Solve (7+(1)/(9))\div(64)/(81)-3 | Equatix

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Question

$$(7+ \frac{ 1 }{ 9 } ) \div \frac{ 64 }{ 81 } -3$$

Evaluate

$6$

Solution Steps

Convert $7$ to fraction $\frac{63}{9}$.

$$\frac{\frac{63}{9}+\frac{1}{9}}{\frac{64}{81}}-3$$

Since $\frac{63}{9}$ and $\frac{1}{9}$ have the same denominator, add them by adding their numerators.

$$\frac{\frac{63+1}{9}}{\frac{64}{81}}-3$$

Add $63$ and $1$ to get $64$.

$$\frac{\frac{64}{9}}{\frac{64}{81}}-3$$

Divide $\frac{64}{9}$ by $\frac{64}{81}$ by multiplying $\frac{64}{9}$ by the reciprocal of $\frac{64}{81}$.

$$\frac{64}{9}\times \frac{81}{64}-3$$

Multiply $\frac{64}{9}$ times $\frac{81}{64}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{64\times 81}{9\times 64}-3$$

Cancel out $64$ in both numerator and denominator.

$$\frac{81}{9}-3$$

Divide $81$ by $9$ to get $9$.

$$9-3$$

Subtract $3$ from $9$ to get $6$.

$$6$$

Show Solution

Factor

$2\times 3$