NCERT Exemplar Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables

# NCERT Exemplar Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables

Edited By Ravindra Pindel | Updated on Sep 05, 2022 12:47 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 3 will help the students to solve pair of linear equations in two variables in an easy way. This chapter provides a detailed understanding of Linear equations in two variables. The NCERT exemplar Class 10 Maths chapter 3 solutions clarifies the difficulties faced by the students while practicing exemplar and prove to be sufficient material to study NCERT Class 10 Maths.

These NCERT exemplar Class 10 Maths chapter 3 solutions are highly detailed and explanatory which covers the concepts related to linear equations in two variables. The NCERT exemplar Class 10 Maths solutions chapter 3 explores the CBSE Syllabus for Class 10 of this specific chapter along with additional interesting practice problems of NCERT.

Question:1

Choose the correct answer from the given four options:Graphically, the pair of equations 6x – 3y + 10 = 0, 2x – y + 9 =0 represents two lines which are
(A) intersecting at exactly one point.
(B) intersecting at exactly two points.
(C) coincident.
(D) parallel.

Solution:
In equation
6x – 3y + 10 = 0
a1 = 6, b1 = –3, c1 = 10
In equation
2x – y + 9 = 0
a2 = 2, b2 = –1, c2 = 9
$\frac{a_{1}}{a_{2}}= \frac{6}{2}= 3$
$\frac{b_{1}}{b_{2}}= \frac{-3}{-1}= 3$

$\frac{c_{1}}{c_{2}}= \frac{10}{9}$
Here, $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
The lines are parallel.

Question:2

The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have
(A) a unique solution
(B) exactly two solutions
(C) infinitely many solutions
(D) no solution

Solution:

In equation
x + 2y + 5 = 0
a1 = 1, b1 = 2, c1 = 5
In equation
–3x –6y + 1 = 0
a2 =–3, b2 = –6, c2 = 1
$\frac{a_{1}}{a_{2}}= \frac{1}{-3}$
$\frac{b_{1}}{b_{2}}= \frac{2}{-6}= \frac{-1}{3}$
$\frac{c_{1}}{c_{2}}= \frac{5}{1}= 5$
Here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Hence these lines are parallel to each other and have no solution.

Question:3

If a pair of linear equations is consistent, then the lines will be
(A) parallel
(B) always coincident
(C) intersecting or coincident
(D) always intersecting

If the pair of linear equations is consistent then they have unique or infinitely many solutions.
In unique solution, $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
Lines are intersecting
In infinitely many solution,
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
Lines are coincident
Hence the pair of lines are intersecting or coincident.

Question:4

The pair of equations y = 0 and y = –7 has
(A) one solution
(B) two solutions
(C) infinitely many solutions
(D) no solution

Solution:

Line y = 0 and y = –7 are parallel to each other.
Hence, they never intersect. Hence the pair of equations have no solution.

Question:5

The pair of equations x = a and y = b graphically represents lines which are
(A) parallel
(B) intersecting at (b, a)
(C) coincident
(D) intersecting at (a, b)

Solution:

The lines x = a and y = b are intersecting at (a, b)

Question:6

For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines?
(A) $\frac{1}{2}$
(B) –$-\frac{1}{2}$
(C) 2
(D) –2

Solution:

Equations are
3x – y + 8 = 0
6x – ky + 16 = 0
In equation
3x – y + 8 = 0
a1 = 3, b1 = –1, c1 = 8
In equation
6x – ky + 16 =0
a2 = 6, b2 = –k, c2 = 16
$\frac{a_{1}}{a_{2}}= \frac{3}{6}= \frac{1}{2}$
$\frac{b_{1}}{b_{2}}= \frac{-1}{-k}= \frac{1}{k}$
$\frac{c_{1}}{c_{2}}= \frac{8}{16}= \frac{1}{2}$
Since lines are coincident
Hence $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
$\frac{1}{2}= \frac{1}{k}= \frac{1}{2}$….(i)
1/k = 1/2 ( from equation (1))
k = 2 Hence the value of k is 2.

Question:7

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(A)$\frac{-5}{4}$
(B)$\frac{2}{5}$
(C)$\frac{15}{4}$|
(D)$\frac{3}{2}$

Solution:

Equations are
3x + 2ky = 2
2x + 5y + 1 = 0
In equation
3x + 2ky – 2 = 0
a1 = 3, b1 = 2k, c1 = –2
In equation
2x + 5y + 1 = 0
a2 = 2, b2 = 5, c2 = 1
$\frac{a_{1}}{a_{2}}= \frac{3}{2}$
$\frac{b_{1}}{b_{2}}= \frac{2k}{5}$
$\frac{c_{1}}{c_{2}}= \frac{-2}{1}=-2$
Since lines are parallel hence
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}$
$\frac{3}{2}=\frac{2k}{5}\neq-2$ ……(1)
$\frac{3}{2}=\frac{2k}{5}$ ( from equation (1))
$2k=\frac{15}{2}\Rightarrow k=\frac{15}{4}$
Hence the value of k is $\frac{15}{4}$ .

Question:8

The value of c for which the pair of linear equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is
(A) 3
(B) – 3
(C) –12
(D) no value

Solution:

Equations are
cx – y – 2 = 0, 6x – 2y – 3 = 0
In equation
cx – y – 2 = 0
a1 = c, b1 = –1, c1 = –2
In equation
6x – 2y – 3 = 0
a2 = 6, b2 = –2, c2 = –3
Since equations have infinitely many solutions
Hence $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
$\frac{c}{6}= \frac{-1}{-2}\neq \frac{-2}{-3}$
Here we find that $\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ i.e., equations are inconsistent
So, we can not assign any value to c.

Question:9

One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be
(A) 10x + 14y + 4 = 0
(B) –10x – 14y + 4 = 0
(C) –10x + 14y + 4 = 0
(D) 10x – 14y = –4

Solution:

Given equation is
–5x + 7y – 2 = 0
a1 = –5, b1 = 7, c1 = –2
For dependent linear equation
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}= \frac{1}{k}$
$\frac{-5}{a_{2}}= \frac{7}{b_{2}}= \frac{-2}{c_{2}}= \frac{1}{k}$
a2 = –5k, b2 = 7k, c2 = –2k
Put k = 2
a2 = –10, b2 = 14, c2 = –4
The revived equation is of the type a2x + b2y + c2 = 0
Put a2 = – 10, b2 = 14, c2= –4
–10x + 14y – 4 = 0
10x – 14 y = –4

Question:10

A pair of linear equations which has a unique solution x = 2, y = –3 is
(A) x + y = –1 ; 2x – 3y = –5
(B) 2x + 5y = –11 ; 4x + 10y = –22
(C) 2x – y = 1 ; 3x + 2y = 0
(D) x– 4y –14 = 0 ; 5x – y – 13 = 0

Solution:

A pair which have unique solution x = 2, y = –3 will satisfy L.H.S. = R.H.S. if we put x = 2 and y = –3
Put x = 2, y = –3 in option (A)
x + y = –1 ; 2x – 3y = –5
2 – 3 = –1 ; 2(2) – 3 (–3) = –5
–1 = –1 (True) ; 4 + 9 = –5
13 = – 5 (False)
Put x = 2, y = –3 in option (B)
2x + 5y = –11 ; 4x + 10y = –22
2(2) + 5(–3) = –11 ; 4(2) + 10 (–3) = –22
4 – 15 = –11 ; 8 – 30 = –22
–11 = –11 (True) –22 = –22 (True)
Put x = 2, y = – 3 in option (C)
2x – y = 1 ; 3x + 2y = 0
2(2) – (–3) = 1 ; 3(2) + 3 (–3) = 0
4 + 3 = 1 ; 6 – 6 = 0
7 = 1 (False) ; 0 = 0 (True)
Put x = 2, y = – 3 in option (D)
x – 4y – 14 = 0 ; 5x – y – 13 = 0
2 – 4 (–3) – 14 = 0 ; 5(2) – (–3) – 13 = 0
2 + 12 – 14 = 0 ; 10 + 3 – 13 = 0
0 = 0 (True) ; 13 – 13 = 0
0 = 0 (True)
By putting the value of x = 2, y = –3 in A, B, C, D; two options B, D satisfy L.H.S = R.H.S
Hence both (B, D) are correct

