Solve the equation. − 5 x + 4 − 3 = x − 1 x + 4 \frac{-5}{x+4}-3=\frac{x-1}{x+4} x + 4 − 5 − 3 = x + 4 x − 1

Hey, everyone. Let's take a look at this practice problem. So here I have negative 5 over x+4 minuteus3 is equal to x minus 1 over x plus 4, and we wanna solve this rational equation. So let's go ahead and get started with step 1, which is to determine our restriction by setting our denominator equal to 0. Now I have 2 denominators here. I have x+4 and x+4. So that means that I can just take that x+4 and set it equal to 0 to determine what my So this So this tells me that if x was equal to negative 4, that would make my denominators 0, which is what I don't want. So I know that my restriction is x cannot be equal to negative 4. Okay. So we've completed step 1. We know what x can't be. Let's go ahead and start with step 2, which is to multiply by our least common denominator. Now, again, both of my denominators here are x+4, so that means they share the common factor of x plus 4. That means that my least common denominator is simply x+4. So moving this down here to expand it, I can go ahead and write x plus 4 times my first term, negative 5 over x+4. Then my least common denominator again, and my second term, which is negative 3. And then that's equal to my least common denominator again times my last term, x minus 1 over x+4. Okay. It looks a little bit crazy right now. Let's go ahead and cancel some stuff out. So my very first term, my x+4 is going to cancel leaving me with just a negative 5. Then I'm gonna go ahead and move this 3 over to the front of those parentheses just to make it easier for me. Negative three times x+4, and that is equal to on my right side, I have x+4 canceling out again, leaving me with x minus 1. So step 2 is done and I am just left with a linear equation. Let's go ahead and do step 3, solving our linear equation. So I need to go ahead and distribute this negative 3 into those parentheses. So I have negative 5 and then negative 3 times x gives me negative 3x, and then negative 3 times 4 is going to give me negative 12. Make sure you don't lose that negative. And that's equal to my right side x minus 1. Now I do have some like terms that need to get combined here. I have this negative 5 and this negative 12, which is going to combine to give me negative 17. Then pulling my 3 x down, I have 7 negative 17 minus 3 x is equal to x minus 1. Nothing needs to get combined over there. And now we can move all of our x terms to one side, all of our constants to the other. So I wanna go ahead and move this x over here, move my 17 over here. So in order to do that, I can subtract x, letting it cancel, and then add 17 to both sides. It will, of course, cancel there. I'm left with negative 3x minus x, which will give me negative 4x, And that's equal to negative 1 +17, which will give me 16. Now my very last step is to isolate x. And to do that, I need to divide both sides by negative 4. So my negative 4 will cancel there, and I'm left with x is equal to 16 divided by negative 4 which is negative 4 and here is my solution. So step 3 is done but we cannot forget step 4 which is to check our solution with our restriction. Now my restriction was that x could not be equal to negative 4, and my solution that I got was that x equals negative 4. So this tells me that this is not my solution, and I actually have no solution because my answer was equal to my restriction which we know is what it can't be. So that's all for this one, I'll see you in the next one.