$$(40\times b)\div12+14-(7\times16)+13\times8)$$
$\frac{10b}{3}+6$
$$\frac{10}{3}b+14-7\times 16+13\times 8$$
$$\frac{10}{3}b+14-112+13\times 8$$
$$\frac{10}{3}b-98+13\times 8$$
$$\frac{10}{3}b-98+104$$
$$\frac{10}{3}b+6$$
Show Solution
Hide Solution
$\frac{2\left(5b+9\right)}{3}$
$$\frac{2\left(5b+9\right)}{3}$$
$$5b+9$$