Question:11

If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then the values of a and b are, respectively
(A) 3 and 5
(B) 5 and 3
(C) 3 and 1
(D) –1 and –3

Solution:

Equation are x – y = 2 … (1)
x + y = 4 … (2)
x – y = 2
x + y = 4
2x = 6
x = 3
Put x = 3 in equation (2)
3 + y = 4
y = 4 – 3
y = 1
it is given that a = x, b = y
Hence a = 3, b = 1

Question:12

Aruna has only Rs 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs 1 and Rs 2 coins are, respectively
(A) 35 and 15
(B) 35 and 20
(C) 15 and 35
(D) 25 and 25

Solution:

Let Rs. 1 coins = x
Rs. 2 coins = y
As per the question:
x + y = 50 …(1)
x + 2y = 75 … (2)
Find value of x, y using elimination method.
x + y = 50
x + 2y = 75
y = 25 (subtract equation 1 from equation 2)
put y = 25 in equation (1)
x + 25 = 50
x = 25
x = 25, y = 25
Hence, Rs. 1 coins = 25
Rs. 2 coins = 25

Question:13

The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively
(A) 4 and 24
(B) 5 and 30
(C) 6 and 36
(D) 3 and 24

Solution:

According to questions at present father’s age is six times than son’s age.
Let son age = x
Father age = y
y = 6x … (1)
After 4 years Father’s age will be four times than son’s age. That is
y + 4 = 4 (x + 4)
y + 4 = 4x + 16
4x – y + 12 = 0 …(2)
Put y = 6x in equation (2)
4x – 6x + 12 = 0
+ 2x = + 12
x = $\frac{12}{2}$ = 6
Put x = 6 in equation (1)
y = 6 (6) = 36
y = 36
Hence present age of father is 36 and son is 6.

Question:1

Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x + 4y = 3 ; 12y + 6x = 6
(ii) x = 2y; y = 2x
(iii) 3x + y – 3 = 0 ;$2x+\frac{2}{3}y= 2$

(i)Solution:
Equations are 2x + 4y = 3
12y + 6x = 6
In equation 2x + 4y – 3 = 0
a1 = 2, b1 = 4, c1 = –3
In equation 12y + 6x – 6 = 0
a2 = 6, b2 = 12, c2 = –6
$\frac{a_{1}}{b_{2}}= \frac{2}{6}= \frac{1}{3},\frac{b_{1}}{b_{2}}= \frac{4}{12}= \frac{1}{3},\frac{c_{1}}{c_{2}}= \frac{-3}{-6}= \frac{1}{2}$
For no solution $\frac{a_{1}}{b_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ also, Here we have $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

Hence the given equation have no solution.
(ii)Solution:
Equations are x – 2y = 0
–2x + y = 0
In equation
x – 2y = 0
a1 = 1, b1 = –2, c1 = 0
In equation
–2x + y = 0
a2 = –2, b2 = 1, c2 = 0
$\frac{a_{1}}{b_{2}}= \frac{1}{-2},\frac{b_{1}}{b_{2}}= \frac{-2}{1}= \frac{c_{1}}{c_{2}}= \frac{0}{0}$
For no solution but$\frac{a_{1}}{b_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ here

$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Hence the given pair of equations a unique solution.
(iii)Solution:
Equations are 3x + y = 3
$2x+\frac{2}{3}y= 2$
In equation
3x + y – 3 = 0
a1 = 3, b1 = 1, c1 = –3
In equation
$2x+\frac{2}{3}y-2= 0$
a2 = 2, b2= $\frac{2}{3}$, c2 = –2
$\frac{a_{1}}{b_{2}}= \frac{3}{2},\frac{b_{1}}{b_{2}}= \frac{3}{2}= \frac{c_{1}}{c_{2}}= \frac{-3}{-2}= \frac{3}{2}$
For no solution $\frac{a_{1}}{b_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ but here
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
Hence the given pair of equations have infinity many solutions.

Question:2

Do the following equations represent a pair of coincident lines? Justify your answer.

(i)$3x+\frac{1}{7}y= 3$; 7x + 3y = 7
(ii) –2x – 3y = 1 ; 6y + 4x = –2
(iii)$\frac{x}{2}+y+\frac{2}{5}= 0;4x+8y+\frac{5}{6}= 0$

(i)Solution:
Equation are 3x + $\frac{1}{7}$y = 3
7x + 3y = 7
In equation
3x + $\frac{1}{7}$y – 3 = 0
a1 = 3 ; b1 = $\frac{1}{7}$ ; c1 = –3
In equation
7x + 3y – 7 = 0
a2 = 7 ; b2 = 3; c2 = –7
$\frac{a_{1}}{a_{2}}= \frac{3}{7};\frac{b_{1}}{b_{2}}= \frac{1}{21};\frac{c_{1}}{c_{2}}= \frac{-3}{-7}= \frac{3}{7}$
For coincident lines $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ but here $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
Hence the given pair of equations does not represent a pair of coincident lines.

(ii)Solution:
Equation are –2x – 3y = 1
4x + 6y = –2
In equation
–2x – 3y – 1 = 0
a1 = –2 ; b1 = –3 ; c1 = –1
In equation
4x + 6y + 2 = 0
a2 = 4 ; b2 = 6; c2 = 2
$\frac{a_{1}}{a_{2}}= \frac{-2}{4}= \frac{-1}{2};\frac{b_{1}}{b_{2}}= \frac{-3}{6}= \frac{-1}{2},\frac{c_{1}}{c_{2}}= \frac{-1}{2}$
For coincident lines $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ also here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
Hence the given pair of equation represents a pair of coincident lines.
Solution: (iii)
Equation are $\frac{x}{2}+y+\frac{2}{5}= 0$
4x + 8y + $\frac{5}{16}$= 0
In equation
$\frac{x}{2}+y+\frac{2}{5}= 0$
a1 = 1/2 ; b1 = 1 ; c1 = $\frac{2}{5}$
In equation
4x + 8y + $\frac{5}{16}$= 0
a2 = 4 ; b2 = 8; c2 = $\frac{5}{16}$
$\frac{a_{1}}{a_{2}}= \frac{1}{8};\frac{b_{1}}{b_{2}}= \frac{1}{8};\frac{c_{1}}{c_{2}}\Rightarrow \frac{2}{5}\times \frac{16}{5}\Rightarrow \frac{32}{25}$
For coincident lines $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$but here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Hence the given pair of equations does not represent a pair of coincident lines.

Question:3

(i) –3x– 4y = 12 ; 4y + 3x = 12
(ii)$\frac{3}{5}x-y= \frac{1}{2};\frac{1}{5}x-3y= \frac{1}{6}$ ;
(iii) 2ax + by = a; 4ax + 2by = 2a ; a, b ≠ 0
(iv) x + 3y = 11 ; 4x + 12y = 22

Solution: (i)
Equation are : –3x – 4y = 12
4y + 3x = 12
In equation
–3x – 4y – 12 = 0
a1 = –3; b1 = –4 ; c1 = –12
In equation
4y + 3x – 12 = 0
a2 = 3 ; b2 = 4; c2 = –12
$\frac{a_{1}}{a_{2}}= \frac{-3}{3}= -1;\frac{b_{1}}{b_{2}}= \frac{-4}{4}= -1,\frac{c_{1}}{c_{2}}\Rightarrow \frac{-12}{-12}= \frac{1}{1}$
For consistency either $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ but here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

