How to solve 3 1/5 + (- 7/8)

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 3 1/5 + (- 7/8) | Equatix

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

$\frac{93}{40}=2.325$

Multiply $3$ and $5$ to get $15$.

$$\frac{15+1}{5}+\frac{-7}{8}$$

Add $15$ and $1$ to get $16$.

$$\frac{16}{5}+\frac{-7}{8}$$

Fraction $\frac{-7}{8}$ can be rewritten as $-\frac{7}{8}$ by extracting the negative sign.

$$\frac{16}{5}-\frac{7}{8}$$

Least common multiple of $5$ and $8$ is $40$. Convert $\frac{16}{5}$ and $\frac{7}{8}$ to fractions with denominator $40$.

$$\frac{128}{40}-\frac{35}{40}$$

Since $\frac{128}{40}$ and $\frac{35}{40}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{128-35}{40}$$

Subtract $35$ from $128$ to get $93$.

$$\frac{93}{40}$$

$\frac{3 \cdot 31}{2 ^ {3} \cdot 5} = 2\frac{13}{40} = 2.325$