# NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5

Pair of Linear Equations in Two Variables

## Chapter 3 Ex.3.5 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**

**Video 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 \(y=-1\)

## Chapter 3 Ex.3.5 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**

**Video 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.

## Chapter 3 Ex.3.5 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**

**Video 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

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

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

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

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

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

## Chapter 3 Ex.3.5 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**

**Video 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 2^{nd} 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.