In Exercises 67–70, find all values of x such that y = 0.

y = 1/...

Question

In Exercises 67–70, find all values of x such that y = 0.

y = 1/(5x + 5) - 3/(x + 1) + 7/5

Accepted Answer

So this problem tells us the five values of X. Such status satisfies Y equals zero. So we look at an equation, we're gonna set everything equal to zero. So say we have zero Equals X -9 Over four x months 32 -5 over X -8 minus 3/5. So you want to try and find at least common denominator so we can get rid of all these fractions. The first thing I've noticed actually is we have a four x plus 32. We can actually factor out a four out of that bottom part. So five x -9 Over four times X -8. I have the rest of the stuff. five X -8 -3/5 going to find the LCD. So our L. C. D. An easy way to find this is to multiply all of our denominators together. So we actually have an x minus eight accounted for already. We have an x minus eight here and we have a five here. I'm gonna say my L. C. D. Is five times x minus eight. Now that we have that we're gonna multiply everything by five times X -8. So zero equals x minus 9/4 times X minus eight, multiplied By five times X -8 -5 over X -8 Times five X -8. And finally native 3/5 Times five times x -8. So we can actually look at things that can cancel now notice here we will have an x months eight on the top of refraction. And x men say on the bottom that will cancel out. So those will be gone Now here we have an X on the bottom on the top those will cancel out again. Finally over here we have 3/5. So fives on the bottom and the fives on the top those will cancel out. We can now rewrite this zero equals X -9/4. multiplied by five -5 times five which is 25 And -3 times X -8. Not to simplify even further. We have zero equals R5 to distribute to the top of our fraction. Where we get five X -45 over four -25. And now our -3. Well this should be to this equation -3 x plus 24. Now I can simplify it further we have zero equals five X -45/4. And we can combine our 25 and 24 25. That turns into -1 and the -3 x. I go even more. I'm gonna move my 1 -1 and -3 X to the other side. So I will add one to both sides and add three X. This gets us 1-plus 3 x equals five X minus 45/4, adding plus one and plus three X on the other side. Now if you notice we have a four on the bottom of this fraction we're gonna multiply both sides by four. That will get rid of it on the side. So now go back over here You have four plus 12 x. Equals five X -45. So I can move all of my constants to one side and my exes to the other now. So I'm gonna go ahead and move my five X. 1% different color B -5 X -5 X. And the other side we're gonna move our four minus four. If I go over here write it out. We now have you have 12 minus five X. Which is seven X. On one side. The other side we have negative 45 minus four which is negative 49. Finally we divide both sides by seven. Do that. Meeting up with x equals -7. Is the answer to our solution when y go zero is negative seven. Making this answer a. Okay I hope to help you solve the problem. Thank you for watching. Goodbye.

In Exercises 67–70, find all values of x such that y = 0. y = 1/(5x + 5) - 3/(x + 1) + 7/5

Key Concepts

Rational Functions

Finding Roots of an Equation

Domain Restrictions

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