NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

# NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Edited By Ramraj Saini | Updated on Nov 13, 2023 02:54 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3

NCERT Solutions for class 10 maths ex 3.3 Pair of Linear Equations in two variables is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 3.3 class 10 deals with the graphical and algebraic representations of linear equations in two variables and their different type of solutions such as a unique solution, no solutions or infinitely many solutions, using the five methods namely the Graphical method, substitution method, cross Multiplication method, elimination method and determinant method.

These class 10 maths ex 3.3 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.3

$\\x + y = 14\\ x - y = 4$

Given, two equations,

$\\x + y = 14.......(1)\\ x - y = 4........(2)$

Now, from (1), we have

$y=14-x........(3)$

Substituting this in (2), we get

$x-(14-x)=4$

$\Rightarrow x-14+x=4$

$\Rightarrow 2x=4+14=18$

$\Rightarrow x=9$

Substituting this value of x in (3)

$\Rightarrow y=14-x=14-9=5$

Hence, Solution of the given equations is x = 9 and y = 5.

$\\s - t = 3\\ \frac{s}{3} + \frac{t}{2} = 6$

Given, two equations,

$\\s - t = 3.........(1)\\ \frac{s}{3} + \frac{t}{2} = 6.......(2)$

Now, from (1), we have

$s=t+3........(3)$

Substituting this in (2), we get

$\frac{t+3}{3}+\frac{t}{2}=6$

$\Rightarrow \frac{2t+6+3t}{6}=6$

$\Rightarrow 5t+6=36$

$\Rightarrow 5t=30$

$\Rightarrow t=6$

Substituting this value of t in (3)

$\Rightarrow s=t+3 = 6+3=9$

Hence, Solution of the given equations is s = 9 and t = 6.

$\\ 3 x - y = 3\\ 9x - 3y = 9$

Given, two equations,

$\\ 3 x - y = 3......(1)\\ 9x - 3y = 9.....(2)$

Now, from (1), we have

$y=3x-3........(3)$

Substituting this in (2), we get

$9x-3(3x-3)=9$

$\Rightarrow 9x-9x+9=9$

$\Rightarrow 9=9$

This is always true, and hence this pair of the equation has infinite solutions.

As we have

$y=3x-3$ ,

One of many possible solutions is $x=1,\:and\:y=0$ .

$\\0.2x + 0.3y = 1.3\\ 0.4x + 0.5y = 2.3$

Given, two equations,

$\\0.2x + 0.3y = 1.3\\ 0.4x + 0.5y = 2.3$

Now, from (1), we have

$y=\frac{1.3-0.2x}{0.3}........(3)$

Substituting this in (2), we get

$0.4x+0.5\left(\frac{1.3-0.2x}{0.3}\right)=2.3$

$\Rightarrow 0.12x+0.65-0.1x=0.69$

$\Rightarrow 0.02x=0.69-0.65=0.04$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=\left ( \frac{1.3-0.2x}{0.3} \right )=\left ( \frac{1.3-0.4}{0.3} \right )=\frac{0.9}{0.3}=3$

Hence, Solution of the given equations is,

$x=2\:and\:y=3$ .

$\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

Given, two equations,

$\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0$

Now, from (1), we have

$y=-\frac{\sqrt{2}x}{\sqrt{3}}........(3)$

Substituting this in (2), we get

$\sqrt{3}x-\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )=0$

$\Rightarrow \sqrt{3}x=\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )$

$\Rightarrow \3x=-4x$

$\Rightarrow7x=0$

$\Rightarrow x=0$

Substituting this value of x in (3)

$\Rightarrow y=-\frac{\sqrt{2}x}{\sqrt{3}}=0$

Hence, Solution of the given equations is,

$x=0,\:and \:y=0$ .

