NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables - Exercise 3.3

Exercise 3.3

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

(i) \[x+y=14;x-y=4\]

Ans: The given equations are:

\[x+y=14\]           …… (i)

\[x-y=4\]             …… (ii)

From equation (i):

\[x=14-y\]          …… (iii)

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

\[\left( 14-y \right)-y=4\]

\[14-2y=4\]

\[10=2y\]

\[y=5\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=9\]

Therefore, \[x=9\] and \[y=5\].

(ii) \[s-t=3;\frac{s}{3}+\frac{t}{2}=6\]

Ans: The given equations are:

\[s-t=3\]           …… (i)

\[\frac{s}{3}+\frac{t}{2}=6\]       …… (ii)

From equation (i):

\[s=t+3\]          …… (iii)

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

\[\frac{t+3}{3}+\frac{t}{2}=6\]

\[2t+6+3t=36\]

\[5t=30\]

\[t=6\]                  …… (iv)

Substituting (iv) in (iii), we get

\[s=9\]

Therefore, \[s=9\] and \[t=6\].

(iii) \[3x-y=3;9x-3y=9\]

Ans: The given equations are:

\[3x-y=3\]           …… (i)

\[9x-3y=9\]             …… (ii)

From equation (i):

\[y=3x-3\]          …… (iii)

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

\[9x-3\left( 3x-3 \right)=9\]

\[9x-9x+9=9\]

\[9=9\]

For all \[x\] and \[y\].

Therefore, the given equations have infinite solutions. One of the solution is \[x=1,y=0\].

(iv) \[0.2x+0.3y=1.3;0.4x+0.5y=2.3\]

Ans: The given equations are:

\[0.2x+0.3y=1.3\]           …… (i)

\[0.4x+0.5y=2.3\]             …… (ii)

From equation (i):

\[x=\frac{1.3-0.3y}{0.2}\]          …… (iii)

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

\[0.4\left( \frac{1.3-0.3y}{0.2} \right)+0.5y=2.3\]

\[2.6-0.6y+0.5y=2.3\]

\[2.6-2.3=0.1y\]

\[y=3\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=\frac{1.3-0.3\left( 3 \right)}{0.2}\]

\[x=2\]

Therefore, \[x=2\] and \[y=3\].

(v) \[\sqrt{2}x+\sqrt{3}y=0;\sqrt{3}x-\sqrt{8}y=0\]

Ans: The given equations are:

\[\sqrt{2}x+\sqrt{3}y=0\]           …… (i)

\[\sqrt{3}x-\sqrt{8}y=0\]             …… (ii)

From equation (i):

\[x=\frac{-\sqrt{3}y}{\sqrt{2}}\]          …… (iii)

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

\[\sqrt{3}\left( \frac{-\sqrt{3}y}{\sqrt{2}} \right)-\sqrt{8}y=0\]

$(\dfrac{-3} { \sqrt{2}}) y-\sqrt{8} y=0$

\[y=0\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=0\]

Therefore, \[x=0\] and \[y=0\].

(vi) \[\frac{3x}{2}-\frac{5y}{3}=-2;\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\]

Ans: The given equations are:

\[\frac{3x}{2}-\frac{5y}{3}=-2\]           …… (i)

\[\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\]             …… (ii)

From equation (i):

\[x=\frac{-12+10y}{9}\]          …… (iii)

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

\[\frac{\left( \frac{-12+10y}{9} \right)}{3}+\frac{y}{2}=\frac{13}{6}\]

\[\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}\]

\[\frac{-24+20y+27y}{54}=\frac{13}{6}\]

\[47y=141\]

\[y=3\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=2\]

Therefore, \[x=2\] and \[y=3\].

2. Solve \[\mathbf{2x}+\mathbf{3y}=\mathbf{11}\] and \[\mathbf{2x}-\mathbf{4y}=-\mathbf{24}\] and hence find the value of ‘\[m\]’ for which \[\mathbf{y}=\mathbf{mx}+\mathbf{3}\].

Ans: The given equations are:

\[2x+3y=11\]           …… (i)

\[2x-4y=-24\]             …… (ii)

From equation (i):

\[x=\frac{11-3y}{2}\]          …… (iii)

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

\[2\left( \frac{11-3y}{2} \right)-4y=-24\]

\[11-3y-4y=-24\]

\[-7y=-35\]

\[y=5\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=-2\]

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

Calculating the value of \[m\]:

\[y=mx+3\]

\[5=-2m+3\]

\[m=-1\]

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 \[\mathbf{26}\] and one number is three times the other. Find them.

