prove that


If $\frac{5 x}{3}-4=\frac{2 x}{5}$, then the numerical value of $2 x-7$ is

(a) $\frac{19}{13}$

(b) $\frac{-13}{19}$

(c) 0

(d) $\frac{13}{19}$



Given, $\frac{5 x}{3}-4=\frac{2 x}{5}$

$\Rightarrow$ $\frac{5 x}{3}-\frac{2 x}{5}=4$ $\left[\right.$ transposing $\frac{2 x}{5}$ to LHS and $-4$ to RHS $]$

$\Rightarrow$ $\frac{25 x-6 x}{15}=4$ [taking LCM in LHS]

$\Rightarrow$ $19 x=60$

$\Rightarrow$ $\frac{19 x}{19}=\frac{60}{19}$ [dividing both sides by 19 ]

$\therefore$ $x=\frac{60}{19}$ 

Now, $2 x-7=2 \times \frac{60}{19}-7$ [putting the value of $x$ ]

$=\frac{120-133}{19}=-\frac{13}{19}$ [taking LCM]

Hence, the numerical value of $2 x-7$ is $-\frac{13}{19}$


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