# Linear Equations

## Exercise 2.6 Part 1

#### Solve the following equations:

Question 1: (8x-3)/(3x)=2

Solution: Given (8x-3)/(3x)=2

Multiplying both sides by 3x we get:

(8x-3)/(3x)xx3x=2xx3x

Or, 8x-3=6x

After transposing 6x to LHS we get:

8x-6x-3=0

Or, 2x-3=0

After transposing -3 to RHS we get:

2x=3

After dividing both sides by 2 we get:

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

Or, x=3/2

Question 2: (9x)/(7-6x)=15

Solution: Given (9x)/(7-6x)=15

Multiplying both sides with 7-6x we get:

(9x)/(7-6x)=15(7-6x)

Or, 9x=105-90x

After transposing -90x to LHS we get:

9x+90x=105

Or, 99x=105

After dividing both sides by 99 we get:

(99x)/(99)=(105)/(99)

Or, x=(35)/(33)

Question 3: (z)/(z+15)=4/9

Solution: Given (z)/(z+15)=4/9

Multiplying both sides by z+15 we get:

(z)/(z+15)xx(z+15)=4/9(z+15)

Or, z=(4z+60)(9)

Or, z=(4z)/(9)+(60)/(9)

Or, z=(4z)/(9)+(20)/(3)

After transposing (4z)/(9) to LHS we get:

z-(4z)/(9)=(20)/(3)

Or, (9z-4z)/(9)=(20)/(3)

Or, (5z)/(9)=(20)/(3)

After multiplying both sides by 9 we get:

(5z)/(9)xx9=(20)/(3)xx9

Or, 5z=20xx3=60

By dividing both sides by 5 we get:

(5z)/(5)=(60)/(5)

Or, z=12

Question 4: (3y+4)/(2-6y)=(-2)/(5)

Solution: Given (3y+4)/(2-6y)=(-2)/(5)

Multiplying both sides by 2-6y we get:

(3y+4)/(2-6y)xx(2-6y)=(-2)/(5)xx(2-6y)

Or, 3y+4=(-2)/(5)xx2+(2)/(5)xx6y

Or, 3y+4=(-4)/(5)+(12y)/(5)

After transposing 4 to RHS and (12y)/(5) to LHS we get:

3y-(12y)/(5)=(-4)/(5)-4

Or, (15y-12y)/(5)=(-4-20)/(5)

Or, (3y)/(5)=(-24)/(5)

Multiplying both sides by 5 we get:

(3y)/(5)xx5=(-24)/(5)xx5

Or, 3y=-24

After dividing both sides by 3 we get:

(3y)/(3)=(-24)/(3)

Or, y=-8