CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
NCERT Solutions for class 10 maths ex 3.5 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 3.5 class 10 deals with the cross multiplication method which is one of the methods to solve the pair of linear equations. In exercise 3.5 Class 10 Maths, pair of linear equations can also be solved by many non-graphical methods such as elimination method, substitution method and cross multiplication method but the NCERT book Class 10 Maths chapter 3 exercise 3.5 focuses on solving the equations only by cross multiplication method it is the easiest and straight forward method.
These class 10 maths ex 3.5 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.
Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.5
$\\x - 3y -3 = 0\\ 3x - 9y -2 = 0$
Answer:
Given, two equations,
$\\x - 3y -3 = 0.........(1)\\ 3x - 9y -2 = 0........(2)$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}$
$\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}$
As we can see,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Hence, the pair of equations has no solution.
$\\2x + y = 5 \\ 3x + 2y = 8$
Answer:
Given, two equations,
$\\2x + y = 5.........(1) \\ 3x + 2y = 8..........(2)$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{2}{3}$
$\frac{b_1}{b_2}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{5}{8}$
As we can see,
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Hence, the pair of equations has exactly one solution.
By Cross multiplication method,
$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$
$\frac{x}{(1)(-8)-(2)(-5)}=\frac{y}{(-5)(3)-(-8)(2)}=\frac{1}{(2)(2)-(3)(1)}$
$\frac{x}{2}=\frac{y}{1}=\frac{1}{1}$
$x=2,\:and\:y=1$
$\\3x -5 y = 20\\ 6x - 10y = 40$
Answer:
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{20}{40}=\frac{1}{2}$
As we can see,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Hence, the pair of equations has infinitely many solutions.
$\\x - 3y -7 = 0\\ 3x -3y -15 =0$
Answer:
Given the equations,
$\\x - 3y -7 = 0.........(1)\\ 3x -3y -15 =0........(2)$
Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{a_1}{a_2}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{-3}{-3}=1$
$\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$
As we can see,
$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Hence, the pair of equations has exactly one solution.
By Cross multiplication method,
$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$
$\frac{x}{(-3)(-15)-(-3)(-7)}=\frac{y}{(-7)(3)-(-15)(1)}=\frac{1}{(1)(-3)-(3)(-3)}$
$\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$
$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$
$x=\frac{24}{6}=4,\:and\:y=-1$
Answer:
Given equations,
$\\2x + 3y = 7 \\(a - b) x + (a + b) y = 3a + b - 2$
As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have an infinite solution is
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3a+b-2}$
From here we get,
$\frac{2}{a-b}=\frac{3}{a+b}$
$\Rightarrow 2(a+b)=3(a-b)$
$\Rightarrow 2a+2b=3a-3b$
$\Rightarrow a-5b=0.........(1)$
Also,
$\frac{2}{a-b}=\frac{7}{3a+b-2}$
$\Rightarrow 2(3a+b-2)=7(a-b)$
$\Rightarrow 6a+2b-4=7a-7b$
$\Rightarrow a-9b+4=0...........(2)$
Now, Subtracting (2) from (1) we get
$\Rightarrow 4b-4=0$
$\Rightarrow b=1$
Substituting this value in (1)
$\Rightarrow a-5(1)=0$
$\Rightarrow a=5$
Hence, $a=5\:and\:b=1$ .
Answer:
Given, the equations,
$\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$
As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have no solution is
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get
$\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{1}{2k+1}$
From here we get,
$\frac{3}{2k-1}=\frac{1}{k-1}$
$\Rightarrow 3(k-1)=2k-1$
$\Rightarrow 3k-3=2k-1$
$\Rightarrow 3k-2k=3-1$
$\Rightarrow k=2$
Hence, the value of K is 2.
Answer:
Given the equations
$\\8x + 5y = 9........(1) \\3x + 2y = 4........(2)$
By Substitution Method,
From (1) we have
$y=\frac{9-8x}{5}.........(3)$
Substituting this in (2),
$3x+2\left ( \frac{9-8x}{5} \right )=4$
$\Rightarrow 15x+18-16x=20$
$\Rightarrow -x=20-18$
$\Rightarrow x=-2$
Substituting this in (3)
$y=\frac{9-8x}{5}=\frac{9-8(-2)}{5}=\frac{25}{5}=5$
Hence $x=-2\:and\:y=5$ .
