}_{2}}} \\\end{align}

Therefore, the given sets of lines are parallel to each other and will not intersect each other thus, there will be no solution for these equations.

Steps:

\begin{align}(\rm{ii}) \quad 2x + y &= 5\\3x + 2y &= 8\end{align}

\begin{align} 2x+y-5&=0 \\ 3x+2y-8&=0 \end{align}

\begin{align}& \frac{{{a}_{1}}~}{{{a}_{2}}}=\frac{2}{3},\quad\frac{{{b}_{1}}~}{{{b}_{2}}}=\frac{1}{2~}\,,\quad\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-5}{-8}=\frac{5}{8} \\ & \frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}} \\\end{align}

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

\begin{align} \frac{x}{{{b}_{1}}{{c}_{2}}\!-\!{{b}_{2}}{{c}_{1}}}&\!=\!\frac{y}{{{c}_{1}}{{a}_{2}}\!-\!{{c}_{2}}{{a}_{1}}}\!=\!\frac{1}{{{a}_{1}}{{b}_{2}}\!-\!{{a}_{2}}{{b}_{1}}} \\ \frac{x}{-8+10}&=\frac{y}{-15+16}=\frac{1}{4-3} \\ \frac{x}{2}=\frac{y}{1}&=1 \\ \frac{x}{2}\,\,&=\,\,1\;\;\text{ and }\;\;\frac{y}{1}\,\,=\,\,1 \\\end{align}

$$\therefore$$ $$x = 2$$ and  $$y = 1$$

Steps:

\begin{align}( \rm{iii})\quad 3x-5y &= 20\\6x-10y &= 40 \end{align}

\begin{align}3x-5y - 20& = 0\\6x-10y - 40& = 0\end{align}

\begin{align}\frac{{{a_1}}}{{{a_2}}}& = \,\frac{{3}}{6} = \frac{1}{2},\frac{{{b_1}}}{{{b_2}}} = \frac{{5}}{{10}} = \frac{1}{2},\\ \frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 20}}{{ - 40}} = \frac{1}{2}\\\\\frac{{{a_1}}}{{{a_2}}} &= \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

Steps:

\begin{align}(\rm{iv})\quad x-3y-7 &= 0\\3x - 3y - 15 &= 0\end{align}

\begin{align} \frac{{{a}_{1}}}{{{a}_{2}}}&=\frac{1}{3},\quad\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-3}{-3}~=1,\\ \frac{{{c}_{1}}}{{{c}_{2}}}&=\frac{-7}{-15}=\frac{7}{15} \\\\ \frac{{{a}_{1}}}{{{a}_{2}}}\,\,&\ne \,\frac{{{b}_{l}}}{{{b}_{2}}}\end{align}

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

\begin{align}\frac{x}{{{b}_{1}}{{c}_{2}}\!-\!{{b}_{2}}{{c}_{1}}}&\!=\!\frac{y}{{{c}_{1}}{{a}_{2}}\!-\!{{c}_{2}}{{a}_{1}}}\!=\!\frac{1}{{{a}_{1}}{{b}_{2}}\!-\!{{a}_{2}}{{b}_{1}}} \\\frac{x}{45-21}&\!=\!\frac{y}{-21\!-\!(-15)}\!=\!\frac{1}{-3\!-\!(-9)} \\\frac{x}{24}&=\frac{y}{-6}=\frac{1}{6} \\\frac{x}{24}&=\frac{1}{6}\,\quad\text{and}\,\quad\frac{y}{-6}=\frac{1}{6} \\ x&=4\,\quad\text{and}\,\quad y=-1\end{align}

$$\therefore$$  $$x=4,$$ and $$y=-1$$

## Question 2

(i) For which values of $$a$$ and $$b$$ will the following pair of linear equations have an infinite number of solutions?

\begin{align}2x + 3y\, &= 7\\\left( {a-b} \right)x + \left( {a + b} \right)y &= 3\,a\, + \,b\, - \,2\end{align}

(ii) For which value of $$k$$ will the following pair of linear equations have no solution?

\begin{align}3x + y &= 1\\\left( {2k-1} \right)x + \left( {k-1} \right)y &= 2k + 1\end{align}

### Solution

Steps:

