5k=3t-3kt

Rewrite the equation as 3t-3kt=5k.

3t-3kt=5k

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

1,t,1

Since 1,t,1 contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part 1,1,1 then find LCM for the variable part t1.

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.

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

Not prime

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

1

The factor for t1 is t itself.

t1=t

t occurs 1 time.

The LCM of t1 is the result of multiplying all prime factors the greatest number of times they occur in either term.

t

t

Multiply each term in 3t-3kt=5k by t in order to remove all the denominators from the equation.

3t⋅t-3kt⋅t=5k⋅t

Simplify each term.

Multiply t by t by adding the exponents.

Move t.

3(t⋅t)-3kt⋅t=5k⋅t

Multiply t by t.

3t2-3kt⋅t=5k⋅t

3t2-3kt⋅t=5k⋅t

Cancel the common factor of t.

Move the leading negative in -3kt into the numerator.

3t2+-3kt⋅t=5k⋅t

Cancel the common factor.

3t2+-3kt⋅t=5k⋅t

Rewrite the expression.

3t2-3k=5k⋅t

3t2-3k=5k⋅t

3t2-3k=5kt

3t2-3k=5kt

Subtract 5kt from both sides of the equation.

3t2-3k-5kt=0

Use the quadratic formula to find the solutions.

-b±b2-4(ac)2a

Substitute the values a=3, b=-5k, and c=-3k into the quadratic formula and solve for t.

5k±(-5k)2-4⋅(3⋅(-3k))2⋅3

Simplify.

Simplify the numerator.

Apply the product rule to -5k.

t=5k±(-5)2k2-4⋅(3⋅(-3k))2⋅3

Raise -5 to the power of 2.

t=5k±25k2-4⋅(3⋅(-3k))2⋅3

Multiply -3 by 3.

t=5k±25k2-4⋅(-9k)2⋅3

Multiply -9 by -4.

t=5k±25k2+36k2⋅3

Factor k out of 25k2+36k.

Factor k out of 25k2.

t=5k±k(25k)+36k2⋅3

Factor k out of 36k.

t=5k±k(25k)+k⋅362⋅3

Factor k out of k(25k)+k⋅36.

t=5k±k(25k+36)2⋅3

t=5k±k(25k+36)2⋅3

t=5k±k(25k+36)2⋅3

Multiply 2 by 3.

t=5k±k(25k+36)6

t=5k±k(25k+36)6

The final answer is the combination of both solutions.

t=5k+k(25k+36)6

t=5k-k(25k+36)6

t=5k+k(25k+36)6

t=5k-k(25k+36)6

Solve for t 5k=3t-(3k)/t