Question

Can someone please help me with this question?
solutions to rational equations

Answer 1

When I was solving, the only solution I got was 5. But when I plug that into the equation, it comes out undefined. So I'm confused if 5 is an extraneous solution, or if the equation has no solution at all. Explain?

Answer 2

Let's start by factoring the right denominator.

\(\dfrac{1}{r - 5} = \dfrac{5}{5(25 - 5r)} \)

\(\dfrac{1}{r - 5} = \dfrac{5}{5\times 5(5 - r)} \)

\(\dfrac{1}{r - 5} = \dfrac{\cancel{5}~1}{\cancel{5\times }5(5 - r)} \)

\(\dfrac{1}{r - 5} = \dfrac{1}{5(5 - r)} \)

Look at the two denominators now to find restrictions in the domain.

\(r - 5 = 0\) and \(5 - r = 0\)

The restrictions are both the same, \(r \ne 5\)

Answer 3

Now we can solve the equation using cross multiplication.

\(\dfrac{1}{r - 5} = \dfrac{1}{5(5 - r)} \)

\(r - 5 = 5(5 - r) \)

\(r - 5 = 25 - 5r\)

\(6r = 30\)

\(r = 5\)

The only solution we get from the equation is r = 5, but we know r = 5 is a restriction on the domain, so there is no solution to this equation.

You are correct.

Answer 4

r = 5 is extraneous, and there is no solution to this equation.

Answer 5

OK
Thanks!