Solve solve (1 - 2a)/a + (1 - 26)/b + (1 - 2c)/c = 0; (1/9 + 1/b + 1/c) ^ 2 = | 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

$$\frac { 1 - 20 } { 2 } + \frac { 126 } { 6 } + \frac { 126 } { 6 } + \frac { 126 } { 6 } - \frac { 1 } { 6 } + \frac { 1 } { 6 } + \frac { 1 } { 6 } = ?$$

Evaluate

$\frac{161}{3}\approx 53.666666667$

Solution Steps

Subtract $20$ from $1$ to get $-19$.

$$\frac{-19}{2}+\frac{126}{6}+\frac{126}{6}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Fraction $\frac{-19}{2}$ can be rewritten as $-\frac{19}{2}$ by extracting the negative sign.

$$-\frac{19}{2}+\frac{126}{6}+\frac{126}{6}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Divide $126$ by $6$ to get $21$.

$$-\frac{19}{2}+21+\frac{126}{6}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Convert $21$ to fraction $\frac{42}{2}$.

$$-\frac{19}{2}+\frac{42}{2}+\frac{126}{6}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Since $-\frac{19}{2}$ and $\frac{42}{2}$ have the same denominator, add them by adding their numerators.

$$\frac{-19+42}{2}+\frac{126}{6}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Add $-19$ and $42$ to get $23$.

$$\frac{23}{2}+\frac{126}{6}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Divide $126$ by $6$ to get $21$.

$$\frac{23}{2}+21+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Convert $21$ to fraction $\frac{42}{2}$.

$$\frac{23}{2}+\frac{42}{2}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Since $\frac{23}{2}$ and $\frac{42}{2}$ have the same denominator, add them by adding their numerators.

$$\frac{23+42}{2}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Add $23$ and $42$ to get $65$.

$$\frac{65}{2}+\frac{126}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Divide $126$ by $6$ to get $21$.

$$\frac{65}{2}+21-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Convert $21$ to fraction $\frac{42}{2}$.

$$\frac{65}{2}+\frac{42}{2}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Since $\frac{65}{2}$ and $\frac{42}{2}$ have the same denominator, add them by adding their numerators.

$$\frac{65+42}{2}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Add $65$ and $42$ to get $107$.

$$\frac{107}{2}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Least common multiple of $2$ and $6$ is $6$. Convert $\frac{107}{2}$ and $\frac{1}{6}$ to fractions with denominator $6$.

$$\frac{321}{6}-\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$$

Since $\frac{321}{6}$ and $\frac{1}{6}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{321-1}{6}+\frac{1}{6}+\frac{1}{6}$$

Subtract $1$ from $321$ to get $320$.

$$\frac{320}{6}+\frac{1}{6}+\frac{1}{6}$$

Since $\frac{320}{6}$ and $\frac{1}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{320+1}{6}+\frac{1}{6}$$

Add $320$ and $1$ to get $321$.

$$\frac{321}{6}+\frac{1}{6}$$

Since $\frac{321}{6}$ and $\frac{1}{6}$ have the same denominator, add them by adding their numerators.

$$\frac{321+1}{6}$$

Add $321$ and $1$ to get $322$.

$$\frac{322}{6}$$

Reduce the fraction $\frac{322}{6}$ to lowest terms by extracting and canceling out $2$.

$$\frac{161}{3}$$

Factor

$\frac{7 \cdot 23}{3} = 53\frac{2}{3} = 53.666666666666664$