Question

$\:\frac{7x + 14}{3} - \frac{17 - 3x}{5} = 6x - \frac{4x + 2}{3} - 5$ Solve for x ​

Answer 1

$\qquad\leadsto\quad \sf \pink{\:\dfrac{7x + 14}{3} - \dfrac{17 - 3x}{5} = 6x - \dfrac{4x + 2}{3} - 5}$

$\qquad\leadsto\quad \sf \:\dfrac{5(7x + 14) - 3(17 - 3x)}{15} =\dfrac{18x - (4x + 2) - 15}{3}$

$\qquad\leadsto\quad \sf \:\dfrac{35x + 70 - 51 + 9x}{15} =\dfrac{18x - 4x - 2 - 15}{3}$

$\qquad\leadsto\quad \sf \:\dfrac{44x + 19}{15} =\dfrac{14x -17}{3}$

$\qquad\leadsto\quad \sf \:3(44x + 19) = 15(14x - 17)$

$\qquad\leadsto\quad \sf \:44x + 19= 5(14x - 17)$

$\qquad\leadsto\quad \sf \:44x + 19= 70x - 85$

$\qquad\leadsto\quad \sf \: - 26x = - 104$

$\qquad\leadsto\quad \sf \: 26x = 104$

$\qquad\leadsto\quad \sf \:x =\cancel{ \dfrac{104}{26}}$

$\qquad\leadsto\quad \sf \pink{\:x = 4}$

Verification :-

LHS

$\qquad\leadsto\quad \sf \purple{\dfrac{7x + 14}{3} - \dfrac{17 - 3x}{5}}$

On substituting x = 4, we get

$\qquad\leadsto\quad \sf \:  \:\:\dfrac{7(4)+ 14}{3} - \dfrac{17 - 3(4)}{5}$

$\qquad\leadsto\quad \sf   \:\:\dfrac{28+ 14}{3} - \dfrac{17 -12}{5}$

$\qquad\leadsto\quad \sf \:  \:\:\dfrac{42}{3} - \dfrac{5}{5}$

$\qquad\leadsto\quad \sf \:  \:14 - 1$

$\qquad\leadsto\quad \sf \purple{ \:  \:13}$

R.H.S

$\qquad\leadsto\quad \sf \purple{\:6x - \dfrac{4x + 2}{3} - 5}$

On Substituting x = 4, we get

$\qquad\leadsto\quad \sf   \:  \:\:6(4) - \dfrac{4(4) + 2}{3} - 5$

$\qquad\leadsto\quad \sf \:  \:\:24 - \dfrac{16 + 2}{3} - 5$

$\qquad\leadsto\quad \sf \:\:19 - \dfrac{18}{3}$

$\qquad\leadsto\quad \sf \:  \:\:19 -6$

$\qquad\leadsto\quad \sf \purple{\:  \:\:13}$

• Hence,( Verified..!!)