Solve each equation. See Example 2. (4x+3)/(x+1) + 2/x = 1/(x

Question

Solve each equation. See Example 2. (4x+3)/(x+1) + 2/x = 1/(x²+x)

Accepted Answer

Hey everyone welcome back in this problem. We're given the following equation and we're asked to solve for X. We have six X plus five over X plus two plus five over X is equal to 14 over X squared plus two X. Alright so just rewriting what we're getting six X plus five over X plus two plus five over X is equal to 14 over X squared plus two X. Okay. Now what we're given an equation like this and we're asked to solve for X. What we wanna do is we want to try to combine this into one term. We want to write all three of these terms as one term, one rational expression in order to solve so to do that we need a common denominator. So we have a denominator of X plus two on our first term. A denominator of X in our second term. And a denominator of X squared plus two X. On our right hand side. Now X squared plus two X. K. Notice that this is equal to X times X plus two. If we were to factor. Okay so that's gonna be our common denominator X times X plus two. Because there are other terms have one of those in them. Alright, just like finding a common denominator when you have just numbers in your fractions. So this six X plus five. Okay. It has X plus two in the denominator. So what's it missing? Well it's missing that X factor. So we're gonna say six X plus five. Okay now we want to multiply the denominator by X. So that we have that common denominator If we're multiplying the denominator by X. We need to multiply the numerator by X as well because then we're multiplying by X over X. Which is just equal to one. Multiplying by one doesn't change our equation. Alright. Next term five. And again this time we have X in the denominator so we want to multiply by X plus two and the factor that it's missing, we do that in the denominator we have to do it in the numerator. So we get five X plus two over X. Times X. Plus two. And on the right hand side we already have that common denominator of X times X Plus two. So we're going to leave it alone. Alright now we have a common denominator okay? And we can combine all of these things. So let's move everything to the left hand side. Okay? And expand. So we're gonna have six X times X. That's gonna be six X squared. And we have five times X. Which is gonna give us plus five X. And then we have plus five x plus five times two which is 10. And then we're gonna subtract 14 from moving our right hand side term over to the left hand side. And all of this is a one term. Okay divided by X times X. Plus two. That common denominator and it's going to be equal to zero. Alright so let's simplify this numerator we get six X squared plus five X plus five X is going to give us plus X. Let me have 10 minus 14 is going to give us negative four divided by X times X plus two. Now you'll notice that every term in the numerator here Is an even number so it's divisible by two. Okay, so let's go ahead and divide the entire equation by two. If we do that on the left hand side we have to do it on the right hand side as well. So six X squared divided by two. That's gonna be three X squared 10 x divided by 2, 5 x and -4 divided by two gives us -2. Our denominator stays the same X times X plus two. And on the right hand side we have zero divided by two. Which is going to give us zero. Alright, we're getting closer. Okay we have this quadratic in the numerator we have our quadratic. That is factored in the denominator. Okay, in order to solve for this let's go ahead and factor that numerator. And if we factor the numerator you can factor this any way you like, whichever way you like to factor the best, we're gonna have three X minus one Times X-plus two over X times X plus two. All equal to zero. Now you'll notice that we have X plus two divided by X plus two. Okay, so those divide out to give us one. However we have to write a restriction that X cannot be equal to negative two, if X is equal to negative two, we're dividing by zero and that's going to cause problems. Okay, so before we divide out that term and completely get rid of it. We need to write that restriction down so that when we get our possible answers at the end we can double check that they weren't one of our restrictions. Alrighty. So now we have three X - over X is equal to zero. Okay, when is this going to be equal to zero? Okay, well this is going to be equal to zero if the numerator is equal to zero. If we have zero divided by anything, it's going to be equal to zero. So all we need is three X minus one to be equal to zero. However, we have another restriction. Okay, We can't have X equal to zero either because we'll run into the same problem as with negative two, we'll divide by zero, which we know we can't do. Okay, so now we have two restrictions. We have X cannot be negative two and it cannot be zero either. Okay, so we have two restrictions continuing our solution. Okay, this is going to give us X is equal to one third. Okay, If we look at X is equal to one third we see that it's not negative two and it's not zero. So it's not one of our restrictions. So that is in fact a solution. So our solution is X is equal to one third, and if we look at our answer choices, that is going to be answer choice B. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each equation. See Example 2. (4x+3)/(x+1) + 2/x = 1/(x²+x)

Key Concepts

Rational Expressions

Least Common Denominator (LCD)

Solving Rational Equations

Watch next

Solve each equation. See Example 2. (4x+3)/(x+1) + 2/x = 1/(x2+x)

Key Concepts

Rational Expressions

Least Common Denominator (LCD)

Solving Rational Equations

Watch next

Solve each equation. See Example 2. (4x+3)/(x+1) + 2/x = 1/(x²+x)