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

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.

Answer:

Answer: D
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
$\frac{a_1}{a_2}=\frac{6}{2}=3$
$\frac{b_1}{b_2}=\frac{-3}{-1}=3$

$\Rightarrow$$\frac{c_1}{c_2}=\frac{10}{9}$
Here, $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
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

Answer:

Answer: D
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 $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
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

Answer:

Answer: C
Solution:

If the pair of linear equations is consistent then they have unique or infinitely many solutions.
In unique solution, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
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

Answer:

Answer: D
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)

Answer:

Answer: D
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

Answer:

Answer: C
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)$\frac{3}{2}$

Answer:

Answer: C
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
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{3}{2}=\frac{2k}{5}\ne -2$ ……(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

Answer:

Answer: D
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 $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{c}{6}=\frac{-1}{-2}\ne \frac{-2}{-3}$
Here we find that $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ 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

Answer:

Answer: D
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

Answer:

Answer: B, D
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

Answer:

Answer : (C)
Solution:
Equation are x – y = 2 … (1)
x + y = 4 … (2)
Add equation (1) and (2)
x – y = 2
x + y = 4
$\Rightarrow$2x = 6
$\Rightarrow$x = 3
Put x = 3 in equation (2)
$\Rightarrow$3 + y = 4
$\Rightarrow$y = 4 – 3
$\Rightarrow$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

Answer:

Answer:(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)
$\Rightarrow$x + 25 = 50
$\Rightarrow$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

Answer:

Answer:(C) 6 and 36
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
$\Rightarrow$y + 4 = 4 (x + 4)
$\Rightarrow$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)
$\Rightarrow$y = 6 (6) = 36
$\Rightarrow$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 ;$2x+\frac{2}{3}y=2$

Answer:

(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 $\frac{a_1}{b_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ also, Here we have $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

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$\frac{a_1}{b_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ here

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
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 $\frac{a_1}{b_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ 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)$\frac{x}{2}+y+\frac{2}{5}=0;4x+8y+\frac{5}{6}=0$

Answer:

(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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
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 $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
Hence, the given pair of equations does not represent a pair of coincident lines.

Question:3

Are the following pair of linear equations consistent? Justify your answer.
(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

Answer:

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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}or\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ but here $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Hence, the pair of linear equations is not consistent.
Solution: (ii)
Equation are : $\frac{3}{5}x-y=\frac{1}{2}$
$\frac{1}{5}x-3y=\frac{1}{6}$
In equation
$\frac{3}{5}x-y=\frac{1}{2}$
a₁= $\frac{3}{5}$; b₁ = –1 ; c₁ = $-\frac{1}{2}$
In equation
$\frac{1}{5}x-3y=\frac{1}{6}$
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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}or\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ also here $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}or\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}or\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ but here $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
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.

Answer:

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
$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{\lambda }{2}=\frac{3}{6}\ne \frac{-7}{14}$
Here we can see that
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
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.

Answer:

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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ 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.

Answer:

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?

Answer:

(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 $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{a_1}{a_2}=\frac{\lambda }{1}$
$\frac{b_1}{b_2}=\frac{1}{\lambda }$
$\frac{c_1}{c_2}=\frac{-{\lambda }^2}{-1}={\lambda }^2$
$\Rightarrow$$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\Rightarrow$$\frac{\lambda }{1}=\frac{1}{\lambda }\ne {\lambda }^2$
$\Rightarrow$$\frac{\lambda }{1}=\frac{1}{\lambda }$
$\Rightarrow$${\lambda }^2=1$
$\Rightarrow$$\lambda$=1,-1
Hence, the value of $\lambda$ is -1
$\lambda \ne 1$ because in this case $\frac{b_1}{b_2}=\frac{c_1}{c_2}$)
(ii)Solution:
The given equations are
$\lambda$x + y =${\lambda }^2$ , x + $\lambda$y = 1
In equation
$\lambda$x + y – ${\lambda }^2$ = 0
a₁ = $\lambda$, b₁ = 1, c₁ = –${\lambda }^2$
In equation
x + $\lambda$y – 1 = 0
a₂ =1, b₂ = $\lambda$, c₂ = –1
$\frac{a_1}{a_2}=\frac{\lambda }{1};\frac{b_1}{b_2}=\frac{1}{\lambda };\frac{c_1}{c_2}=\frac{-{\lambda }^2}{-1}={\lambda }^2$
For infinite many solution
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{\lambda }{1}=\frac{1}{\lambda }={\lambda }^2$
Only one value of l satisfy all the three equations, that is $\lambda$ = 1
(iii) Solution:
The given equations are
$\lambda$x + y = ${\lambda }^2$, x + $\lambda$y = 1
In equation
$\lambda$x + y – ${\lambda }^2$ = 0
a₁ =$\lambda$ , b₁ = 1, c₁ = -${\lambda }^2$
In equation
x + $\lambda$y = 1
a₂ =1, b₂ = $\lambda$, c₂ =–1
$\frac{a_1}{a_2}=\frac{\lambda }{1};\frac{b_1}{b_2}=\frac{1}{\lambda };\frac{c_1}{c_2}=\frac{-{\lambda }^2}{-1}={\lambda }^2$
For unique solution $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\Rightarrow$$\frac{\lambda }{1}\ne \frac{1}{\lambda }$
$\Rightarrow$${\lambda }^2\ne 1$
$\Rightarrow$$\lambda \ne \pm 1$
All real values of $\lambda$ except $\lambda \ne \pm 1$

