Linear Equations
Exercise 2.6 Part 1
Solution of NCERT Exercise From Question 1 to 4
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`