Linear Equations

Exercise 2.5 Part 1

Solution of NCERT Exercise from Question 1 to 5

Solve the following linear equations:

Question 1: `x/2-1/5=x/3+1/4`

Solution: Given `x/2-1/5=x/3+1/4`

After transposing `x/3` to LHS and `-1/5` to RHS we get:

`x/2-x/3=1/4+1/5`

Or, `(3x-2x)/(6)=(5+4)/(20)`

Or, `x/6=(9)/(20)`

After multiplying both sides with 6, we get:

`x/6xx6=(9)/(20)xx6=(54)/(20)`

Or, `x=(27)/(10)`


Question 2: `n/2-(3n)/(4)+(5n)/(6)=21`

Solution: Given `n/2-(3n)/(4)+(5n)/(6)=21`

Or, `(6n-9n+10n)/(12)=21`

Or, `(-3n+10n)/(12)=21`

Or, `(7n)/(12)=21`

After multiplying both sides by 12, we get:

`(7n)/(12)xx12=21xx12`

Or, `7n=252`

Now, after dividing both sides by 7, we get:

`(7n)/(7)=(252)/(7)`

Or, `n=36`


Question 3: `x+7-(8x)/(3)=(17)/(6)-(5x)/(2)`

Solution: Given `x+7-(8x)/(3)=(17)/(6)-(5x)/(2)`

Or, `x-(8x)/(3)+7=(17)/(6)-(5x)/(2)`

After transposing 7 to RHS and `-(5x)/(2)` to LHS we get:

`x-(8x)/(3)+(5x)/(2)=(17)/(6)-7`

Or, `(6x-16x+15x)/(6)=(17-42)/(6)`

Or, `(5x)/(6)=-(25)/(6)`

After multiplying both sides with 6, we get:

`(5x)/(6)xx6=-(25)/(6)xx6`

Or, `5x=-25`

After dividing both sides by 5, we get:

`(5x)/(5)=-(25)/(5)`

Or, `x=-5`

Question 4: `(x-5)/(3)=(x-3)/(5)`

Solution: Given, `(x-5)/(3)=(x-3)/(5)`

After multiplying both sides with 3, we get:

`(x-5)/(3)xx3=(x-3)/(5)xx3`

Or, `(x-5)=((x-3)3)/(5)`

After multiplying both sides with 5, we get:

`(x-5)xx5=((x-3)3)/(5)xx5`

Or, `(x-5)5=(x-3)3`

After removing the brackets from both sides we get:

`5x-25=3x-9`

After transposing 3x to LHS and -25 to RHS we get:

`5x-3x=-9+25`

Or, `2x=16`

After dividing both sides by 2, we get:

`(2x)/(2)=(16)/(2)`

Or, `x=8`

Question 5: `(3t-2)/(4)-(2t+3)/(3)=2/3-t`

Solution: Given `(3t-2)/(4)-(2t+3)/(3)=2/3-t`

After transposing `-t` to LHS, we get:

`(3t-2)/(4)-(2t+3)/(3)+t=2/3`

Or, `(3(3t-2)-4(2t+3)+12t)/(12)=2/3`

After removing the brackets, we get:

`(9t-6-8t-12+12t)/(12)=2/3`

Or, `(9t-8t+12t-6-12)/(12)=2/3`

Or, `(13t-18)/(12)=2/3`

After multiplying both sides by 12, we get:

`(13t-18)/(12)xx12=2/3xx12`

Or, `13t-18=(24)/(3)`

After transposing `-18` to RHS, we get:

`13t=(24)/(3)+18`

Or, `13t=(24+54)/(3)=(78)/(3)=26`

Or, `13t=26`

By dividing both sides by 13, we get:

`(13t)/(13)=(26)/(13)`

Or, `t=2`



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