Solve [(5 + ((9/3) / 6) / 4)(9/4 / (4/9) + 3 + 1/5) - [7 - 9] * 5] | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$[(5+9\div3\div6\div4)(\frac{9}{4}\div\frac{4}{9}+3+\frac{1}{5})-[7-9]5]$$

Evaluate

$\frac{33501}{640}=52.3453125$

Solution Steps

Express $\frac{\frac{\frac{9}{3}}{6}}{4}$ as a single fraction.

$$\left(5+\frac{\frac{9}{3}}{6\times 4}\right)\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Divide $9$ by $3$ to get $3$.

$$\left(5+\frac{3}{6\times 4}\right)\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Multiply $6$ and $4$ to get $24$.

$$\left(5+\frac{3}{24}\right)\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Reduce the fraction $\frac{3}{24}$ to lowest terms by extracting and canceling out $3$.

$$\left(5+\frac{1}{8}\right)\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Convert $5$ to fraction $\frac{40}{8}$.

$$\left(\frac{40}{8}+\frac{1}{8}\right)\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Since $\frac{40}{8}$ and $\frac{1}{8}$ have the same denominator, add them by adding their numerators.

$$\frac{40+1}{8}\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Add $40$ and $1$ to get $41$.

$$\frac{41}{8}\left(\frac{\frac{9}{4}}{\frac{4}{9}}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Divide $\frac{9}{4}$ by $\frac{4}{9}$ by multiplying $\frac{9}{4}$ by the reciprocal of $\frac{4}{9}$.

$$\frac{41}{8}\left(\frac{9}{4}\times \frac{9}{4}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Multiply $\frac{9}{4}$ times $\frac{9}{4}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{41}{8}\left(\frac{9\times 9}{4\times 4}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Do the multiplications in the fraction $\frac{9\times 9}{4\times 4}$.

$$\frac{41}{8}\left(\frac{81}{16}+3+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Convert $3$ to fraction $\frac{48}{16}$.

$$\frac{41}{8}\left(\frac{81}{16}+\frac{48}{16}+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Since $\frac{81}{16}$ and $\frac{48}{16}$ have the same denominator, add them by adding their numerators.

$$\frac{41}{8}\left(\frac{81+48}{16}+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Add $81$ and $48$ to get $129$.

$$\frac{41}{8}\left(\frac{129}{16}+\frac{1}{5}\right)-\left(7-9\right)\times 5$$

Least common multiple of $16$ and $5$ is $80$. Convert $\frac{129}{16}$ and $\frac{1}{5}$ to fractions with denominator $80$.

$$\frac{41}{8}\left(\frac{645}{80}+\frac{16}{80}\right)-\left(7-9\right)\times 5$$

Since $\frac{645}{80}$ and $\frac{16}{80}$ have the same denominator, add them by adding their numerators.

$$\frac{41}{8}\times \frac{645+16}{80}-\left(7-9\right)\times 5$$

Add $645$ and $16$ to get $661$.

$$\frac{41}{8}\times \frac{661}{80}-\left(7-9\right)\times 5$$

Multiply $\frac{41}{8}$ times $\frac{661}{80}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{41\times 661}{8\times 80}-\left(7-9\right)\times 5$$

Do the multiplications in the fraction $\frac{41\times 661}{8\times 80}$.

$$\frac{27101}{640}-\left(7-9\right)\times 5$$

Subtract $9$ from $7$ to get $-2$.

$$\frac{27101}{640}-\left(-2\times 5\right)$$

Multiply $-2$ and $5$ to get $-10$.

$$\frac{27101}{640}-\left(-10\right)$$

The opposite of $-10$ is $10$.

$$\frac{27101}{640}+10$$

Convert $10$ to fraction $\frac{6400}{640}$.

$$\frac{27101}{640}+\frac{6400}{640}$$

Since $\frac{27101}{640}$ and $\frac{6400}{640}$ have the same denominator, add them by adding their numerators.

$$\frac{27101+6400}{640}$$

Add $27101$ and $6400$ to get $33501$.

$$\frac{33501}{640}$$

Factor

$\frac{3 \cdot 13 \cdot 859}{2 ^ {7} \cdot 5} = 52\frac{221}{640} = 52.3453125$