Inequalities – Inequality with absolute value – Exercise 1862

Exercise

Solve the inequality:

$|2x-1|>|x-1|$

Final Answer

Show final answer

Solution

$|2x-1|>|x-1|$

We check when the phrases in the absolute values equal zero:

$2x-1=0 \rightarrow x=\frac{1}{2}$

$x-1=0 \rightarrow x=1$

We divide the x-axis into foreign sections by the points we found. We get three sections:

$x\leq \frac{1}{2}, \frac{1}{2}< x\leq 1, x>1$

In each section, we take the following steps:

1. Choose any number in the section.
2. Get rid of the absolute values ​​by the sign according to the number we chose, and solve the inequality.
3. Intersect result with the original section.

We start with the first section:

$x\leq \frac{1}{2}$

We choose the number x = -4. We set the number in our inequality and remove the absolute values by definition – If the result is positive, we simply remove the absolute value, and if the result is negative, we remove the absolute value and multiply by minus one. Hence, we get

$-(2x-1)>-(x-1)$

We solve the inequality:

$-2x+1>-x+1$

$1-1>2x-x$

$0>x$

Now, intersect this result with the original section. Meaning,

$x<0$

And

$x\leq \frac{1}{2}$

Together we get

$x<0$

Moving on to the second section:

$\frac{1}{2}< x\leq 1$

We choose the number

$x=\frac{3}{4}$

We set the number in our inequality and remove the absolute values by definition – If the result is positive, we simply remove the absolute value, and if the result is negative, we remove the absolute value and multiply by minus one. Hence, we get

$2x-1>-(x-1)$

Solve the inequality:

$2x-1>-x+1$

$3x>2$

$x>\frac{2}{3}$

Now, intersect this result with the original section, meaning

$x>\frac{2}{3}$

and

$\frac{1}{2}< x\leq 1$

together, we get

$\frac{2}{3}< x\leq 1$

Lastly, the third section:

$x>1$

We choose the number x = 4. We set the number in our inequality and remove the absolute values by definition – If the result is positive, we simply remove the absolute value, and if the result is negative, we remove the absolute value and multiply by minus one. Hence, we get

$2x-1>x-1$

Solve the inequality:

$2x-x>1-1$

$x>0$

Now, intersect this result with the original section, meaning

$x>0$

And

$x>1$

Together we get

$x>1$

The final step is to take all the solutions we received and union them, meaning

$x<0$

Or

$\frac{2}{3}< x\leq 1$

Or

$x>1$

Hence, our final answer is

$x<0 \text{ or } x>\frac{2}{3}$

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