# Prove the following

Question:

Express $0.6+0 . \overline{7}+0.4 \overline{7}$ in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0 .$

Solution:

Let $x=0 . \overline{7}=0.777$      $\ldots(\mathrm{i})$

On multiplying both sides of Eq. (i) by 10, we get

$10 x=7.77 \ldots$                ...(ii)

On subtracting Eq. (i) from Eq. (ii), we get

$10 x-x=(7.77 \ldots)-(0.77 \ldots)$

$\Rightarrow$           $9 x=7$

$\therefore$                 $x=\frac{7}{9}$

Now, let $\quad y=0.47=0.4777 \ldots . \quad \ldots$ (iii)

On multiplying both sides of Eq. (iii) by 10 , we get

$10 y=4.777 \ldots$                 $\ldots$ (iv)

On multiplying both sides, Eq. (iv) by 10, we get

$100 y=47.777 \ldots$                    $\ldots(\mathrm{v})$

On subtracting Eq. (iv) from Eq. (v), we get

$(100 y-10 y)=(47.777 \ldots)-(4.777 \ldots)$

$\Rightarrow$                $90 y=43=\frac{43}{90}$

$\therefore$                   $0.6+0 . \overline{7}+0.4 \overline{7}=\frac{6}{10}+\frac{7}{9}+\frac{43}{90}$

$=\frac{54+70+43}{90}=\frac{167}{90}$