Hence the pair of linear equations is not consistent.
Solution: (ii)
Equation are : $\frac{3}{5} x-y=\frac{1}{2}$
$\frac{1}{5} x-3 y=\frac{1}{6}$
In equation
$\frac{3}{5} x-y=\frac{1}{2}$
a1= $\frac{3}{5}$; b1 = –1 ; c1 = $-\frac{1}{2}$
In equation
$\frac{1}{5} x-3 y=\frac{1}{6}$
a2 = 1/5 ; b2 = – 3; c2 = –1/6
$\frac{a_{1}}{a_{2}}= \frac{3}{5}\times \frac{5}{1}= 3;\frac{b_{1}}{b_{2}}= \frac{-1}{-3}= \frac{1}{3};\frac{c_{1}}{c_{2}}= \frac{-1}{-2}\times \frac{6}{1}= 3$
For consistency either $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ also here $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
Hence given equations are consistent.
Solution: (iii)
Equation are: 2ax + by = a
4ax + 2by = 2a
In equation
2ax + by – a = 0
a1 = 2a; b1 = b; c1 = –a
In equation
4ax + 2by – 2a = 0
a2 = 4a ; b2 = 2b; c2 = –2a
$\frac{a_{1}}{a_{2}}= \frac{2a}{4a}= \frac{1}{2};\frac{b_{1}}{b_{2}}= \frac{b}{2b}= \frac{1}{2};\frac{c_{1}}{c_{2}}= \frac{-a}{-2a}= \frac{1}{2}$
For consistency either $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Also, here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Hence given equations are consistent
Solution: (iv)
Given, equations are x + 3 y = 11
4x + 12y =22
In equation
x + 3y = 11
a1 = 1; b1 = 3; c1 = –11
In equation
4x + 12y = 22
a2 = 4 ; b2 = 12; c2 = –22
$\frac{a_{1}}{a_{2}}= \frac{1}{4} ;\frac{b_{1}}{b_{2}}=\frac{3}{12}= \frac{1}{4};\frac{c_{1}}{c_{2}} = \frac{-11}{-22}= \frac{1}{2}$
For consistency either $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ but here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Hence given equations are not consistent.

Question:4

For the pair of equations $\lambda$x + 3y = –7 ,2x + 6y = 14to have infinitely many solutions, the value of $\lambda$ should be 1. Is the statement true?Give reasons.

Solution:
Here equations are $\lambda$ x + 3y = –7
2x + 6y = 14
a1 = $\lambda$ , b1 = 3, c1 = 7
a2 = 2, b2 = 6, c2 = –14
The equation have infinitely many solution
$\therefore \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
$\frac{\lambda }{2}= \frac{3}{6}\neq \frac{-7}{14}$
Here we can see that
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Hence no value of $\lambda$ exist because it is given that equation has infinitely many solutions
Hence the statement is false.

Question:5

For all real values of c, the pair of equations x– 2y = 8, 5x – 10y = c have a unique solution. Justify whether it is true or false.

Solution:
Equation are x – 2y = 8
5x – 10y = c
Here a1 = 1, b1 = –2, c1 = –8
a2 = 5, b2 = –10, c2 = –c
$\frac{a_{1}}{a_{2}}= \frac{1}{5};\frac{b_{1}}{b_{2}}= \frac{-2}{-10}= \frac{1}{5};\frac{c_{1}}{c_{2}}= \frac{-8}{-c}= \frac{8}{c}$
For unique solution $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ but here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$
Hence the given statement is false.

Question:6

The line represented by x = 7 is parallel to the x–axis. Justify whether the statement is true or not.

Solution:

From the above graph it is clear that x = 7 is not parallel to x axis but it is parallel to y-axis.Hence the given statement is false.

Question:1

For which value(s) of $\lambda$ , do the pair of linear equations $\lambda$x + y = $\lambda ^{2}$ and x +$\lambda$y = 1 have
(i) no solution?
(ii) infinitely many solutions?
(iii) a unique solution?

(i)Solution:
The given equations are
$\lambda$x + y =$\lambda ^{2}$ , x + $\lambda$y = 1
In equation
$\lambda$x + y – $\lambda ^{2}$ = 0
a1 = $\lambda$, b1 = 1, c1 = –$\lambda ^{2}$
In equation
x + $\lambda$y – 1 = 0
a2 =1, b2 = $\lambda$, c2 = –1
For no solution $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
$\frac{a_{1}}{a_{2}}= \frac{\lambda }{1}$
$\frac{b_{1}}{b_{2}}= \frac{1}{\lambda }$
$\frac{c_{1}}{c_{2}}= \frac{-\lambda^{2}}{ -1}= \lambda^{2}$
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
$\frac{\lambda }{1}= \frac{1}{\lambda}\neq \lambda^{2}$
$\frac{\lambda }{1}= \frac{1}{\lambda}$
$\lambda ^{2}= 1$
$\lambda$=1,-1
Hence value of $\lambda$ is -1
$\lambda \neq 1$ because in this case $\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$)
(ii)Solution:
The given equations are
$\lambda$x + y =$\lambda ^{2}$ , x + $\lambda$y = 1
In equation
$\lambda$x + y – $\lambda ^{2}$ = 0
a1 = $\lambda$, b1 = 1, c1 = –$\lambda ^{2}$
In equation
x + $\lambda$y – 1 = 0
a2 =1, b2 = $\lambda$, c2 = –1
$\frac{a_{1}}{a_{2}}=\frac{\lambda }{1}; \frac{b_{1}}{b_{2}}= \frac{1}{\lambda };\frac{c_{1}}{c_{2}}=\frac{-\lambda^{2} }{-1}= \lambda ^{2}$
For infinite many solution
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
$\frac{\lambda }{1}= \frac{1}{\lambda }= \lambda ^{2}$
Only one value of l satisfy all the three equations, that is $\lambda$ = 1
(iii) Solution:
The given equations are
$\lambda$x + y = $\lambda^{2}$, x + $\lambda$y = 1
In equation
$\lambda$x + y – $\lambda^{2}$ = 0
a1 =$\lambda$ , b1 = 1, c1 = -$\lambda^{2}$
In equation
x + $\lambda$y = 1
a2 =1, b2 = $\lambda$, c2 =–1
$\frac{a_{1}}{a_{2}}= \frac{\lambda }{1};\frac{b_{1}}{b_{2}}= \frac{1}{\lambda};\frac{c_{1}}{c_{2}}= \frac{-\lambda^{2} }{-1}= \lambda ^{2}$
For unique solution $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
$\frac{\lambda }{1}\neq \frac{1}{\lambda }$
$\lambda ^{2}\neq 1$
$\lambda \neq \pm 1$
All real values of $\lambda$ except $\lambda \neq \pm 1$

Question:2

For which value(s) of k will the pair of equations kx+ 3y = k – 3, 12x + ky= k have no solution?

Solution:
The given equations are
kx + 3y = k –3
12x + ky = k
In equation
kx + 3y = k – 3
a1 = k, b1 = 3, c1 = –k + 3
In equation
12x + ky = k
a2 = 12, b2 = k, c2 = –k
$a_{2}= 12,b_{2}= k,c_{2}= -k$
$\frac{a_{1}}{a_{2}}= \frac{k}{12},\frac{b_{1}}{b_{2}}= \frac{3}{k},\frac{c_{1}}{c_{2}}= \frac{-k+3}{-k}= \frac{k-3}{k}$
We know that if the equations have no solution then
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
$\frac{k}{12}= \frac{3}{k}\equiv \frac{k-3}{k}$
$\frac{k}{12}= \frac{3}{k}$
$k^{2}$ = 36
$k= \pm 36$
($k\neq 6$ because if $k= +6$, then $\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ which is not possible in no solution.)
Hence value of k is –6.

Question:3

For which values of a and b, will the following pair of linear equations have infinitely many solutions?
x + 2y =1, (a – b)x+ (a + b)y = a + b – 2

Solution:
The given equations are
x + 2y = 1
(a – b) x + (a + b)y = a + b – 2
In equation
x + 2y = 1
a1 = 1, b1 = 2, c1 = –1
In equation
(a – b)x (a + b)y = a + b – 2
a2 = (a – b), b2 = (a + b), c2 = – (a + b – 2)
$\frac{a_{1}}{a_{2}}= \frac{1}{\left ( a-b \right )},\frac{b_{1}}{b_{2}}= \frac{2}{\left ( a+b \right )},\frac{c_{1}}{c_{2}}= \frac{-1}{-\left ( a+b-2 \right )}$
For infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$

$\frac{1}{\left ( a-b \right )}= \frac{2}{\left ( a+b \right )}= \frac{1}{a+b-2 }$

 $\frac{1}{\left ( a-b \right )}= \frac{2}{\left ( a+b \right )}\; \;$ a + b = 2a –2b –a + 3b = 0 … (1) $\frac{2}{\left ( a+b \right )}= \frac{1}{a+b-2}$ 2a + 2b – 4 = a + b a + b –4 = 0 …(2)

–a + 3b + a + b = 0 + 4
4b = 4
b = 1
Put b = 1 in equation (1)
–a + 3 (1) = 0
a = 3
Hence a = 3 and b = 1.

