Solve each equation in Exercises 83–108 by the method of your ...

Question

Solve each equation in Exercises 83–108 by the method of your choice. 1/x + 1/(x + 3) = 1/4

Accepted Answer

Hello. Today we're going to be solving this equation using any of our preferred methods. So the equation given to us is one over the quantity two X plus one over x minus four is equal to one half. Now we want to go ahead and eliminate the denominators from every term. So we're going to need to identify a lowest common denominator between the three. Well the denominator is given to us is to x minus four and two X. So the lowest common denominator is going to be a combination of a product of each of the denominators. While in this case here the lowest common denominator is going to be two X times X minus four. And in order to get rid of all these denominators we're gonna multiply every term by the lowest common denominator two X times X minus four. Doing so is going to give us X - Plus one times 2 x Is equal to one times x times x -4. And we're gonna go ahead and simplify things here. So X minus four can just be written as is and one times two X is going to give us two X one times X is just going to give us X. So on the right hand side we have X times the quantity X minus four. Now on the left hand side I'm gonna go ahead and combine like terms so I can combine the two X and the X. To give me three X minus four. And on the right hand side I can distribute X two. The quantity X minus four and that's going to give me x squared minus four X. Now I want to go ahead and set this equation equal to zero because we want to solve for X. And in order to do that I can go ahead and subtract three X to both sides of the equation and go ahead and add four to both sides of the equation. Doing so is going to give me x squared minus seven. X plus four is equal to zero. And we're gonna go ahead and solve this using the quadratic equation. Now let's just quickly recall what the quadratic equation states. It states that X is equal to negative B plus or minus the square root of B squared minus four times a times C. All over two. A and abc are going to be the coefficients of the terms from our tryin Amiel. So A is going to be the coefficient in front of our x square term. So A is going to equal to one. B is the coefficient in front of our X term. So B is equal to negative seven and C is going to be the constant. So C is equal to four. Putting that into the quadratic equation is going to give us X is equal to negative B which is negative negative seven, so positive seven plus or minus the square root of negative seven squared minus four times A which is one times C which is four. All of that is going to be over two times a which is two times one. And now we just need to go ahead and simplify so negative seven squared is going to give us positive 49 negative four times one times four is going to give us negative 16, so X is going to equal to seven plus or minus the square root of 49 minus 16 All of that over two times one, which is going to be two now 49 minus 16 is going to give us 33 so X is going to equal to seven plus or minus the square root of 33/2. Now there's really not much that we can do to go ahead and simplify this any further. But the only thing we can go ahead and do is separate the statements because this is a plus or minus statement so we're going to have two solutions for X. X is going to equal to seven plus the square root of 33/2 and X is going to equal to seven minus the square root of 33/2. And these are gonna be the two solutions to the equation. And since these are the solutions to the equation, the answer to this problem is going to be a So I hope this video helps you in understanding how to solve this type of equation and I'll go ahead and see you all in the next video

Solve each equation in Exercises 83–108 by the method of your choice. 1/x + 1/(x + 3) = 1/4

Key Concepts

Solving Rational Equations

Finding the Least Common Denominator (LCD)

Checking for Extraneous Solutions

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