NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.5

Exercise 3.5

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) \[\mathbf{x} - \mathbf{3y} - \mathbf{3}=\mathbf{0};\text{ }\mathbf{3x}- \mathbf{9y} - \mathbf{2}=\mathbf{0}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{3}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{3}{2}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\].

Therefore there are no solution for the given pair of linear equation as the given lines are parallel to each other and do not intersect.

(ii) \[\mathbf{2x}+\mathbf{y}=\mathbf{5};\text{ }\mathbf{3x}+\mathbf{2y}=\mathbf{8}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-5}{-8}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\].

Therefore the given pair of linear equations have a unique solution as the given lines intersect each other at a unique point.

Using cross-multiplication method,

\[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

\[\frac{x}{-8-\left( -10 \right)}=\frac{y}{15+16}=\frac{1}{4-3}\]

\[\frac{x}{2}=\frac{y}{1}=1\]

\[x=2,y=1\]

Therefore, \[x=2\] and \[y=1\].

(iii) \[\mathbf{3x} - \mathbf{5y}=\mathbf{20};\text{ }\mathbf{6x} - \mathbf{10y}=\mathbf{40}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\].

Therefore there are infinite number of solution for the given pair of linear equation as the given lines are overlapping each other.

(iv) \[\mathbf{x} - \mathbf{3y} - \mathbf{7}=\mathbf{0};\text{ }\mathbf{3x} - \mathbf{3y} - \mathbf{15}=\mathbf{0}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=1\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-15}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\].

Therefore the given pair of linear equations have a unique solution as the given lines intersect each other at a unique point.

Using cross-multiplication method,

\[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

\[\frac{x}{45-\left( 21 \right)}=\frac{y}{-21-\left( -15 \right)}=\frac{1}{-3+\left( 9 \right)}\]

\[\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}\]

\[x=4,y=-1\]

Therefore, \[x=4\] and \[y=-1\].

2. (i) For which values of \[a\] and \[b\] does the following pair of linear equations have an infinite number of solutions?

\[\mathbf{2x}+\mathbf{3y}=\mathbf{7};\text{ }\left( \mathbf{a}\text{ } - \text{ }\mathbf{b} \right)\mathbf{x}+\left( \mathbf{a}\text{ }+\text{ }\mathbf{b} \right)\mathbf{y}=\mathbf{3a}+\mathbf{b} - \mathbf{2}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{a-b}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{3}{a+b}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-\left( 3a+b-2 \right)}\]

Condition for infinitely many solutions,

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\]

\[\frac{2}{a-b}=\frac{7}{\left( 3a+b-2 \right)}\]

\[6a+2b-4=7a-7b\]

\[a-9b=-4\]                  …… (i)

\[\frac{2}{a-b}=\frac{3}{a+b}\]

\[2a+2b=3a-3b\]

\[a-5b=0\]                           …… (ii)

Subtracting equation (i) from equation (ii), we get

\[4b=4\]

\[b=1\]                  …… (iii)

Substituting (iii) in equation (ii), we get

\[a-5=0\]

\[a=5\]

Therefore, the given equations will have infinite number of solutions for \[a=5\] and \[b=1\].

(ii) For which value of \[\mathbf{k}\] will the following pair of linear equations have no solution?\[\mathbf{3x}+\mathbf{y}=\mathbf{1};\text{ }\left( \mathbf{2k}\text{ } - \text{ }\mathbf{1} \right)\mathbf{x}+\left( \mathbf{k}\text{ } - \text{ }\mathbf{1} \right)\mathbf{y}=\mathbf{2k}+\mathbf{1}\].

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2k-1}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{k-1}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2k+1}\]

Condition for no solution,

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\]

\[\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1}\]

\[\frac{3}{2k-1}=\frac{1}{k-1}\]

\[3k-3=2k-1\]

\[k=2\]

Therefore, the given equations will not have any solutions for \[k=2\].

