# Linear Equations

## Exercise 2.3 Part 2

#### Solve the following equations and check your results

Question 6: 8x+4=3(x-1)+7

Solution: Given 8x+4=3(x-1)+7

By removing brackets from RHS we get:

8x+4=3x-3+7

Or, 8x+4=3x+4

By transposing 3x to LHS and 4 to RHS we get:

8x-3x=4-4

Or, 5x=0

After dividing both sides by 5 we get:

(5x)/(5)=0/5

Or, x=0

CHECK: Given equation is 8x+4=3x+4

By substituting the value of x in RHS we get:

8x+4=3xx0+4

Or, 8x+4=0+4=4

By transposing 4 to RHS we get:

8x=4-4

Or, 8x=0

By dividing both sides by 8 we get:

(8x)/(8)=0/8

Or, x=0 proved

Question 7: x=4/5(x+10)

Solution: Given x=4/5(x+10)

After removing the brackets we get:

x=(4x)/(5)+(40)/(5)

Or, x=(4x)/(5)+8

By transposing (4x)/(5) to LHS we get:

x-(4x)/(5)=8

Or, (5x-4x)/(5)=8

Or, x/5=8

By multiplying both sides by 5 we get:

x/5xx5=8xx5

Or, x=40

CHECK: Given equation is: x=4/5(x+10)

By substituting the value of x in RHS we get:

x=4/5(40+10)

Or, x=4/5xx50

Or, x=4xx10=40 proved

Question 8: (2x)/(3)+1=(7x)/(15)+3

Solution: Given (2x)/(3)+1=(7x)/(15)+3

By transposing (7x)/(15) to LHS and 1 to RHS we get:

(2x)/(3)-(7x)/(15)=3-1

Or, (10x-7x)/(15)=2

Or, (3x)/(15)=2

Or, x/5=2

By multiplying both sides by 5 we get:

x/5xx5=2xx5

Or, x=10

CHECK: Given equation is (2x)/(3)+1=(7x)/(15)+3

By substituting the value of x in RHS we get:

(2x)/(3)+1=(7xx10)/(15)+3

Or, (2x)/(3)+1=(14)/(3)+3

By transposing 1 to RHS we get:

(2x)/(3)=(14)/(3)+3-1

Or, (2x)/(3)=(14)/(3)+2

Or, (2x)/(3)=(14+6)/(3)

Or, (2x)/(3)=20/3

By multiplying both sides by 3 we get:

(2x)/(3)xx3=20/3xx3

Or, 2x=20

By dividing both sides by 2 we get:

(2x)/(2)=20/2

Or, x=10 proved

Question 9: 2y+5/3=(26)/(3)-y

Solution: Given 2y+5/3=(26)/(3)-y

By transposing –y to LHS and 5/3 to RHS we get:

2y+y=(26)/(3)-5/3

Or, 3y=(26-5)/(3)

Or, 3y=(21)/(3)=7

By dividing both sides by 3 we get:

(3y)/(3)=7/3

Or, y=7/3

CHECK: Given equation is 2y+5/3=(26)/(3)-y

By substituting the value of y in RHS we get:

2y+5/3=(26)/(3)-7/3

Or, 2y+5/3=(26-7)/(3)

Or, 2y+5/3=(19)/(3)

By transposing 5/3 to RHS we get:

2y=(19)/(3)-5/3

Or, 2y=(19-5)/(3)=(14)/(3)

By dividing both sides by 2 we get:

(2y)/(2)=(14)/(3)÷2

Or, y=7/3 proved

Question 10: 3m=5m-8/5

Solution: Given 3m=5m-8/5

By transposing 5m to LHS we get:

3m-5m=-8/5

Or, -2m=-8/5

Or, 2m=8/5

By dividing both sides by 2 we get:

(2m)/(2)=8/5÷2

Or, m=4/5

CHECK: Given equation is 3m=5m-8/5

By substituting the value of m in RHS we get:

3m=5xx4/5-8/5

Or, 3m=4-8/5

Or, 3m=(20-8)/(5)=(12)/(5)

By dividing both sides by 3 we get:

(3m)/(3)=(12)/(5)÷3

Or, m=4/5

CHECK: Given equation is 3m=5m-8/5

By substituting the value of m in RHS we get:

3m=5xx4/5-8/5

Or, 3m=4-8/5

Or, 3m=(20-8)/(5)

Or, 3m=(12)/(5)

By dividing both sides by 3 we get:

(3m)/(3)=(12)/(5)÷3

Or, m=4/5 proved