I am studying for college finals and I cannot solve the inequation (x-1)/(x-5)<0 using the method the professor taught us, which I have to use in the exam.
This is another inequation using said method:
I noticed that the inequation I posted has a different symbol than the pic, but the first one is the only example of this type the professor gave us with the less than symbol.
This is how I tried to solve it:
I appreciate any help you can give. Thanks in advance.
$\endgroup$ 24 Answers
$\begingroup$If $\dfrac{x-1}{x-5}<0$, that means $x-1$ and $x-5$ have opposite sign.
So $x-1<0$ and $x-5>0$
or $x-1>0$ and $x-5<0$.
Which of those is possible?
$\endgroup$ 3 $\begingroup$Remember if $\frac ab < 0$ then $a$ and $b$ are "different signs".
So if $\frac {x-1}{x-5} < 0$ then either
1) $x - 1 > 0$ and $x -5 < 0$
OR
2) $x-1 < 0$ and $x-5 > 0$.
In case 1) we have $x - 1> 0$ so $x > 1$ and $x-5 < 0$ so $x < 5$. So $x$ is between $1$ and $5$ or $1 < x < 5$
In case 2) we have $x -1 < 0$ and $x< 1$ and $x-5> 0$ so $x > 5$. So $x$ is both less than $1$ and greater than $5$. That's impossible.
So Case 1: is the true case and $1 < x < 5$.
.....
Alternatively. $1 < 5$ always and so $-5 < -1$ always, and $x-5 <x - 1$ always no matter what $x$ is.
So when we know that $x-1$ and $x-5$ are "different signs" we know that $x-5$ must be the negative one (because it is the smaller one) and $x -1$ must be the positive one because it is bigger.
So $x -5 < 0 < x-1$
So $x - 5 < 0$ and $x < 5$. And $0 < x-1$ so $1 < x$. So $1 < x < 5$.
$\endgroup$ $\begingroup$Draw the numbers line.
The quotient $\frac{x-1}{x-5} <0$ when:
(1) $x-1 <0$ AND $x-5 >0$
(2) $x-1 >0$ AND $x-5<0$
Identify the segment where $x-1>0$ and $x-1<0$ on the line.
Identify the segment where $x-5>0$ and $x-5<0$ on the line.
Condition (1) above is shown as the middle segment: $x \in (1,5)$.
Condition (2) above is satisfied by segments (A) and (B). However, $x$ can't be in segments (A) and (B)in the same time, it is impossible.
So $x \in (1,5)$ is your answer.
Always test few values to verify.
First draw a number line and place any values of $x$ that make the left equal the right exactly. This means that $(x-1)(x-5)=0$ and the solutions to this equation are $5$ and $1$.
So mark $5$ and $1$ on your number line.
Notice that your function cannot go above or below the $x$ axis (change sign) anywhere else as it would have to go through zero and the only places where your function equals zero is at $5$ and $1$.
So pick any value in each segment formed by the number line and its crossing (this means one segment is from negative infinity to $1$, the other from $1$ to $5$ and the last one from $5$ to infinity as your function will not change it’s sign on those intervals, since it doesn’t have zeros in those regions).
Now select any number in each section to see if your segment satisfies the general inequality:
For $(-\infty,1)$ select $-3$ and see that it makes our function equal to positive value which is not what we want.
Move on to section $(1,5)$ and let’s plug in $3$. It gives us $-4$ which satisfies our inequality which means that the entire region falls below the $x$ axis so we can include that region.
The final section is $(5,\infty)$ and let’s plug in $7$ which gives a positive value when plugged in and does not satisfy the general inequality.
Now notice that the values that make our function equal zero are not to be included in the answer as we want it to be below zero and equal to zero.
So $(1,5)$ is our range of values that satisfy it.
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