Ans: Assuming one number be \[x\] and another number be \[y\] such that \[y>x\],

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

\[y=3x\]             …… (i)

\[y-x=26\]    …… (ii)

Substituting the value of \[y\] from equation (i) in equation (ii), we get

\[3x-x=26\]

\[2x=26\]

\[x=13\]                  …… (iii)

Substituting (iii) in (i), we get

\[y=39\]

Therefore, numbers are $13$ & $39$.

(ii) The larger of two supplementary angles exceeds the smaller by \[\mathbf{18}\] degrees. Find them.

Ans: Assuming the larger angle be \[x\] and smaller angle be \[y\]. 

The sum of a pair of supplementary angles is always \[{{180}^{\circ }}\].

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

\[x+y=180\]             …… (i)

\[x-y=18\]    …… (ii)

Substituting the value of \[x\] from equation (i) in equation (ii), we get

\[180-y-y=18\]

\[162=2y\]

\[y=81\]                  …… (iii)

Substituting (iii) in (i), we get

\[x=99\]

Therefore, the two angles are  \[x={{99}^{\circ }}\] and \[y={{81}^{\circ }}\].

(iii) The coach of a cricket team buys \[\mathbf{7}\] bats and 6 balls for \[\mathbf{Rs}\text{ }\mathbf{3800}\]. Later, she buys \[\mathbf{3}\] bats and \[\mathbf{5}\] balls for \[\mathbf{Rs}\text{ }\mathbf{1750}\]. Find the cost of each bat and each ball.

Ans: Assuming the cost of a bat is \[x\] and the cost of a ball is \[y\].

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

\[7x+6y=3800\]             …… (i)

\[3x+5y=1750\]    …… (ii)

From equation (i):

\[y=\frac{3800-7x}{6}\]       …… (iii)

Substituting (iii) in equation (ii): 

\[3x+5\left( \frac{3800-7x}{6} \right)=1750\]

\[3x+\frac{9500}{3}-\frac{35x}{6}=1750\]

\[3x-\frac{35x}{6}=1750-\frac{9500}{3}\]

\[\frac{18x-35x}{6}=\frac{5250-9500}{3}\]

\[\frac{-17x}{6}=\frac{-4250}{3}\]

\[x=500\]                  …… (iv)

Substituting (iv) in (iii), we get

\[y=\frac{3800-7\left( 500 \right)}{6}\]

\[y=50\]

Therefore, the bat costs \[Rs\text{ }500\] and the ball costs \[Rs\text{ }50\].

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

Ans: Assuming the fixed charge be \[Rs\text{ }x\] and the per km charge be \[Rs\text{ y}\].

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

\[x+10y=105\]             …… (i)

\[x+15y=155\]    …… (ii)

From equation (i):

\[x=105-10y\]       …… (iii)

Substituting (iii) in equation (ii): 

\[105-10y+15y=155\]

\[5y=50\]

\[y=10\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=105-10\left( 10 \right)\]

\[x=5\]

Therefore, the fixed charge is \[Rs\text{ }5\] and the per km charge is \[Rs\text{ 10}\].

So, charge for \[25\text{ }km\] will be:

\[=Rs\text{ }\left( x+25y \right)\]

\[=Rs\text{ 255}\] 

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

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

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

\[\frac{x+2}{y+2}=\frac{9}{11}\]

\[11x+22=9y+18\]

\[11x-9y=-4\]             …… (i)

\[\frac{x+3}{y+3}=\frac{5}{6}\]

\[6x+18=5y+15\]

\[6x-5y=-3\]    …… (ii)

From equation (i):

\[x=\frac{-4+9y}{11}\]       …… (iii)

Substituting (iii) in equation (ii): 

\[6\left( \frac{-4+9y}{11} \right)-5y=-3\]

\[-24+54y-55y=-33\]

\[y=9\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=\frac{-4+9\left( 9 \right)}{11}\]

\[x=7\]

Therefore, the fraction is \[\frac{7}{9}\].

(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?

Ans: Assuming the age of Jacob be \[x\] and the age of his son be \[y\].

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

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

\[x-3y=10\]             …… (i)

\[\left( x-5 \right)=7\left( y-5 \right)\]

\[x-7y=-30\]    …… (ii)

From equation (i):

\[x=3y+10\]       …… (iii)

Substituting (iii) in equation (ii): 

\[3y+10-7y=-30\]

\[-4y=-40\]

\[y=10\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=3\left( 10 \right)+10\]

\[x=40\]

Therefore, Jacob’s present age is \[40\] years and his son’s present age is \[10\] years.

