How to solve (6/7 + 1/7) 3

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 (6/7 + 1/7) 3 | Equatix

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Question

$$( \frac { 6 } { 7 } + \frac { 1 } { 7 } ) \cdot ( \frac { 4 } { 11 } - \frac { 1 } { 3 } )$$

Evaluate

$\frac{1}{33}\approx 0.03030303$

Solution Steps

Since $\frac{6}{7}$ and $\frac{1}{7}$ have the same denominator, add them by adding their numerators.

$$\frac{6+1}{7}\left(\frac{4}{11}-\frac{1}{3}\right)$$

Add $6$ and $1$ to get $7$.

$$\frac{7}{7}\left(\frac{4}{11}-\frac{1}{3}\right)$$

Divide $7$ by $7$ to get $1$.

$$1\left(\frac{4}{11}-\frac{1}{3}\right)$$

Least common multiple of $11$ and $3$ is $33$. Convert $\frac{4}{11}$ and $\frac{1}{3}$ to fractions with denominator $33$.

$$1\left(\frac{12}{33}-\frac{11}{33}\right)$$

Since $\frac{12}{33}$ and $\frac{11}{33}$ have the same denominator, subtract them by subtracting their numerators.

$$1\times \frac{12-11}{33}$$

Subtract $11$ from $12$ to get $1$.

$$1\times \frac{1}{33}$$

Multiply $1$ and $\frac{1}{33}$ to get $\frac{1}{33}$.

$$\frac{1}{33}$$

Show Solution

Factor

$\frac{1}{3 \cdot 11} = 0.030303030303030304$