Question

\left|\frac{m+3m}{3}\right|-1\le 2

Answer 1

First, let's simplify the expression inside the absolute value:

\left|\frac{4m}{3}\right|-1\le 2

Next, let's get rid of the absolute value by considering two possible cases:

Case 1: $\frac{4m}{3}\ge 0$

In this case, the absolute value is unnecessary because $\frac{4m}{3}\ge 0$ is always non-negative. So we can rewrite the inequality as:

\frac{4m}{3} - 1 \le 2

Next, we can solve for $m$:

\frac{4m}{3} \le 3

Multiplying both sides by $\frac{3}{4}$ (which is positive so the inequality direction doesn't change):

m \le \frac{9}{4}

So in this case, the solution is: m \le \frac{9}{4}.

Case 2: $\frac{4m}{3}< 0$

In this case, the expression inside the absolute value becomes negative, so we need to negate it to make it non-negative:

-\left(\frac{4m}{3}\right) - 1 \le 2

Next, we can solve for $m$:

-\frac{4m}{3} \le 3

Dividing both sides by $-\frac{4}{3}$ (which is negative so the inequality direction reverses):

m \ge -\frac{9}{4}

So in this case, the solution is: m \ge -\frac{9}{4}.

Combining the two cases, we get the final solution:

-\frac{9}{4} \le m \le \frac{9}{4}