22. 1/2+3/t=5/8

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You can multiply both sides by the LCD to get rid of all denominators. What is the LCD of 2, t, and 8?
It's 2t, right? That's what I used and the answer came out weird. Is it 8t?

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No. You need the smallest quantity that is divisible by 2, t, and 8. 2t is divisible by 2 and t, but it is not divisible by 8. What is the smallest quantity that is divisible by 2, t, and 8?
Is it 8t then?
Yes. Now we multiply both sides by 8t. This way we will get rid of all denominators. \(8t \left( \dfrac{1}{2} + \dfrac{3}{t} \right)= 8t \left( \dfrac{5}{8} \right)\) Ok?
Now we distribute the 8t on the left, and we multiply the 8t on the right. \(\dfrac{8t}{1} \left( \dfrac{1}{2} + \dfrac{3}{t} \right)= \dfrac{8t}{1} \left( \dfrac{5}{8} \right)\) \( \dfrac{8t}{2} + \dfrac{8t \times 3}{t} = \dfrac{8t \times 5}{8} \)
Distribute and multiply? How I learned was to cancel out like terms. So in this question, I would cancel out the t in 8t and 3t, and cancel out the 8 in 8t and the 8 in 5/8. Does that end up with the same thing? Or am I doing something wrong?
Your idea of canceling is ok, but you need to be careful how you do it.
Now we cancel out all denominators with factors in the numerators. \( \dfrac{\cancel{8}4t}{\cancel{2}1} + \dfrac{8\cancel{t} \times 3}{\cancel{t} 1} = \dfrac{\cancel{8}t \times 5}{\cancel{8}1} \) \(4t + 24 = 5t\)
Alright, so now you just do basic algebra and solve t, right?
You are correct that the 8t on the left multiplies the fraction 3/t. the t's cancel out leaving just 3 * 8 = 24. You can;t forget to also multiply 8t by the fraction 1/2. That is why it is a good idea to distribute first, then cancel.
Yes.
Okay, thank you!
You're welcome.

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