Class 8 Maths

# Linear Equations

## Exercise 2.3 part 1

### Solve the following equations and check your results

Question 1: 3x=2x+18

Solution: Given 3x=2x+18

By transposing 2x to LHS we get:

3x-2x=18

Or, x=18

CHECK: Given equation is 3x=2x+18

Substituting the value of x in LHS we get:

3x=2xx18+18

Or, 3x=36+18=54

By dividing both sides by 3 we get:

(3x)/(3)=(54)/(3)

Or, x=18 proved

Question 2: 5t-3=3t-5

Solution: Given 5t-3=3t-5

By transposing 3t to LHS and -3 to RHS we get:

5t-3t=-5+3

Or, 2t=-2

By dividing both sides by 2 we get:

(2t)/(2)=-2/2

Or, t=-1

CHECK: Given equation is 5t-3=3t-5

Substituting the value of t in LHS we get:

5xx(-1)-3=3t-5

Or, -5-3=3t-5

Or, -8=3t-5

Transposing -5 to LHS we get:

-8+5=3t

Or, 3t=-3

Dividing both sides by 3 we get:

(3t)/(3)=-3/3

Or, t=-1 proved

Question 3: 5x+9=5+3x

Solution: Given 5x+9=5+3x

By transposing 3x to LHS and 9 to RHS we get:

5x-3x=5-9

Or, 2x=-4

By dividing both sides by 2 we get:

(2x)/(2)=-4/2=-2

Or, x=-2

CHECK: Given equation is 5x+9=5+3x

Substituting the value of x in RHS we get:

5x+9=5+3xx(-2)

Or, 5x+9=5-6=-1

By transposing 9 to RHS we get:

5x=-1-9=-10

By dividing both sides by 5 we get:

(5x)/(5)=(-10)/(5)

Or, x=-2 proved

Question 4: 4x+3=6+2x

Solution: Given 4x+3=6+2x

By transposing 2x to LHS and 3 to RHS we get:

4x-2x=6-3

Or, 2x=3

By dividing both sides by 2 we get:

(2x)/(2)=3/2

Or, x=3/2

CHECK: Given equation is 4x+3=6+2x

Substituting the value of x in RHS we get:

4x+3=6+2xx3/2

Or, 4x+3=6+3=9

By transposing 3 to RHS we get:

4x=9-3=6

By dividing both sides by 4 we get:

(4x)/(4)=6/4

Or, x=3/2 proved

Question 5: 2x-1=14-x

Solution: Given 2x-1=14-x

By transposing –x to LHS and -1 to RHS we get:

2x+x=14+1

Or, 3x=15

By dividing both sides by 3 we get:

(3x)/(3)=(15)/(3)

Or, x=5

CHECK: Given equation is 2x-1=14-x

By substituting the value of x in RHS we get:

2x-1=14-4=9

By transposing -1 to RHS we get:

2x=9+1=10

By dividing both sides by 2 we get:

(2x)/(2)=(10)/(2)

Or, x=5 proved