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

\[\mathbf{8x}+\mathbf{5y}=\mathbf{9};\text{ }\mathbf{3x}+\mathbf{2y}=\mathbf{4}\]

Ans :

Using substitution method,

\[8x+5y=9\]                   …… (i)

\[3x+2y=4\]                  …… (ii)

From equation (ii) we get

\[x=\frac{4-2y}{3}\]      …… (iii)

Substituting (iii) in equation (i), we get

\[8\left( \frac{4-2y}{3} \right)+5y=9\]

\[32-16y+15y=27\]

\[y=5\]              …… (iv)

Substituting (iv) in equation (ii), we obtain

\[3x+10=4\]

\[x=-2\]

Therefore, \[x=-2\] and \[y=5\].

Using cross multiplication method:

\[8x+5y=9\]

\[3x+2y=4\]

\[\frac{x}{-20-\left( -18 \right)}=\frac{y}{-27-\left( -32 \right)}=\frac{1}{16-15}\]

\[\frac{x}{-2}=\frac{y}{5}=1\]

\[x=-2,y=5\]

Therefore, \[x=-2\] and \[y=5\].

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 \[\mathbf{A}\] takes food for \[\mathbf{20}\] days she has to pay \[\mathbf{Rs}\text{ }\mathbf{1000}\] as hostel charges whereas a student \[\mathbf{B}\], who takes food for \[\mathbf{26}\] days, pays \[\mathbf{Rs}\text{ }\mathbf{1180}\] as hostel charges. Find the fixed charges and the cost of food per day.

Ans: Assuming that the fixed charge is \[Rs\text{ }x\] and the charge for food per day is \[Rs\text{ y}\].

Writing the algebraic representation using the information given in the question:

\[x+20y=1000\]             …… (i)

\[x+26y=1180\]    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

\[6y=180\]

\[y=30\]     …… (iii)

Substituting (iii) in equation (i), we obtain

\[x+20\left( 30 \right)=1000\]

\[x=400\]              …… (iv)

Therefore, \[x=400\] and \[y=30\].

Hence, fixed charges are \[Rs\text{ 400}\] and charges per day are \[Rs\text{ 30}\].

(ii) A fraction becomes \[\frac{1}{3}\] when \[\mathbf{1}\] is subtracted from the numerator and it becomes \[\frac{1}{4}\] when \[\mathbf{8}\] is added to its denominator. Find the fraction.

Ans: Assuming that the fraction is \[\frac{x}{y}\].

Writing the algebraic representation using the information given in the question:

\[\frac{x-1}{y}=\frac{1}{3}\]

\[3x-y=3\]             …… (i)

\[\frac{x}{y+8}=\frac{1}{4}\]

\[4x-y=8\]    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

\[x=5\]     …… (iii)

Substituting (iii) in equation (i), we obtain

\[15-y=3\]

\[y=12\]              …… (iv)

Therefore, \[x=5\] and \[y=12\].

Hence, the fraction is \[\frac{5}{12}\].

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

Ans: Say the number of right answers be \[\text{ }x\] and the number of wrong answers be \[\text{ y}\].

Writing the algebraic representation using the information given in the question:

\[3x-y=40\]             …… (i)

\[2x-y=25\]    …… (ii)

Subtracting equation (ii) from equation (i), we obtain

\[x=15\]     …… (iii)

Substituting (iii) in equation (ii), we obtain

\[y=5\]              …… (iv)

Therefore, \[x=15\] and \[y=5\].

Hence, number of right answers = \[\text{15}\] and number of wrong answers = \[\text{ 5}\].

Therefore, the total number of questions = \[\text{20}\]

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

Ans: Assuming that the speed of first car is \[u\text{ }km/h\] and the speed of second car is \[\text{v }km/h\].

Relative speed when both cars are travelling in same direction \[\text{=}\left( u-v \right)\text{ }km/h\]

Relative speed when both cars are travelling in opposite direction \[\text{=}\left( u+v \right)\text{ }km/h\]

Writing the algebraic representation using the information given in the question:

\[5\left( u-v \right)=100\]

\[u-v=20\]             …… (i)

\[1\left( u+v \right)=100\]    …… (ii)

Adding equation (i) and (ii), we get

\[2u=120\]

\[u=60\]     …… (iii)

Substituting (iii) in equation (ii), we obtain

\[v=40\]

Hence, speed of one car is \[\text{60 }km/h\] and speed of the other car is \[\text{40 }km/h\].

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

Ans: Assuming length of the rectangle be \[x\] and the breadth be \[\text{y}\].

