3(f+2)5f=0

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

5f,1

Since 5f,1 contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part 5,1 then find LCM for the variable part f1.

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.

Since 5 has no factors besides 1 and 5.

5 is a prime number

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

Not prime

The LCM of 5,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.

5

The factor for f1 is f itself.

f1=f

f occurs 1 time.

The LCM of f1 is the result of multiplying all prime factors the greatest number of times they occur in either term.

f

The LCM for 5f,1 is the numeric part 5 multiplied by the variable part.

5f

5f

Multiply each term in 3(f+2)5f=0 by 5f in order to remove all the denominators from the equation.

3(f+2)5f⋅(5f)=0⋅(5f)

Simplify 3(f+2)5f⋅(5f).

Rewrite using the commutative property of multiplication.

53(f+2)5ff=0⋅(5f)

Cancel the common factor of 5.

Factor 5 out of 5f.

53(f+2)5(f)f=0⋅(5f)

Cancel the common factor.

53(f+2)5ff=0⋅(5f)

Rewrite the expression.

3(f+2)ff=0⋅(5f)

3(f+2)ff=0⋅(5f)

Cancel the common factor of f.

Cancel the common factor.

3(f+2)ff=0⋅(5f)

Rewrite the expression.

3(f+2)=0⋅(5f)

3(f+2)=0⋅(5f)

Apply the distributive property.

3f+3⋅2=0⋅(5f)

Multiply 3 by 2.

3f+6=0⋅(5f)

3f+6=0⋅(5f)

Multiply 0(5f).

Multiply 5 by 0.

3f+6=0f

Multiply 0 by f.

3f+6=0

3f+6=0

3f+6=0

Subtract 6 from both sides of the equation.

3f=-6

Divide each term by 3 and simplify.

Divide each term in 3f=-6 by 3.

3f3=-63

Cancel the common factor of 3.

Cancel the common factor.

3f3=-63

Divide f by 1.

f=-63

f=-63

Divide -6 by 3.

f=-2

f=-2

f=-2

Simplify (3(f+2))/(5f)=0