Step-by-step explanation:
Given Question :-
Solve for x :-
[tex] \dfrac{2x + 7}{5} - \dfrac{x - 3}{10} = \dfrac{x + 1}{15} [/tex]
[tex] \red{\large\underline{\sf{Solution-}}}[/tex]
Given linear equation is
[tex]\rm :\longmapsto\: \dfrac{2x + 7}{5} - \dfrac{x - 3}{10} = \dfrac{x + 1}{15} [/tex]
[tex]\rm :\longmapsto\: \dfrac{2(2x + 7) - (x - 3)}{10} = \dfrac{x + 1}{15} [/tex]
[tex]\rm :\longmapsto\: \dfrac{4x + 14 - x + 3}{10} = \dfrac{x + 1}{15} [/tex]
[tex]\rm :\longmapsto\: \dfrac{(4x - x) + (14 + 3)}{10} = \dfrac{x + 1}{15} [/tex]
[tex]\rm :\longmapsto\: \dfrac{3x + 17}{10} = \dfrac{x + 1}{15} [/tex]
On multiply by 5 on both sides,
[tex]\rm :\longmapsto\: \dfrac{3x + 17}{2} = \dfrac{x + 1}{3} [/tex]
On cross multiplication, we get
[tex]\rm :\longmapsto\:3(3x + 17) = 2(x + 1)[/tex]
[tex]\rm :\longmapsto\:9x +51 = 2x + 2[/tex]
[tex]\rm :\longmapsto\:9x - 2x = 2 - 51 [/tex]
[tex]\rm :\longmapsto\:7x = - 49[/tex]
[tex]\bf\implies \:x = - 7[/tex]
VERIFICATION
Consider, LHS
[tex]\red{\rm :\longmapsto\: \dfrac{2x + 7}{5} - \dfrac{x - 3}{10}}[/tex]
On substituting the value of x, we get
[tex]\red{\rm \: = \: \dfrac{2( - 7) + 7}{5} - \dfrac{ - 7 - 3}{10}}[/tex]
[tex]\red{\rm \: = \: \dfrac{ - 14 + 7}{5} - \dfrac{ - 10}{10}}[/tex]
[tex]\red{\rm \: = \: \dfrac{ - 7}{5} + 1}[/tex]
[tex]\red{\rm \: = \: \dfrac{ - 7 + 5}{5}}[/tex]
[tex]\red{\rm \: = \: \dfrac{ - 2}{5}}[/tex]
Consider RHS
[tex] \green{\rm :\longmapsto\:\dfrac{x + 1}{15}}[/tex]
On substituting the value of x, we get
[tex] \green{\rm \: = \: \dfrac{ - 7 + 1}{15}}[/tex]
[tex] \green{\rm \: = \: \dfrac{ - 6}{15}}[/tex]
[tex] \green{\rm \: = \: \dfrac{ - 2}{5}}[/tex]
[tex]\rm \implies\:LHS=RHS[/tex]
HENCE, VERIFIED