(i)

\begin{align}&2x + 3y - 7 = 0\\&\left( {a\!-\!b} \right)x \!+\! \left( {a + b} \right)y \!- \!\left( {3a + b-2} \right)\! = \!0\end{align}

\begin{align} &\frac{{{a_1}}}{{{a_2}}} \!=\! \frac{2}{{a \!-\! b}}, \\&\frac{{{b_1}}}{{{b_2}}} \!=\! \frac{3}{{a\! + \! b}},\\& \frac{{{c_1}}}{{{c_2}}} \!=\! \frac{7}{{3a \!+\! b \!- \!2}}\end{align}

For infinitely many solutions,

\begin{align}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}&=\frac{{{c}_{1}}}{{{c}_{2}}} \\ \frac{2}{a-b}&=\frac{7}{3a+b-2} \\ 6a+2b-4&=7a-7b \\ a-9b&=-4 \qquad\; \ldots (1) \\\\ \frac{2}{a-b}&=\frac{3}{a+b} \\ 2a+2b&=3a-3b \\ a-5b&=0\text{ }\,\qquad\;\;\ldots (2) \\\end{align}

Subtracting (1) from (2), we obtain

\begin{align}4b &= 4\\b &= 1\end{align}

Substituting $$b = 1$$ in equation (2), we obtain

\begin{align}a-5\,\, \times \,\,1\,\, &= 0\\a &= 5\end{align}

Hence, $$a = 5$$ and $$b = 1$$ are the values for which the given equations give infinitely many solutions.

Steps:

(ii)

\begin{align}&3x + y - 1 = 0\\&\left( {2k-1} \right)x + \left( {k-1} \right)y - 2k-1 = 0\end{align}

\begin{align}\frac{{{a_1}}}{{{a_2}}}& = \frac{3}{{2k - 1}}, \quad \frac{{{b_1}}}{{{b_2}}}= \frac{1}{{k - 1}},\\\,\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 1}}{{ - 2k - 1}}\,\, = \,\,\frac{1}{{2k + 1}}\end{align}

For no solution,

\begin{align} \frac{{{a}_{1}}}{{{a}_{2}}}&=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}~ \\ \frac{3}{2k-1}&=\frac{1}{k-1}\ne \frac{1}{2k+1} \\ \frac{3}{2k-1}&=\frac{1}{k-1} \\3k-3&=2k-1 \\k&=2\end{align}

Hence, for $$k = 2$$ the given equation has no solution.

## Question 3

Solve the following pair of linear equations by the substitution and cross-multiplication methods:

\begin{align}8x + 5y &= 9\\3x + 2y &= 4\end{align}

### Solution

Steps:

\begin{align} 8x + 5y &= 9 \qquad \ldots (1)\\ 3x + 2y &= 4 \qquad \ldots (2)\end{align}

From equation $$(2)$$, we obtain

\begin{align}3x + 2y &= 4\\3x &= 4 - 2y\\x &= \frac{{4 - 2y}}{3}\, \qquad \ldots (3)\end{align}

Substituting \begin{align}x = \frac{{4 - 2y}}{3}\,\end{align} in equation $$(1),$$ we obtain

\begin{align}8\left( {\frac{{4 - 2y}}{3}} \right) + 5y &= 9\\\frac{{32 - 16y + 15y}}{3} &= 9\\32 - y &= 27\\ y &= 32 - 27\\y &= 5\end{align}

Substituting $$y=5$$ in equation $$(3),$$ we obtain

\begin{align}x &= \frac{{4 - 2 \times 5}}{3}\\x &= \frac{{ - 6}}{3}\\x &= - 2\\\end{align}

