Exercises 41–60 contain rational equations with variables in deno...

Question

Exercises 41–60 contain rational equations with variables in denominators. For each equation, a. write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.

7/2x - 5/3x = 22/3

Accepted Answer

this problem tells us for the following rational equation. If you're able is contained in the denominator and we notice that we have an excess in our denominators solve the equation, indicate the value that makes the denominator zero. Let's go ahead and write this out real quick. We have six over X equals 8/ X plus four. Now I'm gonna try and solve for X. We know sir we have fractions on the bottom of six over X. And 8/3 X. I want to get rid of those fractions. So I'm gonna multiply both sides by the L. C. D. L. C. D. In this case is a number going to multiply to get rid of our fractions. We look here we have an X. And a three X. I can say the X. Is going to fit into the three X. So I'm gonna change my L. C. D to B. Three X. That will get rid of the three as well. In the bottom of that affection. Let's go ahead and multiply everything by our L. C. D. Let's buy that by L. C. D. This fire the L. C. D. And finally there's something left by L. C. D. Now we can cancel some stuff out. You look here we have a three X. On the top X. On the bottom are X's will cancel out on the west side we have a three X. In the top three X. In the bottom those ones will cancel out and finally we have four times three X. Which that would just be multiplied together. So if we write this we have three left here, times six that is 18 equals on the side are three X. And three X canceled out. We just bring down the eights and over here we have it four times three X. So we'll just multiply that ticket 12 X. Now we can go ahead and try and solve for X. I'm gonna subtract both sides by eight to get rid of that eight on the right side we get 18 minus eight which is equals 12 X. Now we divide both sides by 12. Get that We get x equals 10/12. This can be reduced. We can divide both numbers by two meaning we get five or six. So X equals five or six. That is a solution to our equation. Now I want to figure out what value makes. It does not matter Zero. Well go ahead and look over here let's set X equal to zero and three X equal to zero. We solve for X. X equals zero is already solved. And the other side of the bed by three We get x equals zero. So we cannot have our denominator equaling zero means X cannot be equal to zero which means I say X cannot be equal to zero. The answer to our problem is answer A X equals 56 X cannot equal zero. Okay I hope that you solve the problem. Thank you for watching. Goodbye

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