Question

$\frac{4x}{3 } - 1 = \frac{14}{15} x + \frac{19}{5}$
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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=12}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies \frac{4x}{3 } - 1 = \frac{14}{15} x + \frac{19}{5} \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies value \: of \: x =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \frac{4x}{3 } - 1 = \frac{14}{15} x + \frac{19}{5} \\ \\ \tt: \implies \frac{4x}{3} - \frac{14x}{15} = \frac{19}{5} + 1 \\ \\ \tt: \implies \frac{4x \times 5 - 14x}{15} = \frac{19 + 1 \times 5}{5} \\ \\ \tt: \implies \frac{20x - 14x}{15} = \frac{19 + 5}{5} \\ \\ \tt: \implies \frac{6x}{15} = \frac{24}{5} \\ \\ \tt: \implies \frac{6x}{24} = \frac{15}{5} \\ \\ \tt: \implies \frac{x}{4} = 3 \\ \\ \tt: \implies x = 3 \times 4 \\ \\ \green{\tt: \implies x = 12}$

Answer 2

$\LARGE{ \underline{ \purple{ \sf{Required \: answer:}}}}$

Here, we have a equation with variable x and we have to solve for the value of x.

Given Eq.

• $\sf{ \dfrac{4x}{3} - 1 = \dfrac{14}{15}x + \dfrac{19}{5} }$

Isolating x to any one side of the equation. For convenience, I am shifting the term with x to the LHS.

⇛ $\sf{\dfrac{4x}{3} - \dfrac{14}{15}x = \dfrac{19}{5} + 1}$

Taking LCM in both sides,

⇛ $\sf{ \dfrac{20x - 14x}{15} = \dfrac{19 + 5}{5} }$

This is equals to,

⇛ $\sf{ \dfrac{6x}{15} = \dfrac{24}{5} }$

Cross multiplying and dividing 6 from both sides,

⇛ $\sf{x = \dfrac{24 \times 15}{5 \times 6} }$

Cancelling,

⇛ $\sf{x = 4 \times 3}$

And, x = 12 (Answer)