Hence, $$x = - 2, \;y = 5$$

Again, by cross-multiplication method

\begin{align}8x + 5y &= 9\\3x + 2y &= 4\end{align}

\begin{align}8x + 5y-9 &= 0\\3x + 2y-4 &= 0 \end{align}

\begin{align} {a_1} &= 8,\quad{b_1} = 5,\quad{c_1} = - 9\\{a_2} &= 3,\quad{b_2} = 2,\quad{c_2} = - 4 \end{align}

\begin{align}\frac{{x}}{{{b_1}{c_2} - {b_2}{c_1}}} &= \frac{{y}}{{{c_1}{a_2} - {c_2}{a_1}}} \\ &= \frac{1}{{{a_1}{b_2} - {a_2}{b_1}}}\\\frac{x}{{ - 20 - ( - 18)}} &= \frac{y}{{ - 27 - ( - 32)}} \\ &= \frac{1}{{16 - 15}}\\\end{align}

\begin{align} \frac{x}{{ - 2}} &= \frac{y}{5} = 1\\\frac{x}{{ - 2}}& = 1\,\quad {\rm{and}}\quad \frac{y}{5} = 1\\x &= - 2\,\quad {\rm{and}}\quad y = 5\end{align}

## Question 4

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student $$A$$ takes food for $$20$$ days, she has to pay ₹ $$1000$$ as hostel charges whereas a student $$B,$$ who takes food for $$1000$$ days, pays ₹ $$1180$$ as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes $$13$$ when $$1$$ is subtracted from the numerator and it becomes $$14$$ when $$8$$ is added to its denominator. Find the fraction.

(iii) Yash scored $$40$$ marks in a test, getting $$3$$ marks for each right answer and losing $$1$$ mark for each wrong answer. Had $$4$$ marks been awarded for each correct answer and $$2$$ marks been deducted for each incorrect answer, then Yash would have scored $$50$$ marks. How many questions were there in the test?

(iv) Places $$A$$ and $$B$$ are $$100\,\rm{ km}$$ apart on a highway. One car starts from $$A$$ and another from $$B$$ at the same time. If the cars travel in the same direction at different speeds, they meet in $$5$$ hours. If they travel towards each other, they meet in $$1$$ hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by $$9$$ square units, if its length is reduced by $$5$$ units and breadth is increased by $$3$$ units. If we increase the length by $$3$$ units and the breadth by $$2$$ units, the area increases by $$67$$ square units. Find the dimensions of the rectangle.

### Solution

Reasoning:

Assume one variable equal to $$x$$ and another be $$y.$$ Then based on given conditions, two linear equations can be formed which can be easily solved.

Steps:

(i) Let $$x$$ be the fixed charge of the food and $$y$$ be the charge for food per day.

According to the given information,

When a student A, takes food for 20 days, pays ₹ $$1000$$ as hostel charges.

$x+20y=1000\qquad\ldots \left( 1 \right)$

When a student $$B,$$ who takes food for $$26$$ days, pays ₹ $$1180$$ as hostel charges.

$x+26y=1180 \qquad \ldots \left( 2 \right)$

Subtracting equation $$(1)$$ from equation $$(2),$$ we obtain

\begin{align}6y &= 180\\y &= \frac{{180}}{6}\\y &= 30\end{align}

Substituting $$y = 30$$ in equation $$(1),$$ we obtain

\begin{align}x + 20 \times 30 &= 1000\\x &= 1000 - 600\\x &= 400\end{align}

Equations are $$x + 20y = 1000$$ and $$x + 26y = 1180$$ where $$x$$ is the fixed charge of the food and $$y$$ is the charge for food per day

Hence, fixed charge is ₹ $$400$$

And charge per day is ₹ $$30$$

(ii) Let the numerator be $$x$$ and denominator be $$y,$$ thus the fraction be \begin{align} \frac{x}{y} \end{align}

According to the given information,

When $$1$$ is subtracted from the numerator

\begin{align}\frac{{x - 1}}{y}\, &= \frac{1}{3}\\3x - 3 &= y\\3x - y &= 3 \qquad \ldots (1)\end{align}

When $$8$$ is added to the denominator,

\begin{align}\frac{x}{{y + 8}}\, &= \frac{1}{4}\\4x &= y + 8\\4x - y &= 8 \qquad \ldots (2)\end{align}

Subtracting equation $$(1)$$ from equation $$(2),$$ we obtain

$x = 5$

Putting $$x = 5$$ in equation $$(1),$$ we obtain

\begin{align}3 \times 5-y &= 3\\y &= 15 - 3\\y& = 12 \end{align}

Equations are $$3x - y = 3$$ and $$4x - y = 8$$ where the numerator of the fraction is $$x,$$ and denominator is $$y.$$

Hence, the fraction is \begin{align} \frac{5}{{12}} \end{align}

(iii) Let the number of right answers and wrong answers be $$x$$ and $$y$$ respectively.

