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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.
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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.
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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
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.
Hello Aspirant,
Here's how you can find it:
School ID Card: Your registration number is often printed on your school ID card.
Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.
School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.
Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.
Your school is the best place to get this information.
Hello,
It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.
The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.
For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.
Let me know if you need imformation about any exam eligibility criteria.
good wishes for your future!!
Hello Aspirant,
"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).
For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.
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