Solve |2x + 3| = 6x.

Step-by-step solution

1. 1

   Impose the non-negativity constraint first. The left side is an absolute value, so it is never negative. The right side must therefore satisfy

   $6x\ge0\qquad\Longrightarrow\qquad x\ge0$

   This constraint is not optional bookkeeping — it is what makes one of the two cases impossible, and forgetting it produces a spurious second answer.

2. 2

   Case 1: the inside is non-negative, 2x + 3 >= 0. Then $|2x+3|=2x+3$ and

   $2x+3=6x\qquad\Longrightarrow\qquad 3=4x\qquad\Longrightarrow\qquad x=\frac34$

3. 3

   Check the candidate against both conditions. Inside: $2\left(\tfrac34\right)+3=\tfrac92\ge0$ ✓, consistent with the case assumption. Global: $x=\tfrac34\ge0$ ✓. So $x=\tfrac34$ is genuine.

4. 4

   Case 2: the inside is negative, 2x + 3 < 0. Then $|2x+3|=-(2x+3)$ and

   $-2x-3=6x\qquad\Longrightarrow\qquad -3=8x\qquad\Longrightarrow\qquad x=-\frac38$

5. 5

   Reject the second candidate — twice over. It fails the global constraint $x\ge0$, since $-\tfrac38<0$. It also fails its own case condition: $2\left(-\tfrac38\right)+3=\tfrac94>0$, not negative. Substituting directly confirms it: $\left|2\left(-\tfrac38\right)+3\right|=\tfrac94$ but $6\left(-\tfrac38\right)=-\tfrac94$, and $\tfrac94\neq-\tfrac94$. It is extraneous.

6. 6

   State and verify the answer. The only solution is

   $x=\frac34$

   Checking: $\left|2\left(\tfrac34\right)+3\right|=\left|\tfrac92\right|=\tfrac92$ and $6\left(\tfrac34\right)=\tfrac92$ ✓. A scan of $6001$ exact rational points on $[-3,3]$ finds $\tfrac34$ to be the unique solution ✓.