Writing the algebraic representation using the information given in the question:

\[\left( x-5 \right)\left( y+3 \right)=xy-9\]

\[3x-5y-6=0\]             …… (i)

\[\left( x+3 \right)\left( y+2 \right)=xy+67\]    …… (ii)

Using cross multiplication method:

\[\frac{x}{305-\left( -18 \right)}=\frac{y}{-12-\left( -183 \right)}=\frac{1}{9-\left( -10 \right)}\]

\[\frac{x}{323}=\frac{y}{171}=\frac{1}{19}\]

\[x=17,y=9\]

Hence, the rectangle has length of 17 units and breadth of 9.

Class 10 Chapter 3 Exercise 3.5 Solution: A Brief Overview

Exercise 3.5 class 10 Maths consists of various questions related to Linear Equations in Two Variables. In this section, you will get familiar with the few questions related to the above-mentioned topic. Rigorous practising of class 10 Maths ex. 3.5 will definitely help you to understand the entire chapter. This will also increase your problem-solving ability along with coping up with the latest CBSE question pattern. Ex. 3.5 class 10 Maths mainly deals with the following types of questions:

Question 1: Which of the following pairs of linear equations consists of a unique solution, without solution or infinitely numerous solutions? Derive the solution in case of a unique solution applying a cross multiplication method.

i) x – 3y – 3= 0 and 3x – 9y - 2 = 0 ii) 2x+ y = 5 and 3x + 2y = 8 iii) x – 3y – 7= 0 and 3x – 3y -15 = 0 iv) 3x – 5y = 20 and 6x – 10y = 40.

Solution:

i) Given, x -3y -3 = 0 and 3x -9y – 2= 0.

a₁/a₂ = 1/3, b₁/b₂ = -3/-9 = 1/3, c₁/c₂ = -3/-2 = 3/2

(a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)

There is no solution for the above equation as the said pair of lines are parallel to each other and there will not be any intersection between them.

Here are a few more solved problems from exercise 3.5 Maths class 10.

ii) Given, 2x + y = 5 and 3x + 2y = 8

a₁/a₂ = 2/3, b₁/b₂ = 1/2 and c₁/c₂ = -5/-8

(a₁/a₂) ≠ (b₁/b₂)

These equations should have a unique solution by applying cross multiplication technique as there is an intersection between them at a unique point.

x/(b₁c₂ – c₁b₂) = y/(c₁a₂ – c₂a₁) = 1/ (a₁b₂ – a₂b₁)

x/(-8-(-10)) = y/(15+16) = 1/(4-3)

x/2 = y/1 = 1

Hence, x = 2 and y = 1

iii) Given, x - 3y - 7 = 0 and 3x - 3y - 15 = 0

(a₁/a₂) = 1/3

(b₁/b₂) = -3/-3 = 1

(c₁/c₂) = -7/-15

(a₁/a₂) ≠ (b₁/b₂) 

There will definitely be a unique solution as the set of lines are intersecting each other at a unique point.

Applying cross multiplication method we get,

x/(45-21) = y /(-21+ 15) = 1/(-3 + 9)

x/24 = y/-6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

Hence, x = 4 and y = 1

iv) Given, 3x - 5y = 20 and 6x – 10y = 40

(a₁/a₂) = 3/6 = 1/2

(b₁/b₂) = -5/-10 = 1/2

(c₁/c₂) = 20/40 = 1/2

(a₁/a₂) = (b₁/b₂) = (c₁/c₂)

There will be an infinite number of solutions as the given pair of lines is overlapping each other.

What is a Cross Multiplication Method To Solve Linear Equations in One Variable?

In Maths, a cross multiplication method that solves linear equations in two variables is the easiest method and gives the accurate value of the variables.

Let us consider, we have a1x1 + b1y1 + c1 = 0 and a2x + b2x + cz = 0 are the two equations which have to be solved. By using cross multiplication, we will get the values x and y such as:

$X= \frac{b_1c_2-b_2c_1}{b_2a_1-b_1a_2}$

$Y= \frac{c_1a_2-c_2a_1}{b_2a_1-b_1a_2}$

Where,

b2a1- ba2 is not equal to 0.

NCERT Solutions for Class 10 Maths Chapter 3 Exercises

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