In Exercises 91–100, find all values of x satisfying the given co...

Question

In Exercises 91–100, find all values of x satisfying the given conditions.

y1 = 6(2x/(x - 3))^2, y2 = 5(2x/(x - 3)), and y1 exceeds y2 by 6

Accepted Answer

Hello. Today we're going to be solving the following statement. So we are given two equations why one is equal to eight times three X over X plus six squared and Y two is equal to two times three. X over X plus six. And we are told that Y one exceeds Y two by 15. Now what does this statement mean? Why one exceeds Y two by 15? Well what it means is if we were to take the difference of Y one and Y two We are going to be left with 15. So this is gonna be the equation we're going to use to sulfur X. And we're just gonna plug in or given Y one and Y two. So if we plug in our Y one and Y two we are left with eight times the quantity three X over X plus six that quantity squared minus two times three X over X plus six Is equal to 15. Now this is pretty complicated to work with but notice how we have two instances where we have three X over X plus six. We can go ahead and use a substitution. So if I let a variable U equal to three X over X plus six what I am left with is eight times U squared - times you Is equal to 15. And now what I want to go ahead and do is solve for you. Well in order to do that I want to go ahead and set this equation equal to zero and in order to do that I'm going to subtract 15 from both sides of this equation. Doing so is going to give me eight. U squared minus two. U minus 15 is equal to zero. And this is a fact arable trin Amiel because of factors to to u minus three times for you. Plus five is equal to zero. Now using my zero property I can set both of these quantities equal to zero so I have to you minus three is equal to zero and four. U plus five is equal to zero. On the left hand side. I'm going to add three to both sides of this equation and I am left with two. U. Is equal to three. And in order to solve for you I'm going to divide both sides of this equation by two and U. Is equal to 3/2. And that's gonna be my first value of you. Now on the right hand side I'm gonna subtract five from both sides of this equation. So I have four. U. Is equal to negative five. And if I divide both sides of this equation by four I get U. Is equal to negative 5/4. And that's gonna be my second value for you. Now keep in mind we're solving for X. We're not solving for you. So what we're gonna do with both of these quantities of you is go ahead and substitute back our original you quantity which is three X over X plus six. So doing so is going to give me three X over X plus six is equal to 3/2. And we get three X over X plus six is equal to negative 5/4. And now we need to go ahead and solve X for both of these equations. So on the left hand side, if I were to go ahead and cross multiply I get three X times two Is equal to three times x plus six. Well two times three X is going to give me six X. And on the right hand side I'm going to need to distribute this three to the quantity X plus six and that's going to give me three X plus 18. If I subtract three X from both sides of this equation I get three X is equal to 18 and dividing both sides by three is going to give me X is equal to six. And that's gonna be my first solution for X. Now I need to go ahead and do the same thing with the right hand side. If I were to cross multiply I get four times three X is equal to negative five times the quantity X plus six. Four times three X is going to give me 12 X. And I'm gonna have to distribute this negative 52 X plus six. And that gives me negative five X minus 30. If I add five x to both sides of this equation I get x Is equal to negative 30. And dividing both sides of this equation by 17 is going to give me X is equal to negative 30/17, which is my other solution for X. So my solutions for X is six and negative 30/17. And that means that the solution to this problem is going to be B. So I hope this video helps you in understanding how to solve this type of problem, and I'll go ahead and see you all in the next video.

In Exercises 91–100, find all values of x satisfying the given conditions. y1 = 6(2x/(x - 3))^2, y2 = 5(2x/(x - 3)), and y1 exceeds y2 by 6

Key Concepts

Rational Functions

Exceeding Condition

Solving Quadratic Equations

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