Solve for r1 1/35=1/r_1+1/80

$\frac{1}{35} = \frac{1}{r_{1}} + \frac{1}{80}$

Step 1

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$35, r_{1}, 80$

Since $35, r_{1}, 80$ contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $35, 1, 80$ then find LCM for the variable part ${r_{1}}^{1}$ .

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

$35$ has factors of $5$ and $7$ .

$5 \cdot 7$

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The prime factors for $80$ are $2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$ .

$80$ has factors of $2$ and $40$ .

$2 \cdot 40$

$40$ has factors of $2$ and $20$ .

$2 \cdot 2 \cdot 20$

$20$ has factors of $2$ and $10$ .

$2 \cdot 2 \cdot 2 \cdot 10$

$10$ has factors of $2$ and $5$ .

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 5$

The LCM of $35, 1, 80$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 5 \cdot 7$

Multiply $2 \cdot 2 \cdot 2 \cdot 2 \cdot 5 \cdot 7$ .

Multiply $2$ by $2$ .

$4 \cdot 2 \cdot 2 \cdot 5 \cdot 7$

Multiply $4$ by $2$ .

$8 \cdot 2 \cdot 5 \cdot 7$

Multiply $8$ by $2$ .

$16 \cdot 5 \cdot 7$

Multiply $16$ by $5$ .

$80 \cdot 7$

Multiply $80$ by $7$ .

$560$

The factor for ${r_{1}}^{1}$ is $r_{1}$ itself.

${r_{1}}^{1} = r_{1}$

$r_{1}$ occurs $1$ time.

The LCM of ${r_{1}}^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$r_{1}$

The LCM for $35, r_{1}, 80$ is the numeric part $560$ multiplied by the variable part.

$560 r_{1}$

Step 2

Multiply each term in $\frac{1}{35} = \frac{1}{r_{1}} + \frac{1}{80}$ by $560 r_{1}$ to eliminate the fractions.

Multiply each term in $\frac{1}{35} = \frac{1}{r_{1}} + \frac{1}{80}$ by $560 r_{1}$ .

$\frac{1}{35} (560 r_{1}) = \frac{1}{r_{1}} (560 r_{1}) + \frac{1}{80} (560 r_{1})$

Simplify the left side.

Cancel the common factor of $35$ .

Factor $35$ out of $560 r_{1}$ .

$\frac{1}{35} (35 (16 r_{1})) = \frac{1}{r_{1}} (560 r_{1}) + \frac{1}{80} (560 r_{1})$

Cancel the common factor.

$\frac{1}{35} (35 (16 r_{1})) = \frac{1}{r_{1}} (560 r_{1}) + \frac{1}{80} (560 r_{1})$

Rewrite the expression.

$16 r_{1} = \frac{1}{r_{1}} (560 r_{1}) + \frac{1}{80} (560 r_{1})$

Simplify the right side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$16 r_{1} = 560 \frac{1}{r_{1}} r_{1} + \frac{1}{80} (560 r_{1})$

Combine $560$ and $\frac{1}{r_{1}}$ .

$16 r_{1} = \frac{560}{r_{1}} r_{1} + \frac{1}{80} (560 r_{1})$

Cancel the common factor of $r_{1}$ .

Cancel the common factor.

$16 r_{1} = \frac{560}{r_{1}} r_{1} + \frac{1}{80} (560 r_{1})$

Rewrite the expression.

$16 r_{1} = 560 + \frac{1}{80} (560 r_{1})$

Cancel the common factor of $80$ .

Factor $80$ out of $560 r_{1}$ .

$16 r_{1} = 560 + \frac{1}{80} (80 (7 r_{1}))$

Cancel the common factor.

$16 r_{1} = 560 + \frac{1}{80} (80 (7 r_{1}))$

Rewrite the expression.

$16 r_{1} = 560 + 7 r_{1}$

Step 3

Solve the equation.

Move all terms containing $r_{1}$ to the left side of the equation.

Subtract $7 r_{1}$ from both sides of the equation.

$16 r_{1} - 7 r_{1} = 560$

Subtract $7 r_{1}$ from $16 r_{1}$ .

$9 r_{1} = 560$

Divide each term in $9 r_{1} = 560$ by $9$ and simplify.

Divide each term in $9 r_{1} = 560$ by $9$ .

$\frac{9 r_{1}}{9} = \frac{560}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 r_{1}}{9} = \frac{560}{9}$

Divide $r_{1}$ by $1$ .

$r_{1} = \frac{560}{9}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$r_{1} = \frac{560}{9}$

Decimal Form:

$r_{1} = 62.\overline{2}$

Mixed Number Form:

$r_{1} = 62 \frac{2}{9}$

Solve for r1 1/35=1/r_1+1/80

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