You can
put this solution on YOUR website!
(x-5) =
:
The first thing you want to do here is get rid of the denominators
multiply equation by a multiple of both denominators, that would 4
4*
(x-5) = 4*
Cancel out the denominators and you have:
2(x-5) = 1x
:
2x - 10 = x; multiply what's inside the brackets
:
2x - x = +10; basic algebra, add 10 to both sides, subtract x from both side
:
x = 10
:
:
Check in original equation; substitute 10 for x
(10-5) =
(5) =
2.5 = 2.5
You can
put this solution on YOUR website!The thing is, I'm not sure whether your problem is:
or
, but I'm going to assume the former because the latter is trivial.
When dealing with rational functions, usually the best thing is to get everything on the left and leave 0 on the right to start:
Add
to both sides
Now you have the problem of adding two fractions with dissimilar denominators. You need to find a common denominator. Your denominators have a factor of 2 in common, so your LCD is
. So:
Now that we have a common denominator, we can add the numerators, but first it will be convenient to distribute and remove the parentheses from the 2nd numerator. You can leave the denominator as it is for the time being.
Now, there is something to consider before you go any further. Is there a value of x that would make the denominator go to 0? The answer is yes, this function is undefined for x = 5, because that would give us a zero denominator. Therefore, if we arrive at a solution set that contains x = 5, we will have to exclude that solution.
The next thing to realize is that
if and only if
and
. So in order to solve this equation, we don't have to worry about the denominator (except to exclude 5 if it comes up as a solution) and all we have to do is set the numerator equal to zero and solve the quadratic.
Multiply by -1
This one doesn't factor neatly, so the best alternative is the quadratic formula:
We don't have to worry about excluding 5 because we got a conjugate pair of irrational roots.
Since root 33 is larger than 5, one of the roots is negative and the other is positive. We would then expect a graph of the original function to cross the x-axis once to the left of the y-axis, and once to the right of the y-axis. We would also expect to see a vertical asymptote at x = 5.
As advertised.
