Aakash Repeater Courses
Take Aakash iACST and get instant scholarship on coaching programs.
In the previous exercise, it was explained how to show the linear equation using the graphical method, but this method is convenient when the solutions' coordinates are non-integral. There is another method for solving the equation is the arithmetic method for solving linear equations has been explained. In this exercise arithmetic method for solving equations is used to solve the questions. There are many arithmetic methods for solving the equation, namely the Graphical method, substitution method, cross Multiplication method, elimination method and determinant method. In this exercise substitution method has been explained to solve the equation.
This Story also Contains
This exercise deals with the substitution method for solving algebraic equations. Each solutions are explained in detail and crafted in easy-to-understand language. These NCERT solutions are essential for every student to understand the concept. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. These solutions are according to the NCERT Books and cover all the methods according to the latest syllabus.
Q1(i) Solve the following pair of linear equations by the substitution method.
$x+y=14$
$x−y=4$
Answer:
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, the solution of the given equations is x = 9 and y = 5.
Q1(ii) Solve the following pair of linear equations by the substitution method
$s−t=3$
$\frac{s}{3}+\frac{t}{2}=6$
Answer:
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
$\frac{t+3}{3}+\frac{t}{2}=6$
$\Rightarrow 2t+6+3t=36$
$\Rightarrow 5t+6=36$
$\Rightarrow 5t=30$
$\Rightarrow t=6$
Substituting this value of t in (3)
$\Rightarrow s=t+3=6+3=9$
Hence, the solution of the given equations is s = 9 and t = 6.
Q1(iii) Solve the following pair of linear equations by the substitution method.
$3x−y=39$
$9x−3y=9$
Answer:
Given two equations,
$3x−y=3$......(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 equations has infinite solutions.
As we have
$y=3x−3$,
One of many possible solutions is x = 1, and y = 0.
Q1(iv) Solve the following pair of linear equations by the substitution method.
$0.2x+0.3y=1.3$
$0.4x+0.5y=2.3$
Answer:
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
$0.4x+0.5\frac{1.3−0.2x}{0.3}=2.3$
$\Rightarrow 0.12x+0.65−0.1x=0.69$
$\Rightarrow 0.02x=0.69−0.65=0.04$
$\Rightarrow x=2$
Substituting this value of x in (3)
$\Rightarrow y=\frac{1.3−0.2x}{0.3} =\frac{1.3−0.2 × 2}{0.3} =3$
Hence, the solution of the given equations is
x = 2 and y = 3.
Q1(v) Solve the following pair of linear equations by the substitution method.
$\sqrt 2x+ \sqrt 3y=0$
$\sqrt 3x−\sqrt 8y=0$
Answer:
Given two equations,
$\sqrt 2x+ \sqrt 3y=0$
$\sqrt 3x−\sqrt 8y=0$
Now, from (1), we have
$x=\frac{−\sqrt 3y}{\sqrt 2}$........(3)
Substituting this in (2), we get
$\sqrt 3 (\frac{−\sqrt 3y}{\sqrt 2}) - \sqrt 8y=0$
$\Rightarrow \frac{−\sqrt 3y}{\sqrt 2} - 2\sqrt 2y = 0$
$\Rightarrow y(\frac{−\sqrt 3}{\sqrt 2} - 2\sqrt 2) = 0$
$\Rightarrow y=0$
Substituting this value of y in (3)
$\Rightarrow x = 0$
Hence, the solution of the given equations is,
x = 0, and y = 0 .
Q1(vi) Solve the following pair of linear equations by the substitution method.
$\frac{3x}{2}−\frac{5y}{3} =−2$
$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$
Answer:
Given,
$\frac{3x}{2}−\frac{5y}{3} =−2$ ....... (1)
$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$ ........ (2)
From (1) we have,
$x=\frac{(-12 + 10y)}{9}$........(3)
Putting this in (2), we get,
$\frac{\frac{(-12 + 10y)}{9}}{3}+ \frac{y}{2} = \frac{13}{6}$
$\frac{(-12 + 10y)}{27}+ \frac{y}{2} = \frac{13}{6}$
$\frac{(-24 + 20y + 27y)}{54} = \frac{13}{6}$
$47y=141$
$y=3$
Putting this value in (3), we get,
$x=\frac{(-12 + 10× 3)}{9}$
$x =2$
Hence, x = 2 and y = 3.
Q2 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.
Answer:
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
$2x−4(\frac{11−2x}{3})=−24$
$\Rightarrow 6x−44+8x=−72$
$\Rightarrow 14x=44−72$
$\Rightarrow 14x=−28$
$\Rightarrow x=−2$
Substituting this value of x in (3)
$\Rightarrow y=\frac{11−2x}{3}=\frac{11−2×(−2)}{3}=\frac{15}{3}=5$
Hence, the 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, the value of m is -1.
The difference between the two numbers is 26, and one number is three times the other. Find them.
Answer:
Let two numbers be x and y, and the bigger number is y.
Now, according to the question,
$y−x=26$......(1)
And
$y=3x$......(2)
Now, substituting the 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 the two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer:
Let the larger angle be x and the smaller angle be y
Now, as we know, the sum of supplementary angles is 180. so,
$x+y=180^0$.......(1)
Also given in the question,
$x−y=18^0$.......(2)
Now, from (2) we have,
$y=x−18^0$.......(3)
Substituting this value in (1)
$x+x−18^0=180^0$
$\Rightarrow 2x=180^0+18^0$
$\Rightarrow 2x=198^0$
$\Rightarrow x=99^0$
Now, substituting this value of x in (3), we get
$\Rightarrow y=x−18^0=99^0−18^0=81^0$
Hence, the two supplementary angles are
$99^0$ and $81^0$.
Q3 Form the pair of linear equations for the following problems and find their solution by the substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
Answer:
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)
$3x+5\frac{(3800−7x)}{6}=1750$
$\Rightarrow 18x+19000−35x=1750×6$
$\Rightarrow −17x=10500−19000$
$\Rightarrow −17x=−8500$
$\Rightarrow x=\frac{8500}{17}$
$\Rightarrow x=500$
Now, Substituting this value of x in (3)
$y=\frac{(3800−7x)}{6}$
$y =\frac{3800−7×500}{6}$
$y =\frac{3800−3500}{6}=50$
Hence, the cost of one bat is 500 Rs and the cost of one ball is 50 Rs.
Answer:
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.
Answer:
Let the numerator of the fraction be x, and the denominator of the fraction be 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)
$6x−5(\frac{11x+4}{9})=−3$
$\Rightarrow 54x−55x−20=−27$
$\Rightarrow −x=20−27$
$\Rightarrow x=7$
Substituting this value of x in (3)
$y={11x+4}{9}$
$=\frac{11(7)+4}{9}=\frac{81}{9}=9$
Hence, the required fraction is
$x =\frac{7}{9}$
Answer:
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, the present age of Jacob is 40 years and the present age of Jacob's son is 10 years.
Also Read-
Take Aakash iACST and get instant scholarship on coaching programs.
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)
The substitution method is defined as substituting the value of one variable in another equation to obtain the value of another variable
In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.2, there are three questions.
Putting y=2 in 6x-y=10
6x-2=10
6x=10+2
Thus x =12/6=2
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.
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)
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
The General form is ax+by+c=0, where a, b, c are real numbers.
Yes, we can use the substitution method to solve the system of equations in three variables.
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.
Take Aakash iACST and get instant scholarship on coaching programs.
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