Solve -8+(4)/(3)-7\times(-7+(4)/(3)\times(-5+(1)/(3)\times(-5+(1)/(2))) | 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

$$-8+ \frac{ 4 }{ 3 } -7 \times (-7+ \frac{ 4 }{ 3 } \times (-5+ \frac{ 1 }{ 3 } \times (-5+ \frac{ 1 }{ 2 } ))$$

Evaluate

$103$

Solution Steps

Convert $-8$ to fraction $-\frac{24}{3}$.

$$-\frac{24}{3}+\frac{4}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1}{3}\left(-5+\frac{1}{2}\right)\right)\right)$$

Since $-\frac{24}{3}$ and $\frac{4}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{-24+4}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1}{3}\left(-5+\frac{1}{2}\right)\right)\right)$$

Add $-24$ and $4$ to get $-20$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1}{3}\left(-5+\frac{1}{2}\right)\right)\right)$$

Convert $-5$ to fraction $-\frac{10}{2}$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1}{3}\left(-\frac{10}{2}+\frac{1}{2}\right)\right)\right)$$

Since $-\frac{10}{2}$ and $\frac{1}{2}$ have the same denominator, add them by adding their numerators.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1}{3}\times \frac{-10+1}{2}\right)\right)$$

Add $-10$ and $1$ to get $-9$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1}{3}\left(-\frac{9}{2}\right)\right)\right)$$

Multiply $\frac{1}{3}$ times $-\frac{9}{2}$ by multiplying numerator times numerator and denominator times denominator.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{1\left(-9\right)}{3\times 2}\right)\right)$$

Do the multiplications in the fraction $\frac{1\left(-9\right)}{3\times 2}$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5+\frac{-9}{6}\right)\right)$$

Reduce the fraction $\frac{-9}{6}$ to lowest terms by extracting and canceling out $3$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-5-\frac{3}{2}\right)\right)$$

Convert $-5$ to fraction $-\frac{10}{2}$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-\frac{10}{2}-\frac{3}{2}\right)\right)$$

Since $-\frac{10}{2}$ and $\frac{3}{2}$ have the same denominator, subtract them by subtracting their numerators.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\times \frac{-10-3}{2}\right)$$

Subtract $3$ from $-10$ to get $-13$.

$$-\frac{20}{3}-7\left(-7+\frac{4}{3}\left(-\frac{13}{2}\right)\right)$$

Multiply $\frac{4}{3}$ times $-\frac{13}{2}$ by multiplying numerator times numerator and denominator times denominator.

$$-\frac{20}{3}-7\left(-7+\frac{4\left(-13\right)}{3\times 2}\right)$$

Do the multiplications in the fraction $\frac{4\left(-13\right)}{3\times 2}$.

$$-\frac{20}{3}-7\left(-7+\frac{-52}{6}\right)$$

Reduce the fraction $\frac{-52}{6}$ to lowest terms by extracting and canceling out $2$.

$$-\frac{20}{3}-7\left(-7-\frac{26}{3}\right)$$

Convert $-7$ to fraction $-\frac{21}{3}$.

$$-\frac{20}{3}-7\left(-\frac{21}{3}-\frac{26}{3}\right)$$

Since $-\frac{21}{3}$ and $\frac{26}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$-\frac{20}{3}-7\times \frac{-21-26}{3}$$

Subtract $26$ from $-21$ to get $-47$.

$$-\frac{20}{3}-7\left(-\frac{47}{3}\right)$$

Express $7\left(-\frac{47}{3}\right)$ as a single fraction.

$$-\frac{20}{3}-\frac{7\left(-47\right)}{3}$$

Multiply $7$ and $-47$ to get $-329$.

$$-\frac{20}{3}-\frac{-329}{3}$$

Fraction $\frac{-329}{3}$ can be rewritten as $-\frac{329}{3}$ by extracting the negative sign.

$$-\frac{20}{3}-\left(-\frac{329}{3}\right)$$

The opposite of $-\frac{329}{3}$ is $\frac{329}{3}$.

$$-\frac{20}{3}+\frac{329}{3}$$

Since $-\frac{20}{3}$ and $\frac{329}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{-20+329}{3}$$

Add $-20$ and $329$ to get $309$.

$$\frac{309}{3}$$

Divide $309$ by $3$ to get $103$.

$$103$$

Factor

$103$