Class 10 Mathematics

# Linear Equations

## Exercise 3.3 (NCERT) Part 1

##### Solving by substitution method:
• One of the equations is picked and in this, one variable is expressed in terms of another variable.
• The value of one variable (in terms of another variable) is substituted in the second equation to calculate the value of the remaining variable.

Question 1: Solve the following pair of linear equations by the substitution method.

(a) x + y = 14 and x – y = 4

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

x + y = 14
Or, x = 14 – y

Substituting the value of x in second equation we get:

x – y = 4
Or, 14 – y – y = 4
Or, 14 – 2y = 4
Or, 2y = 14 – 4 = 10
Or, y = 5

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

x = 14 – y
Or, x = 14 – 5 = 9
Hence, x = 9 and y = 5

(b) s-t=3 and (s)/(3)+(t)/(2)=6

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

s – t = 3
Or, s = t + 3

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

(s)/(3)+(t)/(2)=6

Or, (t+3)/(3)+(t)/(2)=6

Or, 5t+6=36
Or, 5t=36-6=30
Or, t=30÷5=6

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

s = t + 3
Or, s = 6 + 3 = 9
Hence, s = 9 and t = 6

(c) 3x – y = 3 and 9x – 3y = 9

Solution: Let us use first equation to express one variable in terms of another variable.

3x – y = 3
Or, y = 3x – 3

The second equation is same as the first equation;

9x – 3y = 9

Dividing the equation by 3, we get;

3x – y = 3

Hence, the given pair of linear equations has infinitely many solutions.

(d) 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3

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

0.2x + 0.3y = 1.3
Or, 2x + 3y = 13
Or, 2x = 13 – 3y
Or, x = (13 – 3y)/(2)

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

0.4x+0.5y=2.3
Or, 4x+5y=23
Or, 4((13-3y)/(2))+5y=23

Or, 26-6y+5y=23
Or, 26-y=23
Or, y=26-23=3

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

x=(13-3y)/(2)

=(13-3xx3)/(2)=(13-9)/(2)

=4/2=2

Hence, x = 2 and y = 3

(e) sqrt2x + sqrt3y = 0 and sqrt3x - sqrt8y = 0

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

sqrt2x+sqrt3y=0

Or, sqrt2x=-sqrt3y

Or, x=-(sqrt3)/(sqrt2)y

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

sqrt3x-sqrt8y=0

Or, sqrt3x=sqrt8y

Or, -(sqrt3)/(sqrt2)y\xx\sqrt3=sqrt9y

Or, -(3)/(sqrt2)y=sqrt8y

Or, -3y=sqrt(16)y

Or, -3y=4y
Or, y=-(3)/(4)y

Or, y=-(4)/(3)y

This can be possible only when the value of y is zero. Substituting the value of y in first equation, we get;

x=-(sqrt3)/(sqrt2)y Or, x=0

Hence, x = 0 and y = 0

(f) (3x)/(2)-(5y)/(3)=-2 and x/3+y/2=(13)/(6)

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

(3x)/(2)-(5y)/(3)=-2

Or, (9x-10y)/(6)=-2

Or, 9x-10y=-12
Or, 9x+12=10y
Or, y=(9x+12)/(10)

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

x/3+y/2=(13)/(6)

Or, x/3+(9x+12)/(20)=(13)/(6)

Or, (20x+27x+36)/(60)=(13)/(6)

Or, 47x+36=130
Or, 47x=130-36=94
Or, x=94÷47=2

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

y=(9x+12)/(10)

Or, y=(18+12)/(10)=(30)/(10)=3

Hence, x = 2 and y = 3

Question 2: Solve 2x + 3y = 11 and 2x – 4y = - 24 and hence find the value of ‘m’ for which y = mx + 3.

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

2x + 3y = 11
Or, 2x = 11 – 3y

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

2x – 4y = - 24
Or, 11 – 3y – 4y = - 24
Or, 11 – 7y = - 24
Or, 7y = 11 + 24 = 35
Or, y = 5

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

2x = 11 – 3y
Or, 2x = 11 – 3 xx 5
Or, 2x = 11 – 15 = - 4
Or, x = - 2
Hence, x = - 2 and y = 5

Now; we have to find the value of m in following equation;

y = mx + 3
Or, 5 = m( - 2) + 3
Or, - 2m + 3 = 5
Or, - 2m = 5 – 3 = 2
Or, m = - 1

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Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3 (part 2)

Exercise:3.4 (part 1)

Exercise:3.4 (part 2)

Exercise:3.5 (part 1)

Exercise:3.5 (part 2)

Exercise:3.6 (part 1)

Exercise:3.6 (part 2)

Exercise:3.6 (part 3)