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

Question

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation.

x/3 = x/2 - 2

Accepted Answer

Hi there. This problem says to solve a given linear equation M over five equals M over three minus four. So let me go ahead and rewrite this first and number five equals M over three minus four. I wanna solve for M. But you notice here we have fractions with different denominators. So what we want to do is find a way to divide our denominators to do that. You want to find the least common denominator. So you're gonna find the L. C. D. Now to find the LCD. We want to see which multiples of each of our denominators are the same. So let's start with five for instance five. We can say we have 5 15 2025 and so on. Now let's check. Three. Well three we have 369, 12 15, 18, 21 so on. We're gonna see which numbers are the same in both of our multiples. We look here We have it 15 here and 15 there. There's not any other number that's lower than 15. So our l. c. d. It's 15. So now we're gonna multiply everything we have in our equation 5, 15. Let's multiply the left side first. So by half M. Over five equals and over three minus four. Let's multiply everything by 15, teen and 15. So when I do that I get 15 times in over five equals 15 times M Over three and 15 times negative four which is negative 60. Now we can simplify even further. I would get 15/5 which is three M. 15/3. 5 in. And we just dropped down our negative 60. No to solve our M. One to combine like terms. So we want to get all of RMS together and all of our non M. S on the other side together. Well the only M. S. We have our three M. And five M. Are only non M. Is negative 60. So let's go ahead and move this five em over. We are currently positive five M. We're gonna subtract five since I did it on one side, I have to do it to the other. So we get three M minus five M on the left side. five months 5 m goes away in the bottom. We just go to three M minus five M which is two M. And that's equal to negative 60 with drops down. Now we can go ahead and divide by our negative too because you just want him by itself. We do that we get native to over native to on the left side which just goes to M And we have negative 60 developed at negative two Which is equal to 30. So the solution to our problem is an equals 30 which is answer be okay. I hope they'll solve the problem. Thank you for watching. Goodbye

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

Key Concepts

Linear Equations

Solving for a Variable

Fractions and Common Denominators

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