$$3 \frac { 4 } { 99 } + 4 \frac { 7 } { 33 } - 5 \frac { 9 } { 11 }$$
$\frac{142}{99}\approx 1.434343434$
$$\frac{297+4}{99}+\frac{4\times 33+7}{33}-\frac{5\times 11+9}{11}$$
$$\frac{301}{99}+\frac{4\times 33+7}{33}-\frac{5\times 11+9}{11}$$
$$\frac{301}{99}+\frac{132+7}{33}-\frac{5\times 11+9}{11}$$
$$\frac{301}{99}+\frac{139}{33}-\frac{5\times 11+9}{11}$$
$$\frac{301}{99}+\frac{417}{99}-\frac{5\times 11+9}{11}$$
$$\frac{301+417}{99}-\frac{5\times 11+9}{11}$$
$$\frac{718}{99}-\frac{5\times 11+9}{11}$$
$$\frac{718}{99}-\frac{55+9}{11}$$
$$\frac{718}{99}-\frac{64}{11}$$
$$\frac{718}{99}-\frac{576}{99}$$
$$\frac{718-576}{99}$$
$$\frac{142}{99}$$
Show Solution
Hide Solution
$\frac{2 \cdot 71}{3 ^ {2} \cdot 11} = 1\frac{43}{99} = 1.4343434343434343$