Question:2

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

Answer:

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
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\Rightarrow$$\frac{k}{12}=\frac{3}{k}\equiv \frac{k-3}{k}$
$\Rightarrow$$\frac{k}{12}=\frac{3}{k}$
$\Rightarrow$$k^2$ = 36
$\Rightarrow$$k=\pm 36$
($k\ne 6$ because if $k=+6$, then $\frac{b_1}{b_2}=\frac{c_1}{c_2}$ which is not possible in no solution.)
Hence, the 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

Answer:

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)
$\frac{a_1}{a_2}=\frac{1}{(a-b)},\frac{b_1}{b_2}=\frac{2}{(a+b)},\frac{c_1}{c_2}=\frac{-1}{-(a+b-2)}$
For infinitely many solutions $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{1}{(a-b)}=\frac{2}{(a+b)}=\frac{1}{a+b-2}$
$\Rightarrow$$\frac{1}{(a-b)}=\frac{2}{(a+b)}$
$\Rightarrow$a + b = 2a –2b
$\Rightarrow$–a + 3b = 0 … (1)

$\Rightarrow$$\frac{2}{(a+b)}=\frac{1}{a+b-2}$
$\Rightarrow$2a + 2b – 4 = a + b
$\Rightarrow$ a + b –4 = 0 …(2)

Add equations (1) and (2)
$\Rightarrow$–a + 3b + a + b = 0 + 4
$\Rightarrow$4b = 4
b = 1
Put b = 1 in equation (1)
$\Rightarrow$–a + 3 (1) = 0
$\Rightarrow$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.

Answer:

(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 $\frac{a_1}{a_1}=\frac{b_1}{b_1}\ne \frac{c_1}{c_1}$
$\Rightarrow$$\frac{1}{2}=\frac{1}{2}\ne \frac{5}{p}$
$\Rightarrow$$\frac{1}{2}\ne \frac{5}{p}\Rightarrow p\ne 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 $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\Rightarrow$$\frac{-1}{p}=-p\ne 1$
$\Rightarrow$$\frac{+1}{p}=+p$
$\Rightarrow$$p^2$ = 1
$\Rightarrow$p = ± 1
$\Rightarrow$ p = + 1 ($p\ne -1$ because if p = –1 then $\frac{b_1}{b_2}=\frac{c_1}{c_2}$)
(iii)Solution:
Here the equation are –3x + 5y = 7, 2px – 3y = 1
In equation –3x + 5y = 7
a₁ = –3, b₁ = 5, c₁ = –7
In equation 2px – 3y = 1
a₂ = 2p, b₂ = –3, c₂ = –1
$\frac{a_1}{a_2}=\frac{-3}{2p};\frac{b_1}{b_2}=\frac{5}{-3},\frac{c_1}{c_2}=\frac{-7}{-1}=7$
For unique solution $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\Rightarrow$$\frac{-3}{2p}\ne \frac{-5}{3}$
$\Rightarrow$$-9\ne -10p$
$\Rightarrow$$p\ne \frac{9}{10}$
$\Rightarrow$$p\ne 0\cdot 9$
Hence p can have any real value for unique solution except 0.9
(iv)Solution:
Here the equation are 2x + 3y – 5 = 0
px – 6y –8 = 0
In equation 2x + 3y – 5 = 0
a₁ = 2, b₁ = 3, c₁ = –5
In equation px – 6y – 8 = 0
a₂ = p, b₂ = –6, c₂ = –8
$\frac{a_1}{a_2}=\frac{2}{p};\frac{b_1}{b_2}=\frac{3}{-6}=\frac{-1}{2};\frac{c_1}{c_2}=\frac{-5}{-8}=\frac{5}{8}$
For unique solution $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\Rightarrow$$\frac{2}{p}\ne \frac{-1}{2}$
$\Rightarrow$$4\ne -p$
$\Rightarrow$$p\ne -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 $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow$$\frac{1}{p}=\frac{3}{p+q}=\frac{1}{4}$
$\Rightarrow$$\frac{1}{p}=\frac{1}{4}$
$\Rightarrow$4 = p
$\Rightarrow$p = 4
$\frac{3}{p+q}=\frac{1}{4}$
$\Rightarrow$p + q = 12
Put p =4
$\Rightarrow$q=12-4
$\Rightarrow$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

Answer:

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 $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ but here $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

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?

Answer:

Solution:
Let the equations be
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Put x = –1 and y = 3 in the 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 have x = –1, y = 3 as the solution.

