Class 10 Mathematics

# Linear Equations

## Exercise 3.4 (NCERT) Part 1

##### Elimination Method:

In elimination method, one of the equations is multiplied by a suitable number so that the coefficient of one of the variables in both the equations becomes equal. After that, one equation is subtracted or added to another equation to eliminate one of the variables.

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

(a) `x + y = 5` and `2x – 3y = 4`

**Solution:****Elimination Method:**

Multiply the first equation by 3 as follows:

`3x – 3y = 15`

Now, add second equation to this equation;

`3x+3y+2x-3y=15+4`

Or, `5x=19`

Or, `x=(19)/(5)`

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

`x+y=5`

Or, `(19)/(5)+y=5`

Or, `y=5-(19)/(5)`

`=(35-19)/(5)=6/5`

Hence, `x=(19)/(5)` and `y=6/5`

**Substitution method;**

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

`x + y = 5`

Or, `x = 5 – y`

Substitute the value of x in second equation;

Or, `2(5 – y) – 3y = 4`

Or, `10 – 2y – 3y = 4`

Or, `5y = 10 – 4 = 6`

Or, `y = 6/5`

Substitute the value of y in first equation;

`x+y=5`

Or, `x+(6)/(5)=5`

Or, `x=5-(6)/(5)=(25-6)/(5)=(19)/(5)`

Hence, `x=(19)/(5)` and `y=6/5`

(b) `3x + 4y = 10` and `2x – 2y = 2`

**Solution:**Elimination method:

Multiply second equation by 2;

`4x – 4y = 4`

Add this equation with first equation;

`4x – 4y + 3x + 4y = 4 + 10`

Or, `7x = 14`

Or, `x = 2`

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

`3x + 4y = 10`

Or, `3 x 2 + 4y = 10`

Or, `6 + 4y = 10`

Or, `4y = 10 – 6 = 4`

Or, `y = 1`

Hence, `x = 2` and `y = 1`

**Substitution Method:**

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

`2x – 2y = 2`

Or, `2x = 2y + 2`

Or, `x = y + 1`

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

`3x + 4y = 10`

Or, `3(y + 1) + 4y = 10`

Or, `3y + 3 + 4y = 10`

Or, `7y + 3 = 10`

Or, `7y = 10 – 3 = 7`

Or, `y = 1`

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

`x = y + 1`

Or, `x = 1 + 1 = 2`

Hence, `x = 2 and `y = 1`

(c) `3x – 5y – 4 = 0 and 9x = 2y + 7`

**Solution:** Elimination Method:

Multiply the first equation by 3;

`9x – 15y = 12`

Subtract the second equation from this equation;

`(9x – 15y) – (9x – 2y) = 12 – 7`

Or, `9x – 15y – 9x + 2y = 5`

Or, `- 13y = 5`

Or, `y =(- 5)/(13)`

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

`3x-5y=4`

Or, `3x+5xx(5)/(13)=4`

Or, `3x+(25)/(13)=4`

Or, `3x=4-(25)/(13)`

`=(52-25)/(13)=(27)/(13)`

Or, `x=(27)/(13xx3)=9/13`

Hence, `x=9/13` and `y=-5/13`

Substitution Method;

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

`3x-5y=4`

Or, `3x=5y+4`

Or, `x=(5y+4)/(3)`

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

`9x-2y=7`

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

Or, `15y+12-2y=7`

Or, `13y+12=7`

Or, `13y=7-12=-5`

Or, `y=-5/13`

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

`x=(5y+4)/(3)`

Or, `x=(5xx(-5/13)+4)/(3)`

`=((-25)/(13)+4)/(3)`

`=((-25+52)/(13))/(3)=(27)/(39)=9/13`

Hence, `x=9/13` and `y=-5/13`

(d) `x/2+(2y)/(3)=-1` and `x-y/3=3`

**Solution:** Elimination method:

Equation 1:

`x/2+(2y)/(3)=-1`

Or, `(3x+4y)/(6)=-1`

Or, `3x+4y=-6`

Equation 2:

`x-y/3=3`

Or, `(3x-y)/(3)=3`

Or, `3x-y=9`

Subtracting second equation from first equation, we get;

`3x+4y-(3x-y)=-6-9`

Or, `3x+4y-3x+y=-15`

Or, `5y=-15`

Or, `y=-3`

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

`3x – y = 9`

Or, `3x + 3 = 9`

Or, `3x = 9 – 3 = 6`

Or, `x = 2`

Hence, `x = 2` and `y = - 3`

**Substitution Method:**

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

`3x – y = 9`

Or, `y = 3x – 9`

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

`3x + 4y = - 6`

Or, `3x + 4(3x – 9) = - 6`

Or, `3x + 12x – 36 = - 6`

Or, `15x – 36 = - 6`

Or, `15x = - 6 + 36 = 30`

Or, `x = 2`

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

`y = 3x – 9`

Or, `y = 3 xx 2 – 9`

= `6 – 9 = - 3`

Hence, `x = 2` and `y = - 3`

PrevNext

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 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)