Question

Fill in the blankes:

$\frac{-7}{9}+\cdots =3$

Answer 1

Let the required number be $x$

$\therefore \frac{-7}{9}+x=3$

$\Rightarrow x=3-\left(\frac{-7}{9}\right)$

$=\frac{3}{1}+\frac{7}{9}$

$=\frac{3\times 9}{1\times 9}+\frac{7}{9}$

$=\frac{27}{9}+\frac{7}{9}$

$=\frac{27+7}{9}$

$=\frac{34}{9}$

Hence, the required number is $\frac{34}{9}.$