Solve Date Page - 7 3 + 7 2 -5+[ -3 2 +4x\ -7+ 1 4 *(-3+ 1 3 )^ -

Question

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

Evaluate

$-5$

Solution Steps

Least common multiple of $3$ and $2$ is $6$. Convert $-\frac{7}{3}$ and $\frac{1}{2}$ to fractions with denominator $6$.

$$-\frac{14}{6}+\frac{3}{6}-5-\frac{-3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Since $-\frac{14}{6}$ and $\frac{3}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{-14+3}{6}-5-\frac{-3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Add $-14$ and $3$ to get $-11$.

$$-\frac{11}{6}-5-\frac{-3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Convert $5$ to fraction $\frac{30}{6}$.

$$-\frac{11}{6}-\frac{30}{6}-\frac{-3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Since $-\frac{11}{6}$ and $\frac{30}{6}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-11-30}{6}-\frac{-3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Subtract $30$ from $-11$ to get $-41$.

$$-\frac{41}{6}-\frac{-3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Fraction $\frac{-3}{2}$ can be rewritten as $-\frac{3}{2}$ by extracting the negative sign.

$$-\frac{41}{6}-\left(-\frac{3}{2}\right)+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

The opposite of $-\frac{3}{2}$ is $\frac{3}{2}$.

$$-\frac{41}{6}+\frac{3}{2}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Least common multiple of $6$ and $2$ is $6$. Convert $-\frac{41}{6}$ and $\frac{3}{2}$ to fractions with denominator $6$.

$$-\frac{41}{6}+\frac{9}{6}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Since $-\frac{41}{6}$ and $\frac{9}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{-41+9}{6}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Add $-41$ and $9$ to get $-32$.

$$\frac{-32}{6}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Reduce the fraction $\frac{-32}{6}$ to lowest terms by extracting and canceling out $2$.

$$-\frac{16}{3}+4\times 2-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Multiply $4$ and $2$ to get $8$.

$$-\frac{16}{3}+8-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

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

$$-\frac{16}{3}+\frac{24}{3}-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Since $-\frac{16}{3}$ and $\frac{24}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{-16+24}{3}-7+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Add $-16$ and $24$ to get $8$.

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

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

$$\frac{8}{3}-\frac{21}{3}+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Since $\frac{8}{3}$ and $\frac{21}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{8-21}{3}+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Subtract $21$ from $8$ to get $-13$.

$$-\frac{13}{3}+\frac{1}{4}\left(-3+\frac{1}{3}\right)$$

Convert $-3$ to fraction $-\frac{9}{3}$.

$$-\frac{13}{3}+\frac{1}{4}\left(-\frac{9}{3}+\frac{1}{3}\right)$$

Since $-\frac{9}{3}$ and $\frac{1}{3}$ have the same denominator, add them by adding their numerators.

$$-\frac{13}{3}+\frac{1}{4}\times \frac{-9+1}{3}$$

Add $-9$ and $1$ to get $-8$.

$$-\frac{13}{3}+\frac{1}{4}\left(-\frac{8}{3}\right)$$

Multiply $\frac{1}{4}$ times $-\frac{8}{3}$ by multiplying numerator times numerator and denominator times denominator.

$$-\frac{13}{3}+\frac{1\left(-8\right)}{4\times 3}$$

Do the multiplications in the fraction $\frac{1\left(-8\right)}{4\times 3}$.

$$-\frac{13}{3}+\frac{-8}{12}$$

Reduce the fraction $\frac{-8}{12}$ to lowest terms by extracting and canceling out $4$.

$$-\frac{13}{3}-\frac{2}{3}$$

Since $-\frac{13}{3}$ and $\frac{2}{3}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-13-2}{3}$$

Subtract $2$ from $-13$ to get $-15$.

$$\frac{-15}{3}$$

Divide $-15$ by $3$ to get $-5$.

$$-5$$

Show Solution

Factor

$-5$

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