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

Question

$$- \frac { 7 } { 3 } + \frac { 1 } { 2 } - 5 + [ \frac { - 3 } { 2 } + 4 \times 2 ] - \frac { 1 + 1 } { 4 } \times ( - 0 + \frac { 1 } { 3 } ) ]$$

Evaluate

$-\frac{1}{2}=-0.5$

Solution Steps

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

$$-\frac{14}{6}+\frac{3}{6}-5+\frac{-3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$\frac{-14+3}{6}-5+\frac{-3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{11}{6}-5+\frac{-3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{11}{6}-\frac{30}{6}+\frac{-3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$\frac{-11-30}{6}+\frac{-3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{41}{6}+\frac{-3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{41}{6}-\frac{3}{2}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{41}{6}-\frac{9}{6}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$\frac{-41-9}{6}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

Subtract $9$ from $-41$ to get $-50$.

$$\frac{-50}{6}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

Reduce the fraction $\frac{-50}{6}$ to lowest terms by extracting and canceling out $2$.

$$-\frac{25}{3}+4\times 2-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{25}{3}+8-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

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

$$-\frac{25}{3}+\frac{24}{3}-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

Since $-\frac{25}{3}$ and $\frac{24}{3}$ have the same denominator, add them by adding their numerators.

$$\frac{-25+24}{3}-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

Add $-25$ and $24$ to get $-1$.

$$-\frac{1}{3}-\frac{1+1}{4}\left(0+\frac{1}{3}\right)$$

Add $1$ and $1$ to get $2$.

$$-\frac{1}{3}-\frac{2}{4}\left(0+\frac{1}{3}\right)$$

Reduce the fraction $\frac{2}{4}$ to lowest terms by extracting and canceling out $2$.

$$-\frac{1}{3}-\frac{1}{2}\left(0+\frac{1}{3}\right)$$

Add $0$ and $\frac{1}{3}$ to get $\frac{1}{3}$.

$$-\frac{1}{3}-\frac{1}{2}\times \frac{1}{3}$$

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

$$-\frac{1}{3}-\frac{1\times 1}{2\times 3}$$

Do the multiplications in the fraction $\frac{1\times 1}{2\times 3}$.

$$-\frac{1}{3}-\frac{1}{6}$$

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

$$-\frac{2}{6}-\frac{1}{6}$$

Since $-\frac{2}{6}$ and $\frac{1}{6}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-2-1}{6}$$

Subtract $1$ from $-2$ to get $-3$.

$$\frac{-3}{6}$$

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

$$-\frac{1}{2}$$

Show Solution

Factor

$-\frac{1}{2} = -0.5$

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