Exercises 27–40 contain linear equations with constants in denomi...

Question

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation.3x/5 - (x - 3)/2 = (x + 2)/3

Accepted Answer

So this problem tells solve the given linear equation. So let's go ahead and write our equation. It is K plus 7/ equals 1/6 plus k minus 2/15. So to solve this we want to try and get rid of our fractions to get rid of our fractions. We have to find the L. C. D. Of all of our fractions. So let's go ahead and take a look here. We have a two over here. A six over here and a 15 over here. We're gonna find what these fractions have in common. Let's find the multiples of each for fractions. Well let's say we look at the two we have 2468 10 and so on. Until we get to a certain even number that has both our other factions now. So I'm gonna stop at 30. Now let's look at our six. Well for six we have 6 12 18, 24 and 30. Finally we have our 15 Chart 15 30 and so on Notice here the least common multiple is 30 for all of these. Which means that we're gonna set our l. c. d. equal to 30 now that we have that we can multiply everything by our L. C. D. So if you go back I'm gonna go ahead. And if you write this K plus 7/2 equals 1/ plus k minus 2/15. Multiply everything by L. c. d. multiply this by 30. This here by 30 And this here by we do that Now we're gonna have a 30 over to cancel those out. This goes to 15. This part we have a 30/6. That goes to five and finally here 30/ that just goes to two. Now we can rewrite this once again 15 is left times K-plus seven. That's equal to five times one is just five Plus two times K -2. Now we can go ahead and distribute r. 15 and R. two. Sorry 15 will get distributed to the K. and the 7th. Give us 15 times K. S. 15 K. 15 times seven is 105. Now the other side we have a five Plus two times K. and two times negative two two times K. Is two K. Two times negative two is negative four. Now I'm gonna go ahead and simplify everything we have here. Yeah 15 K. +105. That's equal to five minus four which is just one plus two K. Now I can simplify even further notice here we can move over R. Two K. So I'll move over the two K. To the left And I'm the move over by 105. So I will get 15 -2. Just 13 K. That's equal to 1 -105 which is negative 104. Finally we have 13 divided by 13 K. Equals one of four. So actually divide both sides by 13. Well use a calculator. You have 13 K. 30K over 13. It's just K. And we have native one and 4/13, which should just be negative. Eight being the solution to this problem is K equals negative eight. What's his answer? B. Alright. I hope that you solve the problem. Thank you for watching. Goodbye.

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Exercises 27–40 contain linear equations with constants in denominators. Solve each equation.3x/5 - (x - 3)/2 = (x + 2)/3

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