How to solve ( h) ( +

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve ( h) ( + | Equatix

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Question

$$( \frac { 3 } { 4 } + 5 \frac { 9 } { 10 } ) + ( 7 \frac { 2 } { 3 } - \frac { 1 } { 8 } )$$

Evaluate

$\frac{1703}{120}\approx 14.191666667$

Solution Steps

Multiply $5$ and $10$ to get $50$.

$$\frac{3}{4}+\frac{50+9}{10}+\frac{7\times 3+2}{3}-\frac{1}{8}$$

Add $50$ and $9$ to get $59$.

$$\frac{3}{4}+\frac{59}{10}+\frac{7\times 3+2}{3}-\frac{1}{8}$$

Least common multiple of $4$ and $10$ is $20$. Convert $\frac{3}{4}$ and $\frac{59}{10}$ to fractions with denominator $20$.

$$\frac{15}{20}+\frac{118}{20}+\frac{7\times 3+2}{3}-\frac{1}{8}$$

Since $\frac{15}{20}$ and $\frac{118}{20}$ have the same denominator, add them by adding their numerators.

$$\frac{15+118}{20}+\frac{7\times 3+2}{3}-\frac{1}{8}$$

Add $15$ and $118$ to get $133$.

$$\frac{133}{20}+\frac{7\times 3+2}{3}-\frac{1}{8}$$

Multiply $7$ and $3$ to get $21$.

$$\frac{133}{20}+\frac{21+2}{3}-\frac{1}{8}$$

Add $21$ and $2$ to get $23$.

$$\frac{133}{20}+\frac{23}{3}-\frac{1}{8}$$

Least common multiple of $20$ and $3$ is $60$. Convert $\frac{133}{20}$ and $\frac{23}{3}$ to fractions with denominator $60$.

$$\frac{399}{60}+\frac{460}{60}-\frac{1}{8}$$

Since $\frac{399}{60}$ and $\frac{460}{60}$ have the same denominator, add them by adding their numerators.

$$\frac{399+460}{60}-\frac{1}{8}$$

Add $399$ and $460$ to get $859$.

$$\frac{859}{60}-\frac{1}{8}$$

Least common multiple of $60$ and $8$ is $120$. Convert $\frac{859}{60}$ and $\frac{1}{8}$ to fractions with denominator $120$.

$$\frac{1718}{120}-\frac{15}{120}$$

Since $\frac{1718}{120}$ and $\frac{15}{120}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{1718-15}{120}$$

Subtract $15$ from $1718$ to get $1703$.

$$\frac{1703}{120}$$

Show Solution

Factor

$\frac{13 \cdot 131}{2 ^ {3} \cdot 3 \cdot 5} = 14\frac{23}{120} = 14.191666666666666$