I am trying to solve the following inequality:
$$|3-5x| \le x$$
I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:
$$-x \le 3-5x \le x$$
Then I solved for each side separately,as follows:
$$3-5x \le x$$
$$ x \ge (1/2)$$
$$-x \le 3-5x$$
$$x \le \frac34$$
I know my solution is incorrect and that it actually lies between $\frac12$ and $\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.
Thanks,
$\endgroup$4 Answers
$\begingroup$This is how I learned it back in Algebra I.
You can split it into two equations. Since the sign is $\le$, it will be that Equation 1 and Equation 2 are true.
Here are the steps:
$$|3-5x| \le x$$
$$3-5x \le x \;\;\;\;\;| \;\;-3+5x\le x $$
$$6x \ge 3 \;\;\;\;\;\;\;| \;\;-4x \ge -3$$
$$x \ge \frac 12 \;\;\;| \;\;\;\;x \le \frac 34$$
$$\frac 12 \le x \le \frac 34$$
$\endgroup$ 5 $\begingroup$Divide $ -x \leq 3-5x \leq x $ by $x$ and go from there. Although I think what you have is correct, the solution IS the interval $ \dfrac{1}{2} \leq x \leq \dfrac{3}{4}$.
$\endgroup$ $\begingroup$You need to break up the absolute value into its intervals:
$$ |x| = \begin{cases} x & x > 0 \\ -x & x < 0 \\ 0 & x = 0 \end{cases} $$
Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:
$$ |3-5x| = \begin{cases} 3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarrow x < \frac{3}{5}\\ 5x - 3 & 3 - 5x < 0 \rightarrow 3 <5x \rightarrow x > \frac{3}{5} \\ 0 & 3 - 5x = 0 \rightarrow 3 = 5x \rightarrow x = \frac{3}{5} \end{cases} $$
Now you solve the inequality in each case:
- $x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$ $$ 3 - 5x \leq x \\ 3 \leq 6x \\ x \geq \frac{1}{2} $$
- $x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$ $$ 5x - 3 \leq x \\ 4x \leq 3 \\ x \leq \frac{3}{4} $$
- $x = \frac{3}{5} \rightarrow |3 - 5x| = 0$ $$ 0 \leq x \\ x \geq 0 $$
Now you need to analyze each case:
- $x < \frac{3}{5} \wedge x \geq \frac{1}{2}$
It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$. Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.
- $x > \frac{3}{5} \wedge x \leq \frac{3}{4}$
Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4} = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.
- $x = \frac{3}{5} \wedge x \geq 0$
$\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \frac{3}{5}$ is allowed.
When we combine these results, we find that $\frac{1}{2} \leq x \leq \frac{3}{4}$.
PS Edit:
It's probably easier to use:
$$ |x| = \begin{cases} x & x \geq 0 \\ -x & x \leq 0 \\ \end{cases} $$
In that case you get the two intervals:
$$ \frac{1}{2} \leq x \leq \frac{3}{5} $$
and
$$ \frac{3}{5} \leq x \leq \frac{3}{4} $$
Which clearly combines to give: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
$\endgroup$ $\begingroup$Your solution was correct except you needed to use the word AND correctly. The inequality $-x \le 3-5x \le x$ is equivalent to the inequalities
\begin{align} -x \le 3-5x \quad &\text{AND} \quad 3-5x \le x \\ 4x \le 3 \quad &\text{AND} \quad 3 \le 6x \\ x \le \frac 34 \quad &\text{AND} \quad \frac 12 \le x \\ \frac 12 \le x \quad &\text{AND} \quad x \le \frac 34 \\ x &\in\left[\frac 12, \frac 34 \right] \end{align}
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