Absolute value inequality with variable on both sides

Question

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:

$$3-5x \le x$$

$$ x \ge (1/2)$$

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$$-x \le 3-5x$$

$$x \le \frac34$$

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,

Answer 1

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:

$$|3-5x| \le x$$

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$$3-5x \le x \;\;\;\;\;| \;\;-3+5x\le x $$

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$$6x \ge 3 \;\;\;\;\;\;\;| \;\;-4x \ge -3$$

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$$x \ge \frac 12 \;\;\;| \;\;\;\;x \le \frac 34$$

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$$\frac 12 \le x \le \frac 34$$

Answer 2

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}$.

Answer 3

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:

1. $x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$ $$ 3 - 5x \leq x \\ 3 \leq 6x \\ x \geq \frac{1}{2} $$
2. $x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$ $$ 5x - 3 \leq x \\ 4x \leq 3 \\ x \leq \frac{3}{4} $$
3. $x = \frac{3}{5} \rightarrow |3 - 5x| = 0$ $$ 0 \leq x \\ x \geq 0 $$

Now you need to analyze each case:

1. $x < \frac{3}{5} \wedge x \geq \frac{1}{2}$

It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$. Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.

2. $x > \frac{3}{5} \wedge x \leq \frac{3}{4}$

Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4} = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.

3. $x = \frac{3}{5} \wedge x \geq 0$

$\frac{3}{5} > 0$ 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: $\frac{1}{2} \leq x \leq \frac{3}{4}$.