Solving $|x+1|>|x-3|$ using a distance approach

Question

I know how to solve it squaring both sides of the inequality, but I can't figure out how to solve it "thinking about distances on the real line".

Answer 1

The inequality $|x-(-1)|>|x-3|$ says that $x$ is farther from $-1$ than from $3$.

What point is equidistant from $-1$ and $3$?

Now can you say which points satisfy the inequality?

$$x>1$$

Answer 2

The midpoint $M$of two numbers $a$ and $b$ on the real line is the unique point equidistant from the two numbers. The midpoint $M$ It is given by the formula:

$M = \frac{a+b}{2}$.

Therefore, $M_{-1,3} = \frac{-1+3}{2} = 1$.

Furthermore, the distance between two real numbers $x$ and $y$ is $|x-y|$.

So the original inequality $|x+1| > |x-3|$ is the same as $|x-(-1)| > |x-3|$, which reads, "The distance between $x$ and $-1$ is greater than the distance between $x$ and $3$."

You should also draw a diagram to see what's going on.

Answer 3

For a point to be at a greater distance from A than B, it should be to the right of the midpoint of A and B, i.e., to the right of C($\frac{3-1}{2}=1$)(obvious from the image)

$\therefore $ the inequality holds good for $x>1$