Question:7

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

Answer:

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
$\Rightarrow$6x = 42
$\Rightarrow$ x = $\frac{42}{6}=7$
Put x = 7 in equation (1) we get
$\Rightarrow$2(7) + y = 23
$\Rightarrow$y = 23 – 14
$\Rightarrow$y = 9
Now 5y – 2x = 5(9) – 2 (7)
$\Rightarrow$ 45 – 14
$\Rightarrow$ 31
$\Rightarrow$$\frac{y}{x}-2=\frac{9}{7}-2$
$\Rightarrow$$\frac{y-14}{7}=\frac{-5}{7}$

Question:8

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

Answer:

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
-___-___-_______
$\Rightarrow$8x = 8
$\Rightarrow$ x = $\frac{8}{8}$ = 1
Put x = 1 in equation (1)
$\Rightarrow$3(1) + y = 7
$\Rightarrow$3 + y = 7
$\Rightarrow$y = 7 – 3
$\Rightarrow$ y = 4

Question:9

Solve the following pairs of equations:
(i) x+ y = 3.3 ;$\frac{0\cdot 6}{3x-2y}=-1,3x-2y\ne 0$
(ii)$\frac{x}{3}+\frac{y}{4}=4$ , $\frac{5x}{6}-\frac{y}{8}=4$
(iii)$4x+\frac{6}{y}=15;6x-\frac{8}{y}=14,y\ne 0$
(iv)$\frac{1}{2x}-\frac{1}{y}=-1;\frac{1}{x}+\frac{1}{2y}=8,x,y\ne 0$
(v)43x + 67y = – 24 ; 67x + 43y = 24
(vi)$\frac{x}{a}+\frac{y}{b}=a+b;\frac{x}{a_2}+\frac{y}{b^2}=2(a,b\ne 0)$
(vii)$\frac{2xy}{xy}=\frac{2}{3};\frac{xy}{2x-y}=\frac{-3}{10};x+y\ne 0,2x-y\ne 0$

Answer:

(i)Solution:
Given equations are x + y = 3.3 … (1)
0.6 = 2y – 3x …(2)
Using the elimination method in equations (1) and (2), we get
2x + 2y = 6.6 [multiply eq. (1) by 2]
–3x + 2y = 0.6
+ – –
$\Rightarrow$5x = 6
$\Rightarrow$ x = $\frac{6}{5}$ = 1.2
Put x = 1.2 in (1) we get
$\Rightarrow$1.2 + y = 3.3
$\Rightarrow$y = 3.3 – 1.2
$\Rightarrow$y = 2.1
(ii) 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
$\Rightarrow$4x + 3y + 20x -3y = 48 + 96
$\Rightarrow$24x = 144
$\Rightarrow$$x=\frac{144}{24}=6$
$\Rightarrow$x = 6
Put x = 6 in eq. (1) we have
$\Rightarrow$$\frac{6}{3}+\frac{y}{4}=4$
$\Rightarrow$$2+\frac{y}{4}=4$
$\Rightarrow$$\frac{y}{4}=4-2$
$\Rightarrow$$\frac{y}{4}=2$
$\Rightarrow$y = 8
(iii)Solution:
Equation are: $4x+\frac{6}{y}=15$… (1)
$6x-\frac{8}{y}=14$… (2)
Multiply equation (1) by 6 and equation (2) by 4 and then subtract equation (2) from (1)
24x + $\frac{36}{y}$ = 90
24x – 32/y = 56
– + –
$\Rightarrow$$\frac{36}{y}+\frac{32}{y}=34$
$\Rightarrow$$\frac{36+32}{y}=34$
$\Rightarrow$68 = 34y
$\Rightarrow$ y = $\frac{68}{34}=2$
$\Rightarrow$y = 2
Put y = 2 in equation (1) we get
$\Rightarrow$4x = 15 – 3
$\Rightarrow$4x = 12
$\Rightarrow$ x = $\frac{12}{4}$ = 3
$\Rightarrow$x = 3
(iv)Solution:
Let $\frac{1}{x}$ = u and $\frac{1}{y}$ = v and put in the given equations
u/2 – v = – 1 … (1)
u + $\frac{v}{2}$ = 8 … (2)
Multiply eq. (1) by 2 and subtract from eq. (2) we get
$\Rightarrow$$\frac{v}{2}+2v=8+2$
$\Rightarrow$v + 4v = 10 × 2
$\Rightarrow$5v = 20
$\Rightarrow$ v = $\frac{20}{+5}=4$
$\Rightarrow$v = 4
$\Rightarrow$ v = 1/y = 4$\Rightarrow$
$\Rightarrow$ y = $\frac{1}{4}$
Put v = 4 in equation (1) we get
$\Rightarrow$u/2= –1 + 4
$\Rightarrow$u = 6
$\Rightarrow$$\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
$\Rightarrow$43x + 67(–1) = –24
$\Rightarrow$43x = –24 + 67
$\Rightarrow$43x = 43
$\Rightarrow$x = 1
(vi)Solution:
Given equation are:
$\frac{x}{a}+\frac{y}{b}=a+b$

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

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

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

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