Class 10 Mathematics

# Linear Equations

## Exercise 3.5 (NCERT) Part 1

Question 1: Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(a) x – 3y – 3 = 0 and 3x – 9y – 2 = 0

Solution:

(a_1)/(a_2)=1/3
(b_1)/(b_2)=(-3)/(-9)=1/3
(c_1)/(c_2)=3/2

It is clear that;

(a_1)/(a_2)=(b_1)/(b_2)≠(c_1)/(c_2)

Hence, there will be no solution for the given pair of linear equations.

(b) 2x + y = 5 and 3x + 2y = 8

Solution:

(a_1)/(a_2)=2/3 and (b_1)/(b_2)1/2

It is clear that;

(a_1)/(a_2)≠(b_1)/(b_2)

Hence, there will be unique solution for the given pair of linear equations.

From cross-multiplication method, we know;

(x)/(b_1c_2-b_2c_1)=(y)/(c_1a_2-c_2a_1)=(1)/(a_1b_2-a_2b_1)

Or, (x)/(-8+10)=(y)/(-15+15)=(1)/(4-3)

Or, x/2=y/1=1

Or, x=1 and y=2

(c) 3x – 5y = 20 and 6x – 10y = 40

Solution: a1 = 3, b1 = - 5, c1 = - 20

a2 = 6, b2 = - 10, c2 = - 40

(a_1)/(b_1)=3/6=1/2

(b_1)/(b_2)=(-5)/(-10)=1/2

(c_1)/(c_2)=(-20)/(-40)=1/2

It is clear that;

(a_1)/(a_2)=(b_1)/(b_2)=(c_1)/(c_2)

Hence, there will be infinitely many solutions for the given pair of linear equations.

(d) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Solution: a1 = 1, b1 = - 3, c1 = - 7

a2 = 3, b2 = - 3, c2 = - 15

(a_1)/(a_2)=1/3 and (b_1)/(b_2)=(-3)/(-3)=1

It is clear that;

(a_1)/(a_2)≠(b_1)/(b_2)

Hence, there will be unique solution for the given pair of linear equations.

From cross-multiplication method, we know;

(x)/(b_1c_2-b_2c_1)=(y)/(c_1a_2-c_2a_1)=(1)/(a_1b_2-a_2b_1)

Or, (x)/((-3xx-15)-(-3xx-7))

=(y)/((-7xx3)-(-15xx1))

=(1)/((1xx-3)-(3xx-3)

Or, (x)/(45-21)=(y)/(-21+15)=(1)/(-3+9)

Or, (x)/(24)=(y)/(-6)=(1)/(6)

Or, x=(24)/(6)=4

And y=(-6)/(6)=-1

Question 2: For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7 and (a – b)x + (a + b)y = 3a + b – 2

Solution: a1 = 2, b1 = 3, c1 = - 7

a2 = (a – b), b2 = (a + b), c2 = - (3a + b – 2)

For infinite number of solutions, the equations should fulfill following criterion:

(a_1)/(a_2)=(b_1)/(b_2)=(c_1)/(c_2)

Or, (2)/(a-b)=(3)/(a+b)

=(-7)/(-(3a+b-2))

Or, 2(a + b) = 3(a – b)
Or, 2a + 2b = 3a – 3b
Or, 2a + 5b = 3a
Or, a = 5b --------(1)

Similarly, 6a + 2b – 4 = 7a – 7b
Or, 6a + 2b – 7a + 7b = 4
Or, - a + 9b = 4
Or, a = 9b + 4 --------(2)

From equations (1) and (2), it is clear;

5b = 9b – 4
Or, 4b = 4
Or, b = 1

Substituting the value of b in equation (1), we get;

a = 5b = 5

Hence, a = 5 and b = 1

Question 3: For which value of k will the following pair of linear equations have no solution?

3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1

Solution: a1 = 3, b1 = 1, c1 = 1

a2 = (2k – 1), b2 = (k – 1), c2 = - (2k + 1)

For no solution, the equations should fulfill following criterion:

(a_1)/(a_2)=(b_1)/(b_2)≠(c_1)/(c_2)

Or, (3)/(2k-1)=(1)/(k-1)≠(1)/(2k+1)

Or, 3(k – 1) = 2k – 1
Or, 3k – 3 = 2k – 1
Or, 3k = 2k – 1 + 3
Or, 3k = 2k + 2
Or, k = 2

Question 4: Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9 and 3x + 2y = 4.

Solution: Substitution method:

Let us use the first equation to express one variable in terms of another variable;

8x + 5y = 9

Or, 8x = 9 – 5y
Or, x=(9-5y)/(8)

Substituting the value of x in second equation, we get;

3x+2y=4
Or, 3((9-5y)/(8))+2y=4

Or, (27-15y)/(8)+2y=4

Or, (27-15y+16y)/(8)=4

Or, 27+y=32
Or, y=32-27=5

Substituting the value of y in first equation, we get;

x=(9-5y)/(8)=(9-5xx5)/(8)

=(9-25)/(8)=(-16)/(8)=-2

Hence, x = - 2 and y = 5

#### Cross-multiplication Method:

8x + 5y = 9 and 3x + 2y = 4

a1 = 8, b1 = 5, c1 = - 9

a2 = 3, b2 = 2, c2 = - 4

From cross-multiplication method, we know;

(x)/(b_1c_2-b_2c_1)=(y)/(c_1a_2-c_2a_1)=(1)/(a_1b_2-a_2b_1)

Or, (x)/((5xx-4)-(2xx-9))

=(y)((-9xx3)-(-4xx8))

=(1)/((8xx2)-(3xx5))

Or, (x)/(-20+18)=(y)/(-27+32)=(1)/(16-15)

Or, (x)/(-2)=y/5=1

Or, x = - 2 and y = 5

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Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3(Part 1)

Exercise:3.3 (part 2)

Exercise:3.4 (part 1)

Exercise:3.4 (part 2)

Exercise:3.5 (part 2)

Exercise:3.6 (part 1)

Exercise:3.6 (part 2)

Exercise:3.6 (part 3)