Therefore, total number of questions be $$\left( {x + y} \right)$$

According to the given information,

\begin{align}3x - y &= 40 \qquad \ldots (1)\\\\4x - 2y &= 50\\2x - y &= 25 \qquad \ldots (2)\end{align}

Subtracting equation $$(2)$$ from equation $$(1),$$ we obtain

$x = 15 \qquad \ldots (3)$

Substituting this in equation $$(2),$$ we obtain

\begin{align}2 \times 15 - y &= 25\\y &= 30 - 25\\y &= 5\end{align}

Equations are $$3x - y = 40$$ and $$2x - y = 25$$ where the number of right and wrong answers are $$x$$ and $$y$$ respectively.

number of right answers $$= 15$$ and number of wrong answers $$= 5$$

Hence, Total number of questions $$= 20$$

(iv) Let the speed of $$1^\rm{st}$$ car and $$2^\rm{nd}$$ car be $$u \text{ km/h}$$ and $$v\text{ km/h}$$ respectively.

According to the given information,

When the cars travel in the same direction at different speeds, they meet in $$5$$ hours.

Therefore, distance travelled by $$1^\rm{st}$$ car $$= 5u\,{\rm{km }}$$

and distance travelled by $$2^\rm{nd}$$ car $$= 5v\,{\rm{km}}$$

\begin{align}5u - 5v &= 100\\5\left( {u - v} \right) &= 100\\u - v &= 20 \qquad \ldots (1)\end{align}

When the cars travel towards each other at different speeds, they meet in $$1$$ hour

therefore, distance travelled by $$1^\rm{st}$$ car $$= u\,{\rm{ km}}$$

and distance travelled by 2nd car $$= v\,{\rm{km}}$$

$u + v = 100 \qquad \ldots (2)$

Adding both the equations, we obtain

\begin{align}2u &= 120\\u &= 60\end{align}

Substituting this value in equation (2), we obtain

\begin{align}60 + v &= 100\\v &= 40\end{align}

Equations are $$u - v = 20$$ and $$u+v=100$$ where the speed of $$1^\rm{st}$$ car and $$2^\rm{nd}$$ car be $$u\text{ km/h}$$ and $$v\text{ km/h}$$ respectively.

Hence, speed of the $$1^\rm{st}$$ car $$= 60\,{\rm{km/h}}$$ and speed of the $$2^\rm{nd}$$ car $$= 40\,{\rm{ km/h}}$$

(v) Let length and breadth of rectangle be $$x$$ unit and $$y$$ unit respectively.

Then the area of the rectangle be $$xy$$ square units.

According to the question,

When length is reduced by $$5$$ units and breadth is increased by $$3$$ units, area of the rectangle gets reduced by $$9$$ square units;

\begin{align}\left( {x - 5} \right)\left( {y + 3} \right) &= xy - 9\\xy + 3x - 5y - 15 &= xy - 9\\3x - 5y - 6 &= 0 \qquad \ldots (1)\end{align}

When we increase the length by $$3$$ units and the breadth by $$2$$ units, the area increases by $$67$$ square units;

\begin{align}\left( {x + 3} \right)\left( {y + 2} \right) &= xy + 67\\xy + 2x + 3y + 6 &= xy + 67\\2x + 3y - 61 &= 0 \qquad \ldots (2)\end{align}

By cross-multiplication method, we obtain

\begin{align} \frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}&=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}} \\ & =\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\\frac{x}{305-(-18)}&=\frac{y}{-12-(-183)} \\ & =\frac{1}{9-(-19)} \end{align}

\begin{align} \frac{x}{323}&=\frac{y}{171}=\frac{1}{19} \\\frac{x}{323}&=\frac{1}{19}, \quad \frac{y}{171}=\frac{1}{19} \\ x&=17,\qquad \,y=9\end{align}

Equations are $$3x - 5y - 6 = 0$$ and $$2x + 3y - 61 = 0$$ where length and breadth of the rectangle are $$x$$ and $$y$$ respectively.

Hence, the length and breadth of the rectangle are $$17$$ units and $$9$$ units respectively.