Question

Solve the following equations.

=> $\frac{6m+1}{3} + 1 = \frac{m-3}{6}$

=>5b-2(2b-7)=2(3b-1)+$\frac{7}{2}$

Answer 1

$\bold{Answer :}$

$\bold{(1)}$

Given that,

(6m + 1)/3 + 1 = (m - 3)/6

or, (6m + 1 + 3)/3 = (m - 3)/6

or, (6m + 4)/3 = (m - 3)/6

or, 6 (6m + 4)/3 = m - 3

or, 2 (6m + 4) = m - 3

or, 12m + 8 = m - 3

or, 12m - m = - 3 - 8

or, 11m = - 11

or, m = - 11/11

or, m = - 1

Therefore, the required solution be $\bold{m = -1}$.

$\bold{(2)}$

Given that,

5b - 2 (2b - 7) = 2 (3b - 1) + 7/2

or, 5b - 4b + 14 = 6b - 2 + 7/2

or, b + 14 = 6b - 2 + 7/2

or, 6b - b = 14 + 2 - 7/2

or, 5b = (28 + 4 - 7)/2

or, 5b = 25/2

or, b = 25/(2 × 5)

or, b = 5/2

Therefore, the required solution be $\bold{b = 5/2}$

#$\bold{MarkAsBrainliest}$

Answer 2

$\textbf{Question :- 1 }$

=> $\frac{6m + 1}{3} + 1 = \frac{m - 3}{6}$

$\textbf{Solution :-}$

$\frac{6m + 1}{3} + 1 = \frac{m - 3}{6}$.

=> $\frac{6m + 1 + 3}{3} = \frac{m - 3}{6}$

=> $\frac{6m + 4}{3} = \frac{m - 3}{6}$

=> On multiplying { 6m + 4 } by 6 and { m - 3} by 3

=> we get ,

=> 36m + 24 = 3m - 9

=> 36m - 3m = { - 9 } - 24

=> 33m = { - 33 }

=> m => $\frac{-33}{33}$

=> m = - 1

$\textbf{Hence , the value of " m " is - 1}$

$\textbf{Question : 2}$

♧ 5b - 2 ( 2b - 7 ) = 2( 3b - 1 ) + $\frac{7}{2}$

$\textbf{Solution :-}$

=> 5b - 4b + 14 = ( 6b - 2 ) + $\frac{7}{2}$

On taking L.C.M. of R.H.S.

=> b + 14 = [ 12b - 4 + 7 ] / 2

=> b + 14 = [ 12b + 3 ] / 2

On multiplying { b - 14 } by 2 ,

=> 2b + 28 = 12b + 3

=> 2b - 12b = 3 - 28

=> - 10b = - 25

=> b = 25 / 10

=> b = 5 / 2 or 2.5

$\textbf{Hence , the value of " b " is 5 / 2}$



$\textbf{Thanks !}$☺