Linear Equations

Exercise 2.6 Part 1

Solution of NCERT Exercise From Question 1 to 4

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`



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