Question:4

Find the value(s) of p in for the following pair of equations:
(i) 3x – y – 5 = 0 and 6x – 2y – p = 0, if the lines represented by these equations are parallel.

(ii) – x + py = 1 and px – y = 1,if the pair of equations has no solution.

(iii) – 3x + 5y = 7 and 2px – 3y = 1,
if the lines represented by these equations are intersecting at a unique point.

(iv) 2x + 3y – 5 = 0 and px– 6y – 8 = 0,if the pair of equations has a unique solution.

(v) 2x + 3y = 7 and 2px + py = 28 – qy,if the pair of equations have infinitely many solutions.

(i)Solution:
In equation 3x – y – 5 = 0
a1 = 3, b1 = –1, c1 = –5
In equation 6x – 2y – p = 0
a2 = 6, b2 = –2, c2 = –p
$\frac{a_{1}}{a_{2}}= \frac{3}{6}= \frac{1}{2};\frac{b_{1}}{b_{2}}= \frac{-1}{-2}= \frac{1}{2};\frac{c_{1}}{c_{2}}= \frac{-5}{-p}= \frac{5}{p}$
If the lines are parallel $\frac{a_{1}}{a_{1}}= \frac{b_{1}}{b_{1}}\neq \frac{c_{1}}{c_{1}}$
$\frac{1}{2}= \frac{1}{2}\neq \frac{5}{p}$
$\frac{1}{2}\neq \frac{5}{p}\Rightarrow p\neq 10$
Any real value of p except 10.
(ii)Solution:
Here the equations are
–x + py = 1, px – y = 1
In equation
–x + py = 1
a1 = –1, b1 = p, c1 = –1
In equation
px – y = 1
a2 = p, b2 = –1, c2 = –1
$\frac{a_{1}}{a_{2}}= \frac{-1}{p};\frac{b_{1}}{b_{2}}= \frac{p}{-1}= -p,\frac{c_{1}}{c_{2}}= \frac{-1}{1}= 1$
For no solution $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
$\frac{-1}{p}= -p\neq 1$
$\frac{+1}{p}= +p$
$p^{2}$ = 1
p = ± 1
p = + 1 ($p\neq -1$ because if p = –1 then $\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$)

(iii)Solution:
Here the equation are –3x + 5y = 7, 2px – 3y = 1
In equation –3x + 5y = 7
a1 = –3, b1 = 5, c1 = –7
In equation 2px – 3y = 1
a2 = 2p, b2 = –3, c2 = –1
$\frac{a_{1}}{a_{2}}= \frac{-3}{2p};\frac{b_{1}}{b_{2}}= \frac{5}{-3},\frac{c_{1}}{c_{2}}= \frac{-7}{-1}= 7$
For unique solution $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
$\frac{-3}{2p}\neq \frac{-5}{3}$
$-9\neq -10p$
$p\neq \frac{9}{10}$
$p\neq 0\cdot 9$
Hence p can have any real value for unique solution except 0.9

(iv)Solution:
Here the equation are 2x + 3y – 5 = 0
px – 6y –8 = 0
In equation 2x + 3y – 5 = 0
a1 = 2, b1 = 3, c1 = –5
In equation px – 6y – 8 = 0
a2 = p, b2 = –6, c2 = –8
$\frac{a_{1}}{a_{2}}= \frac{2}{p};\frac{b_{1}}{b_{2}}= \frac{3}{-6}= \frac{-1}{2};\frac{c_{1}}{c_{2}}= \frac{-5}{-8}= \frac{5}{8}$
For unique solution $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
$\frac{2}{p}\neq \frac{-1}{2}$
$4\neq -p$
$p\neq -4$
Hence the equations have unique solution for every real value of p except –4.

(v)Solution:
Here the given equations are 2x + 3y = 7
a1 = 2, b1 = 3, c1 = –7
In equation 2px + py = 28 – qy
2px + py + qy = 28
2px + y(p + q) = 28
a2 = 2p, b2 = p + q, c2 = –28
$\frac{a_{1}}{a_{2}}= \frac{2}{2p}= \frac{1}{p};\frac{b_{1}}{b_{2}}= \frac{3}{p+q};\frac{c_{1}}{c_{2}}= \frac{-7}{-28}= \frac{1}{4}$
For infinite many solution $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
$\frac{1}{p}= \frac{3}{p+q}= \frac{1}{4}$
$\frac{1}{p}= \frac{1}{4}$
4 = p
p = 4
$\frac{3}{p+q}= \frac{1}{4}$
p + q = 12
Put p =4
q=12-4
q=8
For infinitely many solutions, p = 4, q = 8

Question:5

Two straight paths are represented by the equations x – 3y = 2 and –2x + 6y = 5.Check whether the paths cross each other or not

Solution:
Equations are: x – 3y – 2 = 0
–2x + 6y – 5 = 0
Here a1 = 1, b1 = –3, c1 = –2
a2 = –2, b2 = 6, c2 = –5

$\frac{a_{1}}{a_{2}}= -\frac{1}{2},\frac{b_{1}}{b_{2}}= \frac{-3}{6}= -\frac{1}{2},\frac{c_{1}}{c_{2}}= \frac{-2}{-5}= \frac{2}{5}$
As we know that two lines cross each other when $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ but here $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

Hence the paths do not cross each other.

Question:6

Write a pair of linear equations which has the unique solution x = – 1, y = 3. How many such pairs can you write?

Solution:
Let the equations be
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Put x = –1 and y = 3 in given equations.
–a1 + 3b1 + c1 = 0 … (1)
–a2 + 3b2 + c2 = 0 … (2)
Since for different values of a1, b2, c1 and a2, b2, c2 we have different solutions.
Hence infinite many pairs having x = –1, y = 3 as solution.

Question:7

If 2x + y = 23 and 4x – y = 19, find the values of 5y – 2x and $\frac{y}{x}-2$

Solution:
Given equations are
2x + y = 23 … (1)
4x – y = 19 … (2)
Using elimination method in equation (1) and (2) we get
2x + y = 23
4x – y = 19
6x = 42
x = $\frac{42}{6}= 7$
Put x = 7 in equation (1) we get
2(7) + y = 23
y = 23 – 14
y = 9
Now 5y – 2x = 5(9) – 2 (7)
= 45 – 14
= 31
$\frac{y}{x}- 2= \frac{9}{7}-2$
$\frac{y-14}{7}= \frac{-5}{7}$

Question:8

Find the values of x and y in the following rectangle [see Fig. ].

Solution:
We know that opposite sides of a rectangle are equal
3x + y = 7 … (1)
x + 3y = 13 … (2)
Find the value of x and y using elimination method in equation (1) and (2)
9x + 3y = 21 {Multiply (1) by 3}
x + 3y = 13
-___-___-_______
8x = 8
x = $\frac{8}{8}$ = 1
Put x = 1 in equation (1)
3(1) + y = 7
3 + y = 7
y = 7 – 3
y = 4

Question:9

Solve the following pairs of equations:
(i) x+ y = 3.3 ;$\frac{0\cdot 6}{3x-2y}= -1,3x-2y\neq 0$
(ii)$\frac{x}{3}+\frac{y}{4}= 4$ , $\frac{5x}{6}-\frac{y}{8}= 4$
(iii)$4x+\frac{6}{y}= 15;6x-\frac{8}{y}= 14,y\neq 0$
(iv)$\frac{1}{2x}-\frac{1}{y}= -1;\frac{1}{x}+\frac{1}{2y}= 8,x,y\neq 0$
(v)43x + 67y = – 24 ; 67x + 43y = 24
(vi)$\frac{x}{a}+\frac{y}{b}= a+b;\frac{x}{a_{2}}+\frac{y}{b^{2}}= 2 \left ( a,b\neq 0 \right )$