By Cross Multiplication Method
$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$
$\frac{x}{(5)(-4)-(2)(-9)}=\frac{y}{(3)(-9)-(8)(-4)}=\frac{1}{(8)(2)-(3)(5)}$
$\frac{x}{-20+18}=\frac{y}{32-27}=\frac{1}{16-15}$
$\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}$
$x=-2,\:and\:y=5$
Answer:
Let the fixed charge be x and the cost of food per day is y,
Now, According to the question
$x+20y=1000.........(1)$
Also
$x+26y=1180.........(2)$
Now subtracting (1) from (2),
$x+26y-x-20y=1180-100$
$\Rightarrow 6y=180$
$\Rightarrow y=30$
Putting this value in (1)
$x+20(30)=1000$
$\Rightarrow x=1000-600$
$\Rightarrow x=400$
Hence, the Fixed charge is Rs 400 and the cost of food per day is Rs 30.
Answer:
Let numerator of a fraction be x and the denominator is y.
Now, According to the question,
$\frac{x-1}{y}=\frac{1}{3}$
$\Rightarrow 3(x-1)=y$
$\Rightarrow 3x-3=y$
$\Rightarrow 3x-y=3........(1)$
Also,
$\frac{x}{y+8}=\frac{1}{4}$
$\Rightarrow 4x=y+8$
$\Rightarrow 4x-y=8.........(2)$
Now, Subtracting (1) from (2) we get,
$4x-3x=8-3$
$\Rightarrow x=5$
Putting this value in (2) we get,
$4(5)-y=8$
$\Rightarrow y=20-8$
$\Rightarrow y=12$
Hence, the fraction is
$\frac{x}{y}=\frac{5}{12}$ .
Answer:
Let the number of right answer and wrong answer be x and y respectively
Now, According to the question,
$3x-y=40..........(1)$
And
$\\4x-2y=50\\\Rightarrow 2x-y=25..........(2)$
Now, subtracting (2) from (1) we get,
$x=40-25$
$x=15$
Putting this value in (1)
$3(15)-y=40$
$\Rightarrow y=45-40$
$\Rightarrow y=5$
Hence the total number of question is $x+y=15+5=20.$
Answer:
Let the speed of the first car is x and the speed of the second car is y.
Let's solve this problem by using relative motion concept,
the relative speed when they are going in the same direction= x - y
the relative speed when they are going in the opposite direction= x + y
The given relative distance between them = 100 km.
Now, As we know,
Relative distance = Relative speed * time .
So, According to the question,
$5\times(x-y)=100$
$\Rightarrow 5x-5y=100$
$\Rightarrow x-y=20.........(1)$
Also,
$1(x+y)=100$
$\Rightarrow x+y=100........(2)$
Now Adding (1) and (2) we get
$2x=120$
$\Rightarrow x=60$
putting this in (1)
$60-y=20$
$\Rightarrow y=60-20$
$\Rightarrow y=40$
Hence the speeds of the cars are 40 km/hour and 60 km/hour.
Answer:
Let $l$be the length of the rectangle and $b$ be the width,
Now, According to the question,
$(l-5)(b+3)=lb-9$
$\Rightarrow lb+3l-5b-15=lb-9$
$\Rightarrow 3l-5b-6=0..........(1)$
Also,
$(l+3)(b+2)=lb+67$
$\Rightarrow lb+2l+3b+6=lb+67$
$\Rightarrow 2l+3b-61=0..........(2)$
By Cross multiplication method,
$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$
$\frac{x}{(-5)(-61)-(3)(-6)}=\frac{y}{(-6)(2)-(-61)(3)}=\frac{1}{(3)(3)-(2)(-5)}$
$\frac{x}{305+18}=\frac{y}{-12+183}=\frac{1}{9+10}$
$\frac{x}{323}=\frac{y}{171}=\frac{1}{19}$
$x=17,\:and\:y=9$
Hence the length and width of the rectangle are 17 units and 9 units respectively.
Class 10 Maths chapter 3 exercise 3.5: The questions in exercise 3.5 Class 10 Maths, broadly consist of four types of questions. In question one we have to identify whether the pair of equations has a unique solution, no solution, or infinitely many solutions. In question two we have to find the extra variable a and b they can be found with the help of the condition which is applied to the question of equation three is a direct question in which we have to solve both substitution method and cross multiplication method. in question four there are word problems based on real-life application. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.
Also see-
Frequently Asked Questions (FAQs)
We might remark that the solutions to the pair of linear equations are inconsistent.
There are mainly 3 questions that are solved before the Class 10 Maths chapter 3 exercise 3.5
There are 4 questions and question 1 has 4 subparts, question 2 has 2 subparts and question 4 has 5 subparts.
Graphical method
Algebraic method
According to NCERT solutions for Class 10 Maths 1 exercise 3.5 there are three ways to solve a pair of linear equations with a single variable by the algebraic method.
Substitution method
Elimination method
Cross-multiplication Method
Statement with no variable means all the values for that variable is true we can say this when the variable itself get eliminated from the equation leaving the same constant at both sides
The pair of linear equations can be said to have an infinite number of solutions.
On Question asked by student community
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