Solving a Quadratic Inequality

Question

Solve the inequality: $$\dfrac{x-1}{x^2-4x+3}<1.$$

I cancelled the $x-1$ from the numerator and the denominator and I obtained: $$\frac{x-4}{x-3}.$$ Thus, the answer that I got was $$x \in(3,4).$$ But according to WolframAlpha answer should be $$x \in(4,\infty)\cup(1,3)\cup(-\infty,1).$$

What is the best method to solve this inequality and where have I gone wrong?

Answer 1

It's $$\frac{1}{x-3}<1$$ and $x\neq1$ or $$1-\frac{1}{x-3}>0$$ or $$\frac{x-4}{x-3}>0,$$ which by the intervals method gives $x<3$ or $x>4$ and with $x\neq1$ we got the answer: $$(-\infty,1)\cup(1,3)\cup(4,+\infty)$$

Answer 2

First of all, you must check the condition: $x \neq 1$ and $x\neq 3$.

So, you have $$\frac{x-1}{(x-1)(x-3)} < 1,$$ or $$\frac{1}{x-3} < 1.$$

Now, you have

Case 1: $x-3 < 0$ or $x < 3$, but you have $x \neq 1$, then $x < 1$ or $1 < x <3$.

Case 2: $x - 3 > 0$ and $x - 3 > 1$, then $x > 4$.

Answer 3

The quotient $\dfrac{x-1}{x^2-4x+3}$ is defined for all $x$ with $x \ne 1$ and $x \ne 3$. For those $x$ we have

$\dfrac{x-1}{x^2-4x+3}=\dfrac{1}{x-3}$.

For the inequality $\frac{1}{x-3}<1$ you have to distinguish the cases

$x>3$ and $x<3$. Can you proceed ?

Answer 4

$$\frac{(x-1)}{(x-3)(x-1)} < 1$$

Hence $x \neq 1$ and $x \neq 3$.

$$\frac{1}{(x-3)} < 1$$

Multiply $(x-3)^2$ throughout.

$$(x-3) < (x-3)^2$$

$$0< (x-3)(x-3-1)$$

$$0< (x-3)(x-4)$$

Hence $x \in \left((-\infty, 3) \cup (4, \infty)\right) \setminus\{ 1,3\}$

Hence $x \in (-\infty, 1) \cup (1,3) \cup (4, \infty)$.