$\\\frac{3x}{2} - \frac{5y}{3}= - 2\\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Given,

$\\\frac{3x}{2} - \frac{5y}{3}= - 2.........(1)\\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}.............(2)$

From (1) we have,

$x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )........(3)$

Putting this in (2) we get,

$\frac{1}{3}\times\frac{2}{3}\left ( \frac{5y}{3}-2 \right )+\frac{y}{2}=\frac{13}{6}$

$\frac{10y}{27}-\frac{4}{9}+\frac{y}{2}=\frac{13}{6}$

$\frac{20y}{54}-\frac{4}{9}+\frac{27y}{54}=\frac{13}{6}$

$\frac{47y}{54}=\frac{13}{6}+\frac{4}{9}$

$\frac{47y}{54}=\frac{117}{54}+\frac{24}{54}$

$47y=117+24$

$47y=141$

$y=\frac{141}{47}$

$y=3$

putting this value in (3) we get,

$x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )$

$x=\frac{2}{3}\left ( \frac{5\times3}{3}-2 \right )$

$x=\frac{2}{3}\left (5-2 \right )$

$x=\frac{2}{3}\times 3$

$x=2$

Hence $x=2\:\:and\:\:y=3.$

Given, two equations,

$2x + 3y = 11......(1)$

$2x - 4y = -24.......(2)$

Now, from (1), we have

$y=\frac{11-2x}{3}........(3)$

Substituting this in (2), we get

$2x-4\left ( \frac{11-2x}{3} \right )=-24$

$\Rightarrow 6x-44+8x=-72$

$\Rightarrow 14x=44-72$

$\Rightarrow 14x=-28$

$\Rightarrow x=-2$

Substituting this value of x in (3)

$\Rightarrow y=\left ( \frac{11-2x}{3} \right )=\frac{11-2\times(-2)}{3}=\frac{15}{3}=5$

Hence, Solution of the given equations is,

$x=-2,\:and\:y=5.$

Now,

As it satisfies $y=mx+3$ ,

$\Rightarrow 5=m(-2)+3$

$\Rightarrow 2m=3-5$

$\Rightarrow 2m=-2$

$\Rightarrow m=-1$

Hence Value of m is -1.

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

Let two numbers be x and y and let the bigger number is y.

Now, According to the question,

$y-x=26......(1)$

And

$y=3x......(2)$

Now, the substituting value of y from (2) in (1) we get,

$3x-x=26$

$\Rightarrow 2x=26$

$\Rightarrow x=13$

Substituting this in (2)

$\Rightarrow y=3x=3(13)=39$

Hence the two numbers are 13 and 39.

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the larger angle be x and smaller angle be y

Now, As we know the sum of supplementary angles is 180. so,

$x+y=180.......(1)$

Also given in the question,

$x-y=18.......(2)$

Now, From (2) we have,

$y=x-18.......(3)$

Substituting this value in (1)

$x+x-18=180$

$\Rightarrow 2x=180+18$

$\Rightarrow 2x=198$

$\Rightarrow x=99$

Now, Substituting this value of x in (3), we get

$\Rightarrow y=x-18=99-18=81$

Hence the two supplementary angles are

$99^0\:and\:81^0.$

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, According to the question,

$7x+6y=3800......(1)$

$3x+5y=1750......(2)$

Now, From (1) we have

$y=\frac{3800-7x}{6}........(3)$

Substituting this value of y in (2)

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

$\Rightarrow 18x+19000-35x=1750\times6$

$\Rightarrow -17x=10500-19000$

$\Rightarrow -17x=-8500$

$\Rightarrow x=\frac{8500}{17}$

$\Rightarrow x=500$

Now, Substituting this value of x in (3)

$y=\frac{3800-7x}{6}=\frac{3800-7\times500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$

Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.

Let the fixed charge is x and the per km charge is y.

Now According to the question

$x+10y=105.......(1)$

And

$x+15y=155.......(2)$

Now, From (1) we have,

$x=105-10y........(3)$

Substituting this value of x in (2), we have

$105-10y+15y=155$

$\Rightarrow 5y=155-105$

$\Rightarrow 5y=50$

$\Rightarrow y=10$

Now, Substituting this value in (3)

$x=105-10y=105-10(10)=105-100=5$

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair For 25 km :

$\Rightarrow x+25y=5+25(10)=5+250=255$

Hence fair for 25km is 255 Rs.

