Solve -(5)/(3)+2-3\times((4)/(3)-1+(2)/(3)\times(-6+(7)/(2)\times(1-(5)/(3))) | 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

$$- \frac{ 5 }{ 3 } +2-3 \times ( \frac{ 4 }{ 3 } -1+ \frac{ 2 }{ 3 } \times (-6+ \frac{ 7 }{ 2 } \times (1- \frac{ 5 }{ 3 } ))$$

Evaluate

$16$

Solution Steps

Convert $2$ to fraction $\frac{6}{3}$.

$$-\frac{5}{3}+\frac{6}{3}-3\left(\frac{4}{3}-1+\frac{2}{3}\left(-6+\frac{7}{2}\left(1-\frac{5}{3}\right)\right)\right)$$

Since $-\frac{5}{3}$ and $\frac{6}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{-5+6}{3}-3\left(\frac{4}{3}-1+\frac{2}{3}\left(-6+\frac{7}{2}\left(1-\frac{5}{3}\right)\right)\right)$$

Add $-5$ and $6$ to get $1$.

$$\frac{1}{3}-3\left(\frac{4}{3}-1+\frac{2}{3}\left(-6+\frac{7}{2}\left(1-\frac{5}{3}\right)\right)\right)$$

Convert $1$ to fraction $\frac{3}{3}$.

$$\frac{1}{3}-3\left(\frac{4}{3}-\frac{3}{3}+\frac{2}{3}\left(-6+\frac{7}{2}\left(1-\frac{5}{3}\right)\right)\right)$$

Since $\frac{4}{3}$ and $\frac{3}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}-3\left(\frac{4-3}{3}+\frac{2}{3}\left(-6+\frac{7}{2}\left(1-\frac{5}{3}\right)\right)\right)$$

Subtract $3$ from $4$ to get $1$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6+\frac{7}{2}\left(1-\frac{5}{3}\right)\right)\right)$$

Convert $1$ to fraction $\frac{3}{3}$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6+\frac{7}{2}\left(\frac{3}{3}-\frac{5}{3}\right)\right)\right)$$

Since $\frac{3}{3}$ and $\frac{5}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6+\frac{7}{2}\times \frac{3-5}{3}\right)\right)$$

Subtract $5$ from $3$ to get $-2$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6+\frac{7}{2}\left(-\frac{2}{3}\right)\right)\right)$$

Multiply $\frac{7}{2}$ times $-\frac{2}{3}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6+\frac{7\left(-2\right)}{2\times 3}\right)\right)$$

Do the multiplications in the fraction $\frac{7\left(-2\right)}{2\times 3}$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6+\frac{-14}{6}\right)\right)$$

Reduce the fraction $\frac{-14}{6}$ to lowest terms by extracting and canceling out $2$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-6-\frac{7}{3}\right)\right)$$

Convert $-6$ to fraction $-\frac{18}{3}$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-\frac{18}{3}-\frac{7}{3}\right)\right)$$

Since $-\frac{18}{3}$ and $\frac{7}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\times \frac{-18-7}{3}\right)$$

Subtract $7$ from $-18$ to get $-25$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2}{3}\left(-\frac{25}{3}\right)\right)$$

Multiply $\frac{2}{3}$ times $-\frac{25}{3}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{2\left(-25\right)}{3\times 3}\right)$$

Do the multiplications in the fraction $\frac{2\left(-25\right)}{3\times 3}$.

$$\frac{1}{3}-3\left(\frac{1}{3}+\frac{-50}{9}\right)$$

Fraction $\frac{-50}{9}$ can be rewritten as $-\frac{50}{9}$ by extracting the negative sign.

$$\frac{1}{3}-3\left(\frac{1}{3}-\frac{50}{9}\right)$$

Least common multiple of $3$ and $9$ is $9$. Convert $\frac{1}{3}$ and $\frac{50}{9}$ to fractions with denominator $9$.

$$\frac{1}{3}-3\left(\frac{3}{9}-\frac{50}{9}\right)$$

Since $\frac{3}{9}$ and $\frac{50}{9}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}-3\times \frac{3-50}{9}$$

Subtract $50$ from $3$ to get $-47$.

$$\frac{1}{3}-3\left(-\frac{47}{9}\right)$$

Express $3\left(-\frac{47}{9}\right)$ as a single fraction.

$$\frac{1}{3}-\frac{3\left(-47\right)}{9}$$

Multiply $3$ and $-47$ to get $-141$.

$$\frac{1}{3}-\frac{-141}{9}$$

Reduce the fraction $\frac{-141}{9}$ to lowest terms by extracting and canceling out $3$.

$$\frac{1}{3}-\left(-\frac{47}{3}\right)$$

The opposite of $-\frac{47}{3}$ is $\frac{47}{3}$.

$$\frac{1}{3}+\frac{47}{3}$$

Since $\frac{1}{3}$ and $\frac{47}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{1+47}{3}$$

Add $1$ and $47$ to get $48$.

$$\frac{48}{3}$$

Divide $48$ by $3$ to get $16$.

$$16$$

Factor

$2^{4}$