Question

solve the following equations and verify the solutions​

Answer 1

Step-by-step explanation:

ots answer I did my all tried then I found it

Answer 2

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

Solve and verify the solution :

$\rm :\longmapsto\:y + 1 - \dfrac{7}{3} y = \dfrac{11}{9} - \dfrac{5y}{6}$

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

Given equation is

$\rm :\longmapsto\:y + 1 - \dfrac{7}{3} y = \dfrac{11}{9} - \dfrac{5y}{6}$

can be rewritten as

$\rm :\longmapsto\:y + 1 - \dfrac{7y}{3} = \dfrac{11}{9} - \dfrac{5y}{6}$

$\rm :\longmapsto\: \dfrac{3y + 3 - 7y}{3} = \dfrac{22 - 15y}{18}$

$\rm :\longmapsto\: \dfrac{3 - 4y}{3} = \dfrac{22 - 15y}{18}$

$\rm :\longmapsto\:18(3 - 4y) = 3(22 - 15y)$

$\rm :\longmapsto\:54 - 72y =66 - 45y$

$\rm :\longmapsto\:- 72y + 45y=66 -54$

$\rm :\longmapsto\:- 27y =12$

$\rm :\longmapsto\:y = - \: \dfrac{12}{27}$

$\bf :\longmapsto\: \boxed{ \bf{ \: \: y \: = \: - \: \dfrac{4}{9} \: \: }}$

Verification

Consider LHS

$\rm :\longmapsto\:y + 1 - \dfrac{7}{3} y$

$\rm \: = \: - \dfrac{4}{9} + 1 + \dfrac{7}{3} \times \dfrac{4}{9}$

$\rm \: = \: - \dfrac{4}{9} + 1 + \dfrac{28}{27}$

$\rm \: = \: \dfrac{ - 12 + 27 + 28}{9}$

$\rm \: = \: \dfrac{ 15 + 28}{9}$

$\rm \: = \: \dfrac{43}{9}$

Consider RHS

$\rm :\longmapsto\:\dfrac{11}{9} - \dfrac{5y}{6}$

$\rm \: = \:\dfrac{11}{9} + \dfrac{5}{6} \times \dfrac{4}{9}$

$\rm \: = \:\dfrac{11}{9} + \dfrac{10}{27}$

$\rm \: = \:\dfrac{33 + 10}{27}$

$\rm \: = \:\dfrac{43}{27}$

$\bf\implies \:LHS \: = \: RHS$

Hence,

$\bf :\longmapsto\: \boxed{ \bf{ \: \: y \: = \: - \: \dfrac{4}{9} \: \: }}$