(vii)$\frac{2xy}{xy}= \frac{2}{3};\frac{xy}{2x-y}= \frac{-3}{10};x+y\neq 0,2x-y\neq 0$

(i)Solution:
Given equations are x + y = 3.3 … (1)
0.6 = 2y – 3x …(2)
Using elimination method in equation (1) and (2) we get
2x + 2y = 6.6 [multiply eq. (1) by 2]
–3x + 2y = 0.6
+ – –
5x = 6
x = $\frac{6}{5}$ = 1.2
Put x = 1.2 in (1) we get
1.2 + y = 3.3
y = 3.3 – 1.2
y = 2.1
(ii)Answer: y = 6 ; y = 8
Solution:

Given equations are :
$\frac{x}{3}+\frac{y}{4}= 4$ ………(1)

$\frac{5x}{6}-\frac{y}{8}= 4$ ……….(2)
Multiply eq. (1) by 12 and eq. (2) by 24 and then add them.
We get
4x + 3y + 20x -3y = 48 + 96
24x = 144
$x= \frac{144}{24}= 6$
x = 6
Put x = 6 in eq. (1) we have
$\frac{6}{3}+\frac{y}{4}= 4$
$2+\frac{y}{4}= 4$
$\frac{y}{4}= 4-2$
$\frac{y}{4}= 2$
y = 8
(iii)Solution:
Equation are: $4 x+\frac{6}{y}=15 \\$… (1)
$6 x-\frac{8}{y}=14$… (2)
Multiply equation (1) by 6 and equation (2) by 4 and then subtract equation (2) from (1)
24x + $\frac{36}{y}$ = 90
24x – 32/y = 56
– + –

$\frac{36}{y}+\frac{32}{y}= 34$
$\frac{36+32}{y}= 34$
68 = 34y
y = $\frac{68}{34}= 2$
y = 2
Put y = 2 in equation (1) we get

4x = 15 – 3
4x = 12
x = $\frac{12}{4}$ = 3
x = 3
(iv)Solution:
Let $\frac{1}{x}$ = u and $\frac{1}{y}$ = v and put in the given equations
u/2 – v = – 1 … (1)
u + $\frac{v}{2}$ = 8 … (2)
Multiply eq. (1) by 2 and subtract from eq. (2) we get
$\frac{v}{2}+2v= 8+2$
v + 4v = 10 × 2
5v = 20
v = $\frac{20}{+5}= 4$
v = 4
v = 1/y = 4$\Rightarrow$
y = $\frac{1}{4}$
Put v = 4 in equation (1) we get
u/2= –1 + 4
u = 6
$\frac{1}{x} = 6$
(because 1/x = u)
$\Rightarrow x= \frac{1}{6}$
(v)Solution:
Equations are 43x + 67y = –24 … (1)
67x + 43y = 24 … (2)
Multiply equation (1) by 67 and equation (2) by 43 and then subtract equation (2) from (1)
2881x + 4489y = –1608
2881x + 1849y = 1032
– – –
2640y = –2640
y = –1
Put y = –1 in (1) we get
43x + 67(–1) = –24
43x = –24 + 67
43x = 43
x = 1
(vi)Solution:
Given equation are:
$\frac{x}{a}+\frac{y}{b}= a+b$

$\frac{x}{a^{2}}+\frac{y}{b^{2}}= 2$

$\frac{x}{a}+\frac{y}{b}-\left ( a+b \right )= 0$

$\frac{x}{a^{2}}+\frac{y}{b^{2}}-2= 0$
Using cross multiplication, we have

$\frac{x}{\frac{-2}{b}+\frac{a}{b^{2}}+\frac{1}{b}}= \frac{-y}{\frac{-2}{a}+\frac{1}{a}+\frac{b}{b^{2}}}= \frac{1}{\frac{1}{ab^{2}}-\frac{1}{ba^{2}}}$
$\frac{x}{\frac{a-b}{b^{2}}}= \frac{-y}{\frac{-a+b}{a^{2}}}= \frac{1}{\frac{1}{ab^{2}}-\frac{1}{b^{2}}}$
$\frac{x}{\frac{a-b}{b^{2}}}=\frac{1}{\frac{+\left ( a-b \right )}{a^{2}b^{2}}}$
$x= \frac{\left ( a-b \right )\times a^{2} b^{2}}{b^{2}\left ( a-b \right )}$
x = a2
$\frac{-y}{\frac{-\left ( a-b \right )}{a^{2}}}= \frac{1}{\frac{\left ( a-b \right )}{a^{2}b^{2}}}$
y = $\frac{\left ( a-b \right )a^{2}b^{2}}{a^{2}\left ( a-b \right )}$
y = b2
(vii)Solution:
$\frac{2xy}{x+y}= \frac{3}{2}$ … (1)
$\frac{xy}{2x+y}= \frac{-3}{10}$… (2)
Inverse equation (1) & (2) then simplify them we get
$\frac{x}{2xy}+\frac{y}{2xy}= \frac{2}{3}$
$\frac{1}{2y}+\frac{1}{2x}= \frac{2}{3}$ … (3)
$\frac{2x}{xy}-\frac{y}{xy}= \frac{-10}{3}$
$\frac{2}{y}-\frac{1}{x}= \frac{-10}{3}$…(4)
Divide equation (4) by 2 we get
$\frac{2}{2y}-\frac{1}{2x}= \frac{-5}{3}$ … (5)
Add (3) and (5) we get
$\frac{2}{2y}+\frac{1}{2y}= \frac{-5}{3}+\frac{2}{3}$
$\frac{2+1}{2y}= -\frac{5+2}{3}= \frac{-3}{3}= -1$
2 + 1 = –2y
y = $-\frac{3}{2}$
Put y = $-\frac{3}{2}$ in equation (4) we have
$\frac{4}{-3}-\frac{1}{x}= \frac{-10}{3}$
$-\frac{10}{3}+\frac{4}{3}= \frac{-1}{x}$
$-\frac{6}{3}= \frac{-1}{x}$
$x= \frac{3}{6}= \frac{1}{2}$

Question:10

Find the solution of the pair of equation $\frac{x}{10}+\frac{y}{5}-1= 0$ and $\frac{x}{8}+\frac{y}{6}= 15$.
Hence, find $\lambda$, if y = $\lambda$x + 5.

Solution:

$\frac{x}{10}+\frac{y}{5}= 1\Rightarrow \frac{x+2y}{10}= 1\Rightarrow$ x + 2y = 10 …(1)

$\frac{x}{8}+\frac{y}{6}= 15\Rightarrow \frac{3x+4y}{24}= 15\Rightarrow$ 3x + 4y = 360 … (2)
Multiply eq. (1) by 2 and subtract from eq. (2) we get
3x – 2x = 360 – 20.
x = 340
Put x = 340 in eq. 1 we get
340 + 2y = 10
2y = 10 – 340
y = $-\frac{330}{2}$
y = – 165
y = $\lambda$x + 5
Put x = 340 and y = –165 we get
–165 = $\lambda$(340) + 5
–170 = $\lambda$(340)
$\lambda = \frac{-1}{2}$

Question:11

By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them. (i) 3x + y + 4 = 0 ; 6x– 2y + 4 = 0
(ii) x– 2y = 6 ; 3x – 6y = 0
(iii) x + y = 3 ; 3x + 3y = 9

