How to solve 3/8 + 7/5 + 1/7

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 3/8 + 7/5 + 1/7 | Equatix

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$$\frac{3}{8}+\frac{7}{5}+\frac{1}{7}$$

$\frac{537}{280}\approx 1.917857143$

Least common multiple of $8$ and $5$ is $40$. Convert $\frac{3}{8}$ and $\frac{7}{5}$ to fractions with denominator $40$.

$$\frac{15}{40}+\frac{56}{40}+\frac{1}{7}$$

Since $\frac{15}{40}$ and $\frac{56}{40}$ have the same denominator, add them by adding their numerators.

$$\frac{15+56}{40}+\frac{1}{7}$$

Add $15$ and $56$ to get $71$.

$$\frac{71}{40}+\frac{1}{7}$$

Least common multiple of $40$ and $7$ is $280$. Convert $\frac{71}{40}$ and $\frac{1}{7}$ to fractions with denominator $280$.

$$\frac{497}{280}+\frac{40}{280}$$

Since $\frac{497}{280}$ and $\frac{40}{280}$ have the same denominator, add them by adding their numerators.

$$\frac{497+40}{280}$$

Add $497$ and $40$ to get $537$.

$$\frac{537}{280}$$

$\frac{3 \cdot 179}{2 ^ {3} \cdot 5 \cdot 7} = 1\frac{257}{280} = 1.917857142857143$