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`


PrevNext

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)