There are various methods for solving the linear equation. In the previous exercises graphical method and one of the algebraic methods for solving the equation have been discussed. There are many algebraic methods for solving equations, like substitution and elimination methods. Similar to the substitution method elimination method is also used to solve the linear equation. In this method, multiply the equation by numbers to make the coefficient of the equation equal. This will help to eliminate the value and get the solution.
This Story also Contains
These class 10 maths exercise 3.3 solutions are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice is necessary for all these questions to command the concepts, boost confidence and in-depth understanding of concepts. In the exercise, all the NCERT solutions are covered according to the syllabus of the NCERT. These solutions help the students to understand the concept in depth. This exercise included all the questions given in the NCERT Books. In exercise 3.3 of the NCERT, the chapter solutions are explained in detail.
Q1(i) Solve the following pair of linear equations by the elimination method and the substitution method :
$x + y =5 \ \textup{and} \ 2x - 3y = 4$
Answer:
Elimination Method:
Given, equations
$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$
Now, multiplying (1) by 3 we get
$\\3x +3 y =15............(3)$
Now, adding (2) and (3), we get
$\\2x-3y+3x +3 y =4+15$
$\Rightarrow 5x=19$
$\Rightarrow x=\frac{19}{5}$
Substituting this value in (1), we get
$\frac{19}{5}+y=5$
$\Rightarrow y=5-\frac{19}{5}$
$\Rightarrow y=\frac{6}{5}$
Hence,
$x=\frac{19}{5}\:and\:y=\frac{6}{5}$
Substitution method :
Given, equations
$\\x + y =5............(1) \ \textup{and} \\ \ 2x - 3y = 4........(2)$
Now, from (1) we have,
$y=5-x.......(3)$
Substituting this value in (2)
$2x-3(5-x)=4$
$\Rightarrow 2x-15+3x=4$
$\Rightarrow 5x=19$
$\Rightarrow x=\frac{19}{5}$
Substituting this value of x in (3)
$\Rightarrow y=5-x=5-\frac{19}{5}=\frac{6}{5}$
Hence,
$x=\frac{19}{5}\:and\:y=\frac{6}{5}$
Q1(ii) Solve the following pair of linear equations by the elimination method and the substitution method :
$3x + 4 y = 10 \ \textup{and} \ 2x - 2y = 2$
Answer:
Elimination Method:
Given, equations
$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$
Now, multiplying (2) by 2 we get
$\\4x -4 y =4............(3)$
Now, adding (1) and (3), we get
$\\3x+4y+4x -4 y =10+4$
$\Rightarrow 7x=14$
$\Rightarrow x=2$
Putting this value in (2) we get
$2(2)-2y=2$
$\Rightarrow 2y=2$
$\Rightarrow y=1$
Hence,
$x=2\:and\:y=1$
Substitution method :
Given, equations
$\\3x + 4 y = 10............(1) \ \textup{and}\\ \ 2x - 2y = 2..............(2)$
Now, from (2) we have,
$y=\frac{2x-2}{2}=x-1.......(3)$
Substituting this value in (1)
$3x+4(x-1)=10$
$\Rightarrow 3x+4x-4=10$
$\Rightarrow 7x=14$
$\Rightarrow x=2$
Substituting this value of x in (3)
$\Rightarrow y=x-1=2-1=1$
Hence, $x=2\:and\:y=1$
Answer:
Elimination Method:
Given, equations
$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$
$\\\Rightarrow 9x - 2y -7=0........(2)$
Now, multiplying (1) by 3 we get
$\\9x -15 y -12=0............(3)$
Now, subtracting (3) from (2), we get
$9x-2y-7-9x+15y+12=0$
$\Rightarrow 13y+5=0$
$\Rightarrow y=\frac{-5}{13}$
Putting this value in (1), we get
$3x-5(\frac{-5}{13})-4=0$
$\Rightarrow 3x=4-\frac{25}{13}$
$\Rightarrow 3x=\frac{27}{13}$
$\Rightarrow x=\frac{9}{13}$
Hence,
$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$
Substitution method :
Given, equations
$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$
$\\\Rightarrow 9x - 2y -7=0........(2)$
Now, from (2) we have,
$y=\frac{9x-7}{2}.......(3)$
Substituting this value in (1)
$3x-5\left(\frac{9x-7}{2} \right )-4=0$
$\Rightarrow 6x-45x+35-8=0$
$\Rightarrow -39x+27=0$
$\Rightarrow x=\frac{27}{39}=\frac{9}{13}$
Substituting this value of x in (3)
$\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}$
Hence, $x=\frac{9}{13}\:and\:y=-\frac{5}{13}$
Answer:
Elimination Method:
Given, equations
$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$
Now, multiplying (2) by 2, we get
$\\2x - \frac{2y}{3} =6............(3)$
Now, adding (1) and (3), we get
$\\\frac{x}{2}+\frac{2y}{3}+2x-\frac{2y}{3}=-1+6$
$\Rightarrow \frac{5x}{2}=5$
$\Rightarrow x=2$
Putting this value in (2), we get
$2-\frac{y}{3}=3$
$\Rightarrow \frac{y}{3}=-1$
$\Rightarrow y=-3$
Hence,
$x=2\:and\:y=-3$
Substitution method :
Given, equations
$\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)$
Now, from (2) we have,
$y=3(x-3)......(3)$
Substituting this value in (1)
$\frac{x}{2}+\frac{2(3(x-3))}{3}=-1$
$\Rightarrow \frac{x}{2}+2x-6=-1$
$\Rightarrow \frac{5x}{2}=5$
$\Rightarrow x=2$
Substituting this value of x in (3)
$\Rightarrow y=3(x-3)=3(2-1)=-3$
Hence, $x=2\:and\:y=-3$
Answer:
Let the numerator of the fraction be x, and the denominator is y,
Now, according to the question,
$\frac{x+1}{y-1}=1$
$\Rightarrow x+1=y-1$
$\Rightarrow x-y=-2.........(1)$
Also,
$\frac{x}{y+1}=\frac{1}{2}$
$\Rightarrow 2x=y+1$
$\Rightarrow 2x-y=1..........(2)$
Now, subtracting (1) from (2), we get
$x=3$
Putting this value in (1)
$3-y=-2$
$\Rightarrow y=5$
Hence, $x=3\:and\:y=5$
And the fraction is: $\frac{3}{5}$
Answer:
Let the age of Nuri be x and the age of Sonu be y.