(i)Solution:
Given equations are 3x + y + 4 = 0
6x – 2y + 4 = 0
Here a1 = 3, b1 = 1, c1 = 4
a2 = 6, b2 = –2, c2 = 4
$\frac{a_{1}}{a_{2}}= \frac{3}{6}= \frac{1}{2},\frac{b_{1}}{b_{2}}= \frac{-1}{2},\frac{c_{1}}{c_{2}}= \frac{4}{4}= \frac{1}{1}$
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
Hence the given pair of linear equations are intersecting at one point.
Hence given pair of linear equations is consistent
Let as plot the graph of given equations
In equation 3x + y + 4 = 0

 x 0 -1 -2 y -4 -1 2

In equation 6x – 2y + 4 = 0

 x 0 -1 -2 y 2 -1 -4

Here two lines AB and CD intersect at only one point that is E.Hence given pair of linear equation is consistent.(ii)Solution:
Given equations are x –2y –6 = 0
3x – 6y = 0
Here a1 = 1, b1 = –2, c1 = –6
a2 = 3, b2 = –6, c2 = 0
$\frac{a_{1}}{a_{2}}= \frac{1}{3},\frac{b_{1}}{b_{2}}= \frac{-2}{6}= \frac{1}{3},\frac{c_{1}}{c_{2}}= \frac{-6}{0}$
Here $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Here the given pair of linear equations is parallel. Therefore it has no solution Hence given pair of linear equations is inconsistent
(iii)Solution:
The given equations are x + y – 3 = 0
3x +3y – 9 = 0
Here a1 = 1, b1 = 1, c1 = –3
a2 = 3, b2 = 3, c2 = –9
$\frac{a_{1}}{a_{2}}= \frac{1}{3},\frac{b_{1}}{b_{2}}= \frac{1}{3},\frac{c_{1}}{c_{2}}= \frac{-3}{-9}= \frac{1}{3}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$
Therefore the given pair of equations is coincident and have infinitely many solution.Hence, the given pair of linear equations is consistent Let us plot the graph of given equations
x + y = 3

 x 0 1 2 y 3 2 1

3x + 3y = 9

 x 0 1 2 y 3 2 1

Here lines AB and CD is coincident.Therefore the above linear equations have infinitely many solutions.

Question:12

Draw the graph of the pair of equations 2x + y = 4 and 2x – y = 4. Write the vertices of the triangle formed by these lines and the y-axis. Also find the area of this triangle.

Solution:
Given equation are:
2x + y = 4
2x –y = 4
2x + y = 4

 x 0 1 2

 y 4 2 0

2x-y=4

 x 0 1 2 y -4 -2 0

Hence from the above graph it is clear that both the lines and y-axis makes triangle ABC.
Now, area of triangle ABC is
2(area of $\bigtriangleup$AOB)
$2\left ( \frac{1}{2}\times OB\times OA\right )$

$2\left ( \frac{1}{2}\times 2\times 4\right )$
2(4)
8 sq. units.
Hence, required area of triangle is 8 sq. units.

Question:13

Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x – y = 1. How many such lines can we find?

Solution:
Given equations are x + y = 2
2x – y = 1
x + y = 2

 x 1 2 3 y 1 0 -1

2x – y = 1

 x 1 2 3 y 1 3 5

From the graph it is clear that two lines AB and CD intersect and point E (1, 1) Hence infinite lines can pass through the intersecting point E like 4 = x, 2x – y = 1 etc.

Question:14

If x + 1 is a factor of 2x3+ ax2+ 2bx + 1, then find the values of a and b given that 2a – 3b = 4.

Solution:
It is given that (x + 1) is a factor of P(x) = 2x3 + ax2 + 2bx + 1 then according to remainder theorem P(–1) is equal to 0.
P(–1) = 2(–1)3 + a(–1)2 + 2b (–1) + 1
0 = –2 + a – 2b + 1
$\Rightarrow$a – 2b = 1 … (1)
Given 2a – 3b = 4 … (2)
Solving eq. (1) and (2) by substitution method
2a – 4b = 2 {multiply eq. (1) by 2}
2a – 3b = 4
– + –
–b = –2
b = 2
Put b = 2 in equation (1)
a – 2(2) = 1
a – 4 = 1
a = 1 + 4
a = 5

Question:15

The angles of a triangle are x, y and $40^{\circ}$. The difference between the two angles x and y is $30^{\circ}$. Find x and y.

Solution:
Given: x – y = $30^{\circ}$ …(1)

According to sum of interior angles of a triangle’ property
x + y + $40^{\circ}$ = 180
x + y = $140^{\circ}$ … (2)
Solve eq. (1) and eq. (2) using substitution method.
x – y = $30^{\circ}$
x + y = $140^{\circ}$
2x = 170
x = $\frac{170}{2}$
x = $85^{\circ}$
Put x = $85^{\circ}$ in eq. (1) we get
$85^{\circ}$– y =$30^{\circ}$
$85^{\circ}$$30^{\circ}$ = y
$55^{\circ}$ = y

Question:16

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

Solution: Let x is the age of salim
y is the age of his daughter
According to question
(x – 2) = 3(y – 2)
x – 2 = 3y – 6
x – 3y = –4 …(1)
According to second condition
x + 6 = 2(y + 6) + 4
x + 6 = 2y + 12 + 4
x – 2y = 10 … (2)
Solve (1) and (2) by substitution method
x – 3y = – 4
x – 2y = 10
– + –
–y = – 14
y = 14
Put y = 14 in equation (1)
x – 3(14) = – 4
x – 42 = –4
x = –4 + 42
x = 38
Salim’s age = 38 years
His daughter’s age = 14 years

Question:17

The age of the father is twice the sum of the ages of his two children. After 20years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Solution:
Let x be the age of father y and z be the ages of children
According to the question:
x = 2 (y + z) … (1)
After 20 years
x + 20 = (y + 20) + (z + 20)
x + 20 = y + z + 40
x + 20 – 40 = y + z
x – 20 = y + z … (2)
Put the value of (y + z) from equation (2) in equation (1) we get
x = 2 (x – 20)
x = 2x – 40
x – 2x = – 40
–x = – 40
x = 40
Hence father’s age is 40 years

Question:18

Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers

Solution:
Let x and y be two numbers
According to question
x : y = 5 : 6
$\frac{x}{y}= \frac{5}{6}$
x = $\frac{5y}{6}$ … (1)
According to second condition
$\frac{x-8}{y-8}= \frac{4}{5}$
5x – 40 = 4y – 32
5x – 4y = –32 + 40
5x – 4y = 8 … (2)
Put value of x from equation (1) in (2) we get
$5\left ( \frac{5y}{6} \right )-4y= 8$
$\frac{25y}{6}-4y= 8$
$\frac{25y-24y}{6}= 8$
25y – 24y = 48
y = 48
Put y = 48 in eq. (1)
x = $\frac{5\times 48}{6}$
x = 40

Question:19

There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.

Solution:
According to question:
Let x and y are the number of students in class A and B.
x – 10 = y + 10
x – y = 20 … (1)
Second condition is
(x + 20) = 2(y – 20)
x + 20 = 2y – 40
x – 2y = – 60 … (2)
Subtract (1) and (2) we get
y = 20 + 60
y = 80
Put y = 80 in (1)
x – 80 = 20
x = 20 + 80
x = 100
Hence in hall A there are 100 students and in hall B there are 80 students.

Question:20

A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 16 for the book kept for four days. Find the fixed charges and the charge for each extra day.

Solution:
Let x be the fixed charges and y be the extra charges for each day.
According to question equation for latika is
x + 4y = 22 … (1)
For Anand
x + 2y = 16 … (2)
Subtract equation (1) and (2) we get
4y – 2y = 22 – 16
2y = 6
y = $\frac{6}{2}$ = 3
Put y = 3 in equation (1)
x + 4 (3) = 22
x = 22 – 12
x = 10
The fixed charge is = 10 Rs.

Question:21

In a competitive examination, one mark is awarded for each correct answer while $\frac{1}{2}$ mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

Solution:
Let x be the number of correct answer and (120 – x) be the number of wrong answer .
According to question
x – $\frac{\left ( 120-x \right )}{2}= 90$
$\frac{2x-120+x}{2}= 90$
3x – 120 = 180
3x = 180 + 120
3x = 300
$x= \frac{300}{3}$
x = 100
Hence Jayanti answered 100 questions correctly.

Question:22

The angles of a cyclic quadrilateral ABCD are $\angle$A = (6x + 10)°, $\angle$B = (5x)° $\angle$C = (x + y)°, $\angle$D = (3y – 10)°
Find x and y, and hence the values of the four angles.