Let the numerator of the fraction be x and denominator of the fraction is y

Now According to the question,

$\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x+2)=9(y+2)$

$\Rightarrow 11x+22=9y+18$

$\Rightarrow 11x-9y=-4...........(1)$

Also,

$\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x+3)=5(y+3)$

$\Rightarrow 6x+18=5y+15$

$\Rightarrow 6x-5y=-3...........(2)$

Now, From (1) we have

$y=\frac{11x+4}{9}.............(3)$

Substituting this value of y in (2)

$6x-5\left ( \frac{11x+4}{9} \right )=-3$

$\Rightarrow 54x-55x-20=-27$

$\Rightarrow -x=20-27$

$\Rightarrow x=7$

Substituting this value of x in (3)

$y=\frac{11x+4}{9}=\frac{11(7)+4}{9}=\frac{81}{9}=9$

Hence the required fraction is

$\frac{x}{y}=\frac{7}{9}.$

Let x be the age of Jacob and y be the age of Jacob's son,

Now, According to the question

$x+5=3(y+5)$

$\Rightarrow x+5=3y+15$

$\Rightarrow x-3y=10..........(1)$

Also,

$x-5=7(y-5)$

$\Rightarrow x-5=7y-35$

$\Rightarrow x-7y=-30.........(2)$

Now,

From (1) we have,

$x=10+3y...........(3)$

Substituting this value of x in (2)

$10+3y-7y=-30$

$\Rightarrow -4y=-30-10$

$\Rightarrow 4y=40$

$\Rightarrow y=10$

Substituting this value of y in (3),

$x=10+3y=10+3(10)=10+30=40$

Hence, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.

## More About NCERT Solutions for Class 10 Maths Exercise 3.3:

In exercise 3.3 Class 10 Maths, The method of substitution involves three main steps. First, we need to solve one equation for one of the variables. Then we need to substitute the expression (value) into the other equation and solve. Finally, we need to substitute the value into the original equation to find the corresponding variable. The solution to the system of linear equations is written in the form of (x, y). The NCERT solutions for Class 10 Maths exercise 3.3 mainly focused on solving the pair of linear equations in two variables by using a substitution method. All the three questions related to the substitution method are given in exercise 3.3 Class 10 Maths. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 3.3 :

• NCERT solutions for Class 10 Maths exercise 3.3 helps the students who wish to score high in CBSE term exams
• n exercise 3.3 Class 10 Maths, the substitution method is the easy method to solve equations and it takes away an equation to make it easier for you to do and also the method gives you only one equation to work on instead of having two.
• By solving the NCERT solution for Class 10th Maths chapter 3 exercise 3.3 exercises, students can attain a good score in term exams as well as in competitive examinations.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. Find the value of x in 6x-y=10 if y=2 .

Putting y=2 in 6x-y=10

6x-2=10

6x=10+2

Thus x =12/6=2

2. Mothers age is six times her daughter’s age. Six years after the age of mother will be four times the age of her daughter. Find the present age of the mother and her daughter.

Let the present age of mother be x years and the present age of daughter be y years.

Given that, x=6y

Also, x+6=4(y+6)

Now substituting x=6y in x+6=4(y+6)

6y+6=4(y+6)

6y-4y+6-24=0

2y-18=0

Thus y=9

The daughter’s age is 9 years then her mother’s age is 54 years.

3. Solve x-3y=4 and 2x+y=1 by substitution method.

Given, x-3y=4 ⇒ x=4+3y

Now substituting x=4+3y in 2x+y=1

2(4+3y)+y=1

8+6y+y=1

7y=1-8 ⇒ y=-1

Substitute y=-1 in x-3y=4

x-3(-1)=4

x=4-3 ⇒ x=1

Thus (x, y)=(1, -1)

4. Mention the different methods to solve the system of equations linear equations in two variables.

The methods used to find the solution of a pair of linear equations are

• Graphical method

• Algebraic method

• Substitution Method

• Elimination Method

• Cross- multiplication Method

5. What is the general form of linear equation?

The General form is ax+by+c=0, where a, b, c are real numbers.

6. Whether the substitution method can be used to solve the system of equations in three variables?

Yes, we can use the substitution method to solve the system of equations in three variables.

7. Define substitution method according to NCERT solutions for Class 10 Maths chapter 3 exercise 3.3?

The substitution method is defined as substituting the value of one variable in another equation to obtain the value of another variable

8. In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.3, how many questions are there?

In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.3, there are three questions.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

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You can pursue diplomas at various institutions like:

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