Solve -9+(9)/(2)-7\times(6-(1)/(2)\times(-(5)/(2)+3\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

$$-9+ \frac{ 9 }{ 2 } -7 \times (6- \frac{ 1 }{ 2 } \times (- \frac{ 5 }{ 2 } +3 \times (-1+ \frac{ 5 }{ 3 } )))$$

Evaluate

$-\frac{193}{4}=-48.25$

Solution Steps

Convert $-9$ to fraction $-\frac{18}{2}$.

$$-\frac{18}{2}+\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+3\left(-1+\frac{5}{3}\right)\right)\right)$$

Since $-\frac{18}{2}$ and $\frac{9}{2}$ have the same denominator, add them by adding their numerators.

$$\frac{-18+9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+3\left(-1+\frac{5}{3}\right)\right)\right)$$

Add $-18$ and $9$ to get $-9$.

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+3\left(-1+\frac{5}{3}\right)\right)\right)$$

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

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+3\left(-\frac{3}{3}+\frac{5}{3}\right)\right)\right)$$

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

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+3\times \frac{-3+5}{3}\right)\right)$$

Add $-3$ and $5$ to get $2$.

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+3\times \frac{2}{3}\right)\right)$$

Cancel out $3$ and $3$.

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+2\right)\right)$$

Convert $2$ to fraction $\frac{4}{2}$.

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{5}{2}+\frac{4}{2}\right)\right)$$

Since $-\frac{5}{2}$ and $\frac{4}{2}$ have the same denominator, add them by adding their numerators.

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\times \frac{-5+4}{2}\right)$$

Add $-5$ and $4$ to get $-1$.

$$-\frac{9}{2}-7\left(6-\frac{1}{2}\left(-\frac{1}{2}\right)\right)$$

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

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

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

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

Fraction $\frac{-1}{4}$ can be rewritten as $-\frac{1}{4}$ by extracting the negative sign.

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

The opposite of $-\frac{1}{4}$ is $\frac{1}{4}$.

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

Convert $6$ to fraction $\frac{24}{4}$.

$$-\frac{9}{2}-7\left(\frac{24}{4}+\frac{1}{4}\right)$$

Since $\frac{24}{4}$ and $\frac{1}{4}$ have the same denominator, add them by adding their numerators.

$$-\frac{9}{2}-7\times \frac{24+1}{4}$$

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

$$-\frac{9}{2}-7\times \frac{25}{4}$$

Express $7\times \frac{25}{4}$ as a single fraction.

$$-\frac{9}{2}-\frac{7\times 25}{4}$$

Multiply $7$ and $25$ to get $175$.

$$-\frac{9}{2}-\frac{175}{4}$$

Least common multiple of $2$ and $4$ is $4$. Convert $-\frac{9}{2}$ and $\frac{175}{4}$ to fractions with denominator $4$.

$$-\frac{18}{4}-\frac{175}{4}$$

Since $-\frac{18}{4}$ and $\frac{175}{4}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{-18-175}{4}$$

Subtract $175$ from $-18$ to get $-193$.

$$-\frac{193}{4}$$

Factor

$-\frac{193}{4} = -48\frac{1}{4} = -48.25$