Solution:
According to property of cyclic quadrilateral, sum of opposite angles = $180^{\circ}$.
$\angle$A + $\angle$C = $180^{\circ}$
6x + $10^{\circ}$ + x + y = $180^{\circ}$
7x + y = $180^{\circ}$$10^{\circ}$
7x + y = $170^{\circ}$ … (1)
Similarly $\angle$B + $\angle$D = $180^{\circ}$
5x + 3y – $10^{\circ}$ = $180^{\circ}$
5x + 3y = $180^{\circ}$ + $10^{\circ}$
5x + 3y = $190^{\circ}$ … (2)
Solving eq. (1) and eq. (2) we get
21x + 3y = $510^{\circ}$ {Multiply eq. (1) by 3} … (3)
Now subtract equation (2) from (3) then we get
16x = $320^{\circ}$
$x= \frac{320^{\circ}}{16}$
x = $20^{\circ}$
Put x = $20^{\circ}$ in eq. (1) we get
7($20^{\circ}$) + y = $170^{\circ}$
y = $170^{\circ}$$140^{\circ}$
y = $30^{\circ}$
Hence $\angle$A = 6x + $10^{\circ}$
= 6 × $20^{\circ}$ +$10^{\circ}$
=$120^{\circ}$ + $10^{\circ}$
= $130^{\circ}$
$\angle$B = (5x)
= 5 × $20^{\circ}$ = $100^{\circ}$
$\angle$$c= \left ( x+y \right )^{\circ}$
($20^{\circ}$ + $30^{\circ}$) = $50^{\circ}$
$\angle$D = (3y – 10)°
= 3 × $30^{\circ}$$10^{\circ}$
= $90^{\circ}$$10^{\circ}$
= $80^{\circ}$
Hence the value of four angles are:
$130^{\circ}$, $100^{\circ}$, $50^{\circ}$ and $80^{\circ}$ respectively.

Question:1

Graphically, solve the following pair of equations:2x + y = 6 and 2x – y + 2 = 0 Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Solution:
Here the equations are 2x + y = 6, 2x – y = –2
2x + y = 6

 x 0 3 y 6 0

2x – y = –2

 x 0 -1 y 2 0

In graph we find that the lines are intersecting at (1, 4) Hence x = 1, y = 4 is the solution of the equations.The two triangles are ABC and DBE
Area of $\bigtriangleup$ABE = $\frac{1}{2}$ × Base × perpendicular
$= \frac{1}{2}\times 4\times 4= 8$
Area of $\bigtriangleup$ADC = $\frac{1}{2}$ × Base × perpendicular
$= \frac{1}{2}\times 4\times 1$

$= 2$
$\bigtriangleup$ABE area : $\bigtriangleup$ADC Area
4 : 1

Question:2

Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x, x + y = 8

Solution:
y = x
y – x = 0

 x 0 1 y 0 1

3y – x = 0

 x 0 3 y 0 1

x + y = 8

 x 0 8 y 8 0

The require triangle is $\bigtriangleup$ABC with vertices (0, 0), (4, 4), (2, 6)

Question:3

Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x–axis.

Solution:
The given equations are x = 3, x = 5, 2x – y – 4 = 0
2x – y = 4

 x 2 0 y 0 -4

The quadrilateral ABCD is a trapezium.
Area of ABCD (trapezium) = $\frac{1}{2}$ × (a + b)h
a = AD, b = BC, h = AB
= $\frac{1}{2}$× (AD + BC) × AB

= $\frac{1}{2}$× (2 + 6) × 2

Area = 8 sq. units

Question:4

The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Solution:
Let the cost of a pen is x and the cost of a pencil box is y.
As per the question
4x + 4y = 100 … (1)
3x = y + 15 … (2)
Divide equation (1) by 4.
x + y = 25 ………..(3)
By adding equation (1) and (3) we get
4x = 40
x = 10
Put x = 10 in eq. (1)
40 + 4y = 100
4y = 100 – 40
y = 60/4 = 15
x = 10, y = 15
Cost of pen = Rs. 10
Cost of pencil box = Rs.15

Question:5

Determine, algebraically, the vertices of the triangle formed by the lines 3x – y = 3, 2x – 3y = 2 and x + 2y = 8

Solution:
Determine, algebraically, the vertices of the triangle formed by the lines
3x – y = 3 … (1)
2x – 3y = 2 … (2)
x + 2y = 8 ... (3)

For vertices of A, solve equation (1) and (3)
3x – y = 3
x + 2y = 8 × 3
3x + 6y = 24
3x + y = –3
7y = 21
y = 3
Put y = 3 in eq. (1)
3x – 3 = 3
3x = 6
x = 2
point A is (2, 3)
For vertices of B solve equation (1) and (2)
3x – y = 3 × 3
2x – 3y = 2
9x – 3y = 9
– + –
–7x = –7
x = 1
Put x = 1 in equation (1)
3(1) – y = 3
3 – 3 = y
y = 0
point B is (1, 0)
For vertices of C solve (2) and (3)
2x – 3y = 2
x + 2y = 8 × 2
2x + 4y = 16
2x – 3y = 2
– + –
7y = 14
y = 2
Put y = 2 in eq. (2)
2x – 3(2) = 2
2x = 2 + 6
x = $\frac{8}{2}$ = 4
Point C (4, 2)
Hence vertices of triangle ABC are A (2, 3), B (1, 0) and C (4, 2)

Question:6

Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus.On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus.

Solution:
We know that
Speed = $\frac{Distance}{Time}$
Time = Distance / Speed
Let the speed of rickshaw is x and of bus is y.
So, according to question
$\frac{2}{x}+\frac{\left ( 14-2 \right )}{y}= \frac{1}{2}Hr$
$\frac{2}{x}+\frac{12}{y}= \frac{1}{2}\cdots (1)$
$\frac{4}{x}+\frac{10}{y}= \frac{39}{60}\, Hour\cdots (2)$
Solve eq. (1) and (2)
$\left ( \frac{2}{x}+\frac{12}{y} = \frac{1}{2}\right )\times 2$
$\frac{4}{x}+\frac{24}{y}= 1$
$\frac{4}{x}+\frac{10}{y}= \frac{39}{60}$
– – –
$\frac{24}{y}-\frac{10}{y}=1- \frac{39}{60}$
$\frac{24-10}{y}= \frac{60-39}{60}$
$\frac{14}{y}= \frac{21}{60}$
21y = 840
$y= \frac{840}{21}= 40$
Put y = 40 in eq. (1)
$\frac{2}{x}+\frac{12}{40}= \frac{1}{2}$
$\frac{2}{x}= \frac{20-12}{40}$
$\frac{2}{x}= \frac{8}{40}$
40 = 4x
x = 10
Speed of rickshaw = 10 km/h
Speed of bus = 40 km/h

Question:7

A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.

Solution:
According to question,
Person’s speed with upstream = (5 – x) km/h
Person’s speed with downstream = (5 + x) km/h
Time with upstream $= \frac{40}{5-x}\left ( Using\, t= \frac{D}{S} \right )$
Time with downstream $= \frac{40}{5+x}\left ( Using\, t= \frac{D}{S} \right )$
$\frac{40}{5-x}= \frac{3\times 40}{5+x}$
(5 + x)40 = 120 (5 – x)
200 + 40x = 600 - 120 x
160x = 400
x = 2.5
Speed of stream = 2.5 km/h

Question:8

A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Solution:
Let speed of boat in still water = x km/h
Let speed of stream = y km/h
Speed of boat in upstream = (x - y) km/h
Speed of boat in downstream = (x + y) km/h
We know that time =$\frac{distance}{speed}$
So, $\frac{30}{\left ( x-y \right )}+\frac{28}{x+y}= 7$… (1)
$\frac{21}{\left ( x-y \right )}+\frac{21}{x+y}= 5$ … (2)
Put $\frac{1}{\left ( x-y \right )}= u,\frac{1}{\left ( x+y \right )}= v$ in (1), (2)
30u + 28v = 7 … (3)
21u + 21v = 5 … (4)
Solve equation (3), (4)
Multiply equation (3) by 21 and (4) by 30
630u + 588 v = 147
630u + 630v = 150
_-______-____-________
+42v = +3
$v= \frac{3}{42}= \frac{1}{14}$
Put $v= \frac{1}{14}$ in eq. (4)
$21u+21\left ( \frac{1}{14} \right )= 5$
$21u= 5-\frac{3}{2}$
$21u= \frac{10-3}{2}$
21u = $\frac{7}{2}$
$u= \frac{7}{2\times 21}\Rightarrow u= \frac{1}{6}$
We put $u= \frac{1}{x-y}\, and\, v= \frac{1}{x+y}$
So,
$\frac{1}{x-y}= \frac{1}{6}$ | $\frac{1}{x+y}= \frac{1}{14}$
x-y=6..(5) x+y=14...(6)
By adding equation (5) and (6)
2x = 20
x = 10
Put x = 10 in eq. (6)
10 + y = 14
y = 4
Hence speed of the boat in still water = 10 km/hr.
Speed of the stream = 4 km/hr.

