# Which is the proper way to solve the inequality problems?

I am in my high school and I am taught two methods to solve the inequality problems 1) by using the definition of given function in the inequality and taking the according cases. 2) turning point method.

But for many problems I got different solutions by these two methods.

How to solve this inequality?

$$| 3x +2 | \leq 2$$

By using the definition of modulus fn, it is solved to

$$x \text{ belongs to } [-4/3, 0]$$

By using the turning point method, it gives,

$$x \text{ belongs to } [-8/3, 4/3]$$

Which is correct and how to effectively solve such inequalities?

Please also help me understand inequalities involving the greatest integer function and fractional part function.

Thanks.

• You want to solve $$|3x+2|\le 2$$ over the reals? Jul 5 '19 at 12:55
• Welcome to stackexchange. We don't know those :methods by name. If you edit the question to show us the actual steps you followed in each case we can help you find any mistakes. Use mathjax: math.meta.stackexchange.com/questions/5020/… Jul 5 '19 at 12:58
• If you have proposed ranges, it is easy to check simply by checking values. for instance, $1\in [-\frac 83, \frac 43]$ but $|3\times 1 +2|=5$.
– lulu
Jul 5 '19 at 13:04

It's $$-2\leq3x+2\leq2$$ or $$-4\leq3x\leq0$$ or $$-\frac{4}{3}\leq x\leq0.$$

I don't know either of the methods you named, but I use the following method for solving inequalities.

If we have some expression $$X$$, then $$|X|\leq b$$ simply means that $$X$$ is either less than/equal to $$b$$ or greater than/equal to minus $$b$$. In other words, we can write this as $$-b\leq X\leq b$$

We can see this if we look at $$|x|\leq 1$$. We can quite easily see that any value for $$x$$ between $$-1$$ and $$1$$ will satisfy this equation, and thus $$-1\leq x\leq 1$$.

Therefore, you can write your equation as $$-2\leq 3x+2\leq 2$$

Now we solve it as we would with any other equation, noting that whatever you do to one of the three parts of the inequality, you must do to both of the other parts.

So, to start, we subtract $$2$$ from each part, giving us \begin{align}-2\color{red}{-2}&\leq 3x+2\color{red}{-2}&&\leq 2\color{red}{-2}\\ -4&\leq3x&&\leq0\end{align}

Next we divide all three parts by $$3$$

\begin{align}\frac{-4}{\color{red}3}&\leq\frac{3x}{\color{red}3}&\leq\frac{0}{\color{red}3}\\ \frac{-4}{3}&\leq x&\leq 0 \end{align}

Now we have solved for $$x$$