Highlights of NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

There is no other alternative to practice, especially when it comes to a subject like mathematics. This guide will help you to cover all the important portions of Class 10 Maths Chapter 3 Exercise 3.3. Below are the important sections of the chapter:

• The algebraic method for solving linear equations.

• Reducing equations to linear equations in two variables.

• Problems on linear equations like find the value of two variables as given in Exercise 3.3 Class 10 Maths.

In Class 10 Maths Exercise 3.3 solutions, you will find all these concepts comprehensively discussed and problems solved in a step-by-step manner.

Solved linear equations by the substitution method in Class 10th maths Chapter 3 Exercise 3.3:

(i) x + y = 14

x – y = 4

(ii) s – t = 3

(s/3) + (t/2) = 6

(iii) 3x – y = 3

9x – 3y = 9

Solutions:

(i) Two equations are given below:

x + y = 14 ………. (i)

x – y = 4 ……………….. (ii)

From the 1st equation, we get,

x = 14 – y

Now, by substituting the value of x with (14-y), in second equation, we get -

(14 – y) – y = 4

or, 14 – 2y = 4

or, 2y = 10

or, y = 5

Now, by using the value of y i.e. 5, we can find the value of x

∵ x = 14 – y

∴ x = 14 – 5

Or, x = 9

Hence, x = 9 and y = 5.

(ii) Given below are the two equations:

s – t = 3  ……………….. (i)

(s/3) + (t/2) = 6 …………….. (ii)

From the 1st equation, we get,

s = 3 + t ________________(1)

Now, by substituting the value of s with (3 + t) in second equation, we get,

(3+t)/3 + (t/2) = 6

⇒ (2(3+t) + 3t )/6 = 6

⇒ (6+2t+3t)/6 = 6

⇒ (6+5t) = 36

⇒5t = 30

⇒t = 6

Now, we need to substitute t with 6 in equation (1)

s = 3 + 6 = 9

Therefore, s = 9 and t = 6.

(iii) Given below are two equations:

3x – y = 3 ……. (i)

9x – 3y = 9 ……… (ii)

From the 1st equation, we get,

x = (3+y)/3

Now, by substituting x with (3+y)/3 in the 2nd equation, we get,

9(3+y)/3 – 3y = 9

⇒9 +3y -3y = 9

⇒ 9 = 9

Therefore, y is proven to have infinite values.

And, since x = (3+y) /3

So the variable x, too, has infinite values.

The above sums are important to clear the concepts covered in Exercise 3.3 Class 10 maths NCERT solutions.

You will get many more such solved problems in Class 10 Maths Chapter 3 Exercise 3.3 Solution.

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Class 10 Maths Ex 3.3 Solutions by Vedantu has been prepared by some of the best subject matter experts from across the country. So, you get accurate solutions and explanations that strictly adhere to the NCERT standards. Besides,

• You can access the study materials along with Maths NCERT Ex 3.3 Class 10 Solutions at any point of time; it’s available on Vedantu’s platform in an easily downloadable PDF format - free of cost.

• Every concept of the chapter and problem from NCERT Class 10 Maths Chapter 3 Exercise 3.3 is explained and solved comprehensively.

• The problems and solutions given in the study material will give you an idea of the sums that you can expect in the examination.

So, all you need to do to score full marks in linear equation problems, in exams, is to practice with Vedantu’s Maths Class 10 Chapter 3 Exercise 3.3 Solution.

NCERT Solutions Class 10 Maths All Chapters

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solutions Class 10 Maths Chapter 3 Exercises

Important Topics under NCERT Solutions for Class 10 Maths Chapter 3

Chapter 3 of the class 10 maths syllabus includes Linear Equations, which is a very important chapter in mathematics that is covered in class 10. This chapter on Linear Equations includes three major areas, and in order to internalize these ideas properly, students are required to go through the important topics under Linear Equations, individually. The following is a list of the important topics covered under the chapter on Linear Equations. We recommend that students follow every topic minutely to master the concept of Linear Equations.

• Using the algebraic method to solve linear equations

• Reducing the given equations to linear equations in two variables

• Solving problems on linear equations, like finding the value of two variables, and so on

Importance of Class 10 Maths Chapter 3 Linear Equations

Linear equations are equations of degree 1. For instance, ax + by +c = 0, where a and b are not equal to zero. Equations of straight lines are known as linear equations.

They are important because they enable describing the relationship between two variables, calculating rates, making predictions, and carrying out conversions, among other things. We encourage students to learn as much as they can from this chapter to be able to solve problems on linear equations easily in exams.