Solve the equation. 2 x + 4 x − 1 = 5 \frac{2x+4}{x-1}=5 x − 1 2 x + 4 = 5

Hey, everyone. Let's take a look at this practice problem. So here I have 2x+4 over x minus 1 is equal to 5, and I wanna solve this rational equation. So my very first step is going to be to determine my restriction by setting my denominator equal to 0. So the denominator that I have with a variable is x minus 1. So if I take this x minus 1 and set it equal to 0 and add 1 to both sides, I'm left with x equals 1. This tells me that if I plug 1 back in for my denominator, then I'm gonna get a 0, which is what I don't want. So I know that x cannot be equal to 1. Now if you just wanna eyeball this, denominator and see that, oh, if I plugged 1 in here, that would make my denominator 0, that's totally fine too, but you can always set your denominator equal to 0 to find this as well. So we are done with step 1. Step 2 is gonna be to multiply by our least common denominator in order to eliminate fractions. So I'm gonna rewrite my equation over here. I have 2x+xminusoneequals5. So I have this denominator, x minus 1, and I don't have anything else. So that means that this denominator is my least common denominator. So I can go ahead and take my whole equation and multiply it by x minus 1. Now remember when we're multiplying by our LCD, we want to make sure to distribute it to every term in our equation. So here I'm going to get x minus 1 times my first term, 2x+4 over x minus 1 is equal to x minus 1 at times 5. Now simplifying this a little bit, I'm gonna see that this x minus 1 just cancels, leaving me with just 2x+4, and that's equal to I'm just gonna pull this 5 over to the front to make it easier, 5 times x minus 1. So we have completed step 2, and we can go ahead and move on to step 3 because I now just have a linear equation that I can solve. So I'm gonna go ahead and distribute this 5. The left side of my equation is gonna stay the same, 2x+4, and that's equal to 5x minus 5 with that 5 distributed. Okay. So our next step is gonna be to combine any like terms, which I don't have any like terms here. So I can go ahead and move on to the next, which is to take all of my x terms to one side, all of my constants to the other. So I'm gonna go ahead and move this 2 over here and move my negative 5 to the other side. So in order to do that, I need to subtract 2 x to cancel it. Whatever I do to one side, I have to do to the other. And then I'm gonna add 5 to both sides. So 4 +5 is going to give me a 9, and then I have 5x minus 2x x, which is going to give me positive 3x. Now my very last step is gonna be to isolate x, which I can do by dividing both sides by 3. So that 3 will cancel, and I'm left with my solution of x equals 3 because 9 divided by 3 gives me 3. So this is my solution. I've completed step number 3. And then my very last step is, of course, to check my solution with my restriction. I said that x couldn't be equal to 1 and for my solution I got x equals 3 which is not 1, so I'm good to go and my solution is x equals 3. That's all for this one. I'll see you in the next one.