Question

$\frac{16}{9}÷\frac{48}{45}-\frac{2}{7}=$ $1\frac{x}{21}$. Find x.

Answer 1

For the value of $x$, we solve the given equation:-

$\frac{16}{9}÷\frac{48}{45}-\frac{2}{7}=1\frac{x}{21}$

$\Rightarrow \frac{16}{9}\times \frac{45}{48}-\frac{2}{7}=\frac{21+x}{21}$ [Divide $48$ by $16$ we get $3$ & $45$ by $9$ we get $5$]

$\Rightarrow \frac{5}{3}-\frac{2}{7}=\frac{21+x}{21}$

$\Rightarrow \frac{35-6}{21}=\frac{21+x}{21}$ [cancle $21$ both sides]

$\Rightarrow 29=21+x$

$\Rightarrow x=29-21=8$

Hence,

$x=8$.