Question

please give me the correct answer​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Solve for m :-

$\rm :\longmapsto\:\dfrac{3m + 5}{5} - \dfrac{5m - 7}{6} = \dfrac{m - 2}{3}$

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\:\dfrac{3m + 5}{5} - \dfrac{5m - 7}{6} = \dfrac{m - 2}{3}$

On talking LCM on LHS, we get

$\rm :\longmapsto\:\dfrac{6(3m + 5) - 5(5m - 7)}{30} = \dfrac{m - 2}{3}$

$\rm :\longmapsto\:\dfrac{18m + 30- 25m + 35}{30} = \dfrac{m - 2}{3}$

$\rm :\longmapsto\:\dfrac{(18m- 25m) + (30 + 35)}{30} = \dfrac{m - 2}{3}$

$\rm :\longmapsto\:\dfrac{- 7m + 65}{30} = \dfrac{m - 2}{3}$

On multiply by 3, both sides

$\rm :\longmapsto\:\dfrac{- 7m + 65}{10} = \dfrac{m - 2}{1}$

$\rm :\longmapsto\:10m - 20 = - 7m + 65$

$\rm :\longmapsto\:10m + 7m = 20+ 65$

$\rm :\longmapsto\:17m = 85$

$\bf\implies \:m = \cancel \dfrac{85}{17} = 5$

Verification :-

$\rm :\longmapsto\:\dfrac{3m + 5}{5} - \dfrac{5m - 7}{6} = \dfrac{m - 2}{3}$

On substituting m = 5, we get

$\rm :\longmapsto\:\dfrac{3 \times 5 + 5}{5} - \dfrac{5 \times 5 - 7}{6} = \dfrac{5 - 2}{3}$

$\rm :\longmapsto\:\dfrac{15 + 5}{5} - \dfrac{25 - 7}{6} = \dfrac{3}{3}$

$\rm :\longmapsto\:\dfrac{20}{5} - \dfrac{18}{6} = 1$

$\rm :\longmapsto\:4 - 3 = 1$

$\bf\implies \:1 = 1$

Hence, Verified

Answer 2

Answer:

GivenQuestion−

Solve for m :-

\rm :\longmapsto\:\dfrac{3m + 5}{5} - \dfrac{5m - 7}{6} = \dfrac{m - 2}{3}:⟼

5

3m+5

−

6

5m−7

=

3

m−2

\large\underline{\sf{Solution-}}

Solution−

Given equation is

\rm :\longmapsto\:\dfrac{3m + 5}{5} - \dfrac{5m - 7}{6} = \dfrac{m - 2}{3}:⟼

5

3m+5

−

6

5m−7

=

3

m−2

On talking LCM on LHS, we get

\rm :\longmapsto\:\dfrac{6(3m + 5) - 5(5m - 7)}{30} = \dfrac{m - 2}{3}:⟼

30

6(3m+5)−5(5m−7)

=

3

m−2

\rm :\longmapsto\:\dfrac{18m + 30- 25m + 35}{30} = \dfrac{m - 2}{3}:⟼

30

18m+30−25m+35

=

3

m−2

\rm :\longmapsto\:\dfrac{(18m- 25m) + (30 + 35)}{30} = \dfrac{m - 2}{3}:⟼

30

(18m−25m)+(30+35)

=

3

m−2

\rm :\longmapsto\:\dfrac{- 7m + 65}{30} = \dfrac{m - 2}{3}:⟼

30

−7m+65

=

3

m−2

On multiply by 3, both sides

\rm :\longmapsto\:\dfrac{- 7m + 65}{10} = \dfrac{m - 2}{1}:⟼

10

−7m+65

=

1

m−2

\rm :\longmapsto\:10m - 20 = - 7m + 65:⟼10m−20=−7m+65

\rm :\longmapsto\:10m + 7m = 20+ 65:⟼10m+7m=20+65

\rm :\longmapsto\:17m = 85:⟼17m=85

\bf\implies \:m = \cancel \dfrac{85}{17} = 5⟹m=

17

85

=5

Verification :-

\rm :\longmapsto\:\dfrac{3m + 5}{5} - \dfrac{5m - 7}{6} = \dfrac{m - 2}{3}:⟼

5

3m+5

−

6

5m−7

=

3

m−2

On substituting m = 5, we get

\rm :\longmapsto\:\dfrac{3 \times 5 + 5}{5} - \dfrac{5 \times 5 - 7}{6} = \dfrac{5 - 2}{3}:⟼

5

3×5+5

−

6

5×5−7

=

3

5−2

\rm :\longmapsto\:\dfrac{15 + 5}{5} - \dfrac{25 - 7}{6} = \dfrac{3}{3}:⟼

5

15+5

−

6

25−7

=

3

3

\rm :\longmapsto\:\dfrac{20}{5} - \dfrac{18}{6} = 1:⟼

5

20

−

6

18

=1

\rm :\longmapsto\:4 - 3 = 1:⟼4−3=1

\bf\implies \:1 = 1⟹1=1

Hence, Verified