Solve the following.p2p−4+44−p=1\frac{p^2}{p-4}+\frac{4}{4-p}=1

p = 0 p=0 p = 0 and p = 1 p=1 p = 1

Solve the following. p 2 p − 4 + 4 4 − p = 1 \frac{p^2}{p-4}+\frac{4}{4-p}=1

this problem, we are asked to solve the following equation, and in this case, we're trying to solve four p. Now the way that I can solve this is by recognizing something. We are dealing with some rational expressions within this equation. So what I need to do is find a least common denominator. Now to find the LCD, well, that's going to be a product of all the unique prime factors. Now one of the things I notice is one of our denominators is p minus four. The other denominator is four minus p. Now these are both about as factored as they can get, so it doesn't seem like there's a lot I can do to get a similar denominator. However, remember there's this rule of thumb when we have this situation where the terms are basically reversed, p minus four and four minus p. We can't just change the order of them, but there is a way that we can reverse this by changing a sign. Let me show you. So in this first expression I see, I'm gonna keep this entirely the same. We'll keep this as p over p minus four. Now this is gonna be plus four, and then four minus p is the same thing as four plus negative p. And all this is going to equal one. Now from here, what I'm going to do is I am gonna switch the order now because notice how we're no longer dealing with a minus sign, we're now dealing with a plus in the second expression. We have a plus sign now. And for plus signs, you can change the order. So we'll have p squared over p minus four, that's gonna be plus, and then we'll have four over negative p plus four. Now this is all fine and good except for the fact that we still don't have the same the same denominator here. But there is something that I'm going to do. What I'm going to do is I'm going to pull a negative sign out of each of these terms, which is going to change the signs on each of them. So we'll keep this p squared over p minus four, but if I go ahead and pull a negative sign out of this denominator, well, that's gonna be negative, and then we'll have p, and then this will change to a negative minus four. And notice here how we now have a very similar denominator. The only difference is this negative sign right about here. But what I can do is I can take that negative sign and I can put it on the entire rational expression, which is going to change this from plus four over negative p minus four to just a minus four over p minus four. And notice we now have an LCD in common. We have a p minus four. So what I need to do is take both sides of this equation and multiply it by p minus four. Now, distributing this into the brackets, we'll have p minus four times p squared over p minus four. Notice that these terms will cancel completely. And And notice if I distribute it over here, this will also completely cancel with this one. So the p minus four entirely goes away, and all we end up with on the left side of this equation is p squared minus four. Now that's going to equal one, but then I need to multiply this by p minus four, which is just gonna be p minus four. So this is the equation we are dealing with now. Now from here, I'm going to go ahead and add four on both sides of this equation. Notice doing that is going to cancel the fours right about here, and the fours right about there. So it's just gonna give me that p squared is equal to p. Now from here, what I'm going to do is subtract p on both sides of this equation. That's gonna get the p's to cancel right about there, giving me p squared minus p is equal to zero. Now at this point, I can pull a p out of each of these terms. It's gonna give me p times p minus one is equal to zero. So notice I'm gonna end up getting two solutions to my problem. First off, we'll have that p is equal to zero, and then we'll have that p minus one is equal to zero. And if I wanna go ahead and solve for p, well, I'll just add one on both sides of the equation, giving me that p equals one. So p equals zero and p equals one would both be solutions to this equation. Now this problem doesn't tell us to do this, but I can also go ahead and check to see if this would be, this would be a correct solution. So we get p equals zero and p equals one. Let's see if both of these would solve the problem. So I can see that we have p equals zero and p equals one, and let's try this in the equation we have. So I'm gonna take every place that I see p, and I'm gonna replace it with a zero. That's gonna give me zero squared over zero minus four plus four over four minus zero is equal to one. Now anytime I see zero in the numerator, that just goes away completely. The entire term does. So I end up with four over four minus zero, which is four over four, which equals one, and four over four does in fact equal one. So this is indeed a solution. So we now have our solution right about here. And what I can also do is I can check the p equals one. Well, to check the p equals one, that's just replacing every place that I see p with one. So I'll go back to the original equation we had, and that's going to be one squared over one minus four plus and then we'll have four over four minus one is equal to one. One squared is one. One minus four is negative three. You will have plus, and then it's going to be four. And then we have four minus one, which is three. And then that equals one. And now what I need to do is add these together. Well, to add these together, that's going to be this negative one third plus one fourth, which ends up giving me positive three thirds over one, and three over three does in fact equal one. So once again, this solved the problem. So these two are solutions to the problem p equals zero and p equals one, and that is how you can solve this equation. Again, this was the only answer that you needed right about down here, but I went ahead and wanted to check our solution as well just to make sure we did in fact have two solutions. You could do this if you wanted to. It doesn't specifically ask us to check, but it always helps to double check to make sure you solved everything correctly. So hope you found this video helpful. Let me know if you have any questions, and I will see you all in the next one.

Solve the following. p 2 p − 4 + 4 4 − p = 1 \frac{p^2}{p-4}+\frac{4}{4-p}=1

p = 0 p=0 p = 0 and p = 1 p=1 p = 1

p = − 1 p=-1 p = − 1 and p = 1 p=1 p = 1

8. Rational Expressions and Equations

Rational Equations

Beginning & Intermediate Algebra

8. Rational Expressions and Equations

Rational Equations

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Multiple Choice

Solve the following.
$\frac{p^{2}}{p - 4} + \frac{4}{4 - p} = 1$

$p=-1$ and $p=1$

$p=0$ and $p=1$

$p=-1$

$p=1$

Verified step by step guidance

Rewrite the given equation: $\frac{p^2}{p-4} + \frac{4}{4-p} = 1$.

Notice that the denominators $p-4$ and \$4-p$ are opposites. Use the fact that $4-p = -(p-4)$ to rewrite the second fraction as $\frac{4}{4-p} = \frac{4}{-(p-4)} = -\frac{4}{p-4}$.

Substitute this back into the equation to get a single denominator: $\frac{p^2}{p-4} - \frac{4}{p-4} = 1$.

Combine the fractions on the left side since they have the same denominator: $\frac{p^2 - 4}{p-4} = 1$.

Multiply both sides of the equation by $p-4$ (noting $p \neq 4$ to avoid division by zero) to clear the denominator: $p^2 - 4 = 1 \times (p-4)$, then simplify and solve the resulting quadratic equation.

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