Question:9

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Solution:
Let the first digit = x
Let the ten’s digit = y
Therefore the number = 10y + x
As per the question
8(x + y) –5 = 10y + x
8x + 8y – 10y – x = 5
7x – 2y = 5 … (1)
16(y – x) + 3 = 10y + x
16y – 16x – 10y – x = – 3
–17x + 6y = –3
17x – 6y = 3 … (2)
Multiply equation (1) by 3
21x – 6y = 15
17x – 6y = 3
– + –
4x = 12
x = 3
Put x = 3 in eq. (1)
7(3) –2y = 5
–2y = 5 –21
y = $\frac{16}{2}$ = 8
Hence the number is = 10y + x = 10(8) + 3 = 83

Question:10

A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the station A to B costs Rs 2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B costs Rs 3810. Find the full first class fare from station A to B, and also the reservation charges for a ticket.

Solution:
Let the full ticket cost = y Rs.
Let reservation charge = x Rs.
According to questions
x + y = 2530 … (1)
(x + y) + (x + y/2) = 3810 …(2)
Here x + y is for full ticket and x + y/2 is for half.
Solve equation (1) and (2)
x + y + x + $\frac{y}{2}$ = 3810
2x + y + $\frac{y}{2}$ = 3810
4x + 2y + y = 3810 × 2
4x + 3y = 7620 … (3)
Multiply equation (1) by 3
3x + 3y = 7590
4x + 3y = 7620
– – –
–x = – 30
Put x = 30 in eq. (1)
30 + y = 2530
y = 2500
Hence Reservation charge = 30Rs.
Full First class charge = 2500 Rs.

Question:11

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, there by, getting a sum Rs 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs 1028. Find the cost price of the saree and the list price (price before discount) of the sweater.

Solution:
Let the cost price of saree = Rs. x
Let the cost price of sweater = Rs. y
Saree with 8% profit = $(1+\frac{8}{100})x=\frac{108x}{100}$
Sweater with 10% discount = $(1-\frac{10}{100})y$ = $\frac{90y}{100}$
Saree with 10% profit = x + $x\times \frac{10}{100}= \frac{110x}{100}$
Sweater with 8% discount = y – $y\times \frac{8}{100}= \frac{92y}{100}$
According to question:
$\frac{108x}{100}+ \frac{92y}{100}= 1008$
108x + 90y = 100800
6x + 5y = 5600 … (1) (Divide by 18)
$\frac{110x}{100}+ \frac{92y}{100}= 1028$
110x + 92y = 102800
55x + 56y = 51400 … (2) (Divide by 2)
Multiply equation (1) by 46 and eq. (2) by 5
276 x + 230 y = 257600
275 x + 230 y = 257000
– – –
x = 600
Put x = 600 in eq. (1)
6(600) + 5y = 5600
5y = 5600 –3600
5y = 2000
y = 400
Price of saree = Rs. 600
Price of sweater = Rs. 400

Question:12

Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme

Solution:
Let money invested in A and B are x, y respectively.
So, according to question.
0.08x + 0.09y = 1860 … (1)

$x\times \frac{9}{100}+y\times \frac{8}{100}= 1880$
0.09x + 0.08y = 1880 … (2)
Multiply equation (1) by 9 and (2) by 8
0.72x + 0.81y = 16740
0.72x + 0.64y = 15040
– – –
0.17y = 1700

y = $\frac{1700}{0\cdot 7}= 10000$
Put y = 10000 in eq. (1)
0.08x + 900 = 1860
0.08x = 960
x = $\frac{960}{8}\times 100$ = 12000
x = 12000
He invested Rs. 12000 in A and Rs. 10000 in B.

Question:13

Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had.

Solution:
Let bananas in lot A = x
Bananas in lot B = y
According to question:
$\left ( \frac{x}{3} \right )\left ( 2 \right )+y\left ( 1 \right )= 400$
2x + 3y = 400 × 3
2x + 3y = 1200 … (1)
x(1) + $\left ( \frac{y}{5} \right )$(4) = 460
5x + 4y = 460 × 5
5x + 4y = 2300 … (2)
Multiply equation (1) by 5 and eq. (2) by 2
10x + 15y = 6000
10x + 8y = 4600
– – –
7y = 1400
$y= \frac{1400}{7}= 200$
Put y = 200 in eq. (1)
2x + 3(200) = 1200
2x + 600 = 1200
2x = 600
x = 300
Total number of bananas = x + y = 300 + 200 = 500

## NCERT Exemplar Solutions Class 10 Maths Chapter 3 Important Topics:

• In this chapter, not only the algebraic expression of the equation in two variables is explained but also the graphical visualisation is discussed.
• Students will also learn graphical method to solve two linear equations for their common solution.
• In algebraic method for solving equations, techniques like cross multiplication substitution or elimination are discussed in detail.
• NCERT exemplar Class 10 Maths solutions chapter 3 discusses the method to form linear equation in two variables in any given condition of question statement.
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## NCERT Class 10 Exemplar Solutions for Other Subjects:

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## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry & Its Equations Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 12 Surface Areas and Volumes Chapter 13 Statistics and Probability

## Features of NCERT Exemplar Class 10 Maths Solutions Chapter 3:

These Class 10 Maths NCERT exemplar chapter 3 solutions provide a detailed knowledge of linear equations in two variables. The knowledge of this chapter will be useful whether a student will pursue mathematics in higher classes or not. Even for exams like NEET the basic method of solving linear equations is a prerequisite. A good volume of practice problems available in the exemplar provides a student with ample opportunity to clear the concept of Pair of Linear Equations in Two Variables.

Class 10 Maths NCERT exemplar solutions chapter 3 Pair of Linear Equations in Two Variables is a good training practice for the students to solve problems of RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths, NCERT Class 10 Maths, et cetera.

NCERT exemplar Class 10 Maths solutions chapter 3 pdf download is the feature that makes these solutions accessible to the students in an offline scenario and can keep the doubts at bay while practicing NCERT exemplar Class 10 Maths chapter 3.

### Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

### Must, Read NCERT Solution Subject Wise

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

### Also, Check NCERT Books and NCERT Syllabus here

1. Do every pair of linear equations in two variables have a unique solution?

No, a pair of linear equations may have a unique solution or no solution or infinite possible solutions.

2. What will be the shape of the graph for a linear equation in two variables?

For any linear equation in two variables if we try to draw the graph, it will be straight line

3. If two straight lines are parallel to each other Will we get any solution for these two equations?

If two straight lines are parallel to each other it means they don’t have any common point. Therefore, equations representing these two straight lines will not have any solution

4. If two straight lines meet at origin what will be the common solution for these two equations?

As these two straight lines are passing through origin that means (zero, zero) will satisfy both the equations, hence, the common solution will be X =0 and Y =0.

5. Is the chapter Pair of Linear Equations in Two Variables important for Board examinations?

The chapter of Pair of Linear Equations in Two Variables is essential for the board exams as it includes around 10-12% weightage of the whole paper.

6. How many questions can I expect from the Pair of Linear equations in two variables in board examination?

Generally, a total of 3-4 questions of different types appear in the board examinations from this chapter. NCERT exemplar Class 10 Maths solutions chapter 3 is adequate to correctly answer these questions from Pair of Linear equations in two variables.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

1. Starting Point:

• Determine your starting point in Aligarh or the nearby area.

3. By Local Transport:

• Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
• Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
4. Landmarks to Look For:

• As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.

• If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
6. Final Destination:

• Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

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.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9