# Linear Equations

## Exercise 2.5 Part 2

##### Solution of NCERT Exercise from Question 6 to 10

Question 6: m-(m-1)/(2)=1-(m-2)/(3)

Solution: Given, m-(m-1)/(2)=1-(m-2)/(3)

After transposing -(m-2)/(3) to LHS we get:

m-(m-1)/(2)+(m-2)/(3)=1

Or, (6m-3(m-1)+2(m-2))/(6)=1

Or, (6m-3m+3+2m-4)/(6)=1

Or, (3m+2m+3-4)/(6)=1

Or, (5m-1)/(6)=1

Now, after multiplying both sides with 6, we get:

(5m-1)/(6)xx6=1xx6

Or, 5m-1=6

After transposing -1 to RHS we get:

5m=6+1=7

After dividing both sides with 5 we get:

(5m)/(5)=7/5

Or, m=7/5

#### Simplify and solve the following linear equations:

Question 7: 3(t-3)=5(2t+1)

Solution: Given 3(t-3)=5(2t+1)

Or, 3t-9=10t+5

After transposing -9 to RHS and 10t to LHS we get:

3t-10t=5+9

Or, -7t=14

After dividing both sides by 7 we get:

(-7t)/(7)=(14)/(7)

Or, -t=2

Or, t=-2

Question 8: 15(y-4)-2(y-9)+5(y+6)=0

Solution: Given 15(y-4)-2(y-9)+5(y+6)=0

After removing the brackets we get:

15y-60-2y+18+5y+30=0

After arranging the above equation we get:

15y-2y+5y-60+18+30=0

Or, 18y-12=0

After transposing -12 to RHS we get:

18y=12

After dividing both sides by 18 we get:

(18y)/(18)=(12)/(18)

Or, y=2/3

Question 9: 3(5z-7)-2(9z-11)=4(8z-13)-17

Solution: Given 3(5z-7)-2(9z-11)=4(8z-13)-17

After removing the brackets we get:

15z-21-18z+22=32z-52-17

Or, 15z-18z-21+22=32z-69

Or, -3z+1=32z-69

After transposing 1 to RHS and 32z to LHS we get:

-3x-32z=-69-1

Or, -35z=-70

After dividing both sides by 35 we get:

(-35z)/(35)=(-70)/(35)

Or, -z=-2

Or, z=2

Question 10: 0.25(4f-3)=0.05(10f-9)

Solution: Given 0.25(4f-3)=0.05(10f-9)

After removing the brackets we get:

f-0.75=0.5f-0.45

After transposing 0.5f to LHS and -0.75 to RHS we get:

f-0.5f=-0.45+0.75

Or, 0.5f=0.3

After dividing both sides by 0.5 we get:

(0.5f)/(0.5)=(0.3)/(0.5)

Or, f=3/5=0.6