Now, according to the question
$x-5=3(y-5)$
$\Rightarrow x-5=3y-15$
$\Rightarrow x-3y=-10.........(1)$
Also,
$x+10=2 (y+10)$
$\Rightarrow x+10=2y+20$
$\Rightarrow x-2y=10........(2)$
Now, subtracting (1) from (2), we get
$y=20$
Putting this value in (2)
$x-2(20)=10$
$\Rightarrow x=50$
Hence, the age of Nuri is 50 and the age of Nuri is 20.
Answer:
Let the unit digit of the number be x and the 10's digit be y.
Now, according to the question,
$x+y=9.......(1)$
Also
$9(10y+x)=2(10x+y)$
$\Rightarrow 90y+9x=20x+2y$
$\Rightarrow 88y-11x=0$
$\Rightarrow 8y-x=0.........(2)$
Now adding (1) and (2), we get,
$\Rightarrow 9y=9$
$\Rightarrow y=1$
Now putting this value in (1)
$x+1=9$
$\Rightarrow x=8$
Hence, the number is 18.
Answer:
Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.
Now, according to the question,
$x+y=25..........(1)$
And
$50x+100y=2000$
$\Rightarrow x+2y=40.............(2)$
Now, subtracting (1) from (2), we get
$y=15$
Putting this value in (1).
$x+15=25$
$\Rightarrow x=10$
Hence, Meena received 10, 50 Rs notes and 15, 100 Rs notes.
Answer:
Let the fixed charge be x, and per day charge is y.
Now, according to the question,
$x+4y=27...........(1)$
And
$x+2y=21...........(2)$
Now, Subtracting (2) from (1). We get,
$4y-2y=27-21$
$\Rightarrow 2y=6$
$\Rightarrow y=3$
Putting this in (1)
$x+4(3)=27$
$\Rightarrow x=27-12=15$
Hence, the fixed charge is 15 Rs and the per-day charge is 3 Rs.
Also Read:
Also see-
Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.
Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.
Frequently Asked Questions (FAQs)
We multiply both the equations with some non-zero constant to eliminate one of the variables by adding both of the new equations.
Graphical Method
Elimination method
Elimination method as it is time-efficient and not lengthy.
In order to make the coefficient of one of the variables numerically equal but with opposite signs so that they get cancelled when we add them together.
Yes, once we find the value of the first variable, we substitute its value in any of the equation to get the value of the second equation.
It doesn’t provide us the desired point if the values are irrational.
On Question asked by student community
Hrllo,
If you want to view the CM Shri School 2025-26 admission test result, visit the official website http://www.edudel.nic.in/ (http://www.edudel.nic.in/,) ; here, click on the "CM Shri Schools Admission Test - 2025." then select the "CM Shri Admission Test Result 2025 - Merit List" link. Here you need to log in with your credentials and view or download the merit list pdf.
I hope it will clear your query!!
Hello aspirant,
Fairness and equal opportunity are guaranteed by the rigorous merit-based admissions process at CM Shri Schools. The merit list will be made public on edudel.nic.in, the official DOE website. Based on how well they did on the admission exam, students will be shortlisted.
To know the result, you can visit our site through following link:
https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
Thank you
Hello
The CM Shri School Admission Test was conducted for students seeking quality education.
The result and selection list are expected to be released around 20 September 2025.
Students who qualified will get admission into CM Shri Model Schools across the state.
The official list of selected students will be available on the edudel.nic.in website.
Candidates can check their names in the merit list PDF once uploaded.
Selection is based on the student’s performance in the entrance exam.
Keep checking the official site or the school notice board for timely updates.
You can check CM Shri school result here: https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025 Thank You.
Hello! If you are looking for the CM Shree School result for 2025, here is the link provided by Careers360 where you can check your result easily. I’ll be attaching it for your reference so you can access it directly.
https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters