### Instruction: Find all real values of $r$ that satisfy $\frac1r > \frac1{r-1} + \frac1{r-4}.$ (Give your answer in interval notation.).### Response: To solve for $r$, I need to clear the fractions by multiplying both sides by a common denominator.
The lowest common denominator of the three fractions is $r(r-1)(r-4)$, so I multiply both sides by that.
On the left side, the $r$'s cancel out, leaving $(r-1)(r-4)$.
On the right side, I need to distribute the $r(r-1)(r-4)$ to each fraction and cancel out the common factors.
For the first fraction, the $(r-1)$'s cancel out, leaving $r(r-4)$.
For the second fraction, the $(r-4)$'s cancel out, leaving $r(r-1)$.
So the inequality becomes $(r-1)(r-4) > r(r-4) + r(r-1)$.
Now I need to expand the brackets and collect like terms.
On the left side, I get $r^2 - 5r + 4$.
On the right side, I get $2r^2 - 5r$.
Subtracting $2r^2 - 5r$ from both sides, I get $-r^2 + 4 > 0$.
To make it easier to see the solution, I multiply both sides by $-1$ and flip the inequality sign.
I get $r^2 - 4 < 0$.
This means that the square of $r$ is less than 4, or that $r$ is between $-2$ and $2$.
In interval notation, this is $(-2,2)$.