Solve the inequality -2|3m| + 3 < 51.

Step-by-step solution

1. 1

   Isolate the absolute value term. Subtract $3$ from both sides:

   $-2|3m|<48$

2. 2

   Divide by $-2$ and reverse the sign. Division by a negative flips the direction:

   $|3m|>-24$

3. 3

   Stop and read the inequality instead of splitting it. The usual split $A<-b$ or $A>b$ assumes $b>0$. Here the right-hand side is negative, so that template does not apply. Compare the two sides directly:

   $|3m|\ge 0>-24$

4. 4

   Conclude that every real number is a solution. Since $|3m|$ is at least $0$ and $0$ already exceeds $-24$, the inequality holds for every $m$:

   $m\in(-\infty,\infty)$

   Spot-check $m=0$: $-2\cdot 0+3=3<51$ $\checkmark$; and $m=1000$: $-6000+3=-5997<51$ $\checkmark$.

5. 5

   Compare with the near-identical problem that is not trivial. Changing the right-hand side from $51$ to $-51$ gives $|3m|>27$, whose solution is only $m<-9$ or $m>9$. The sign of the number left after the flip is what decides between 'all reals' and a genuine two-branch answer.