Solve the inequality 3x + 9 > 6x + 7.

Step-by-step solution

1. 1

   Choose the side that keeps the coefficient positive. Both sides carry an $x$ term. Subtracting the smaller one, $3x$, leaves $3x$ on the right and dodges the need to divide by a negative number later — the single most error-prone step in inequalities:

   $9>3x+7$

2. 2

   Move the constant. Subtract $7$ from both sides:

   $2>3x$

3. 3

   Divide by 3. The divisor is positive, so the inequality sign is unchanged:

   $\frac{2}{3}>x$

4. 4

   Rewrite with the variable on the left. Flipping the two sides of an inequality also flips the symbol, so $\tfrac23>x$ becomes

   $x<\frac{2}{3}$

   This is a rewrite, not a new operation: both forms say the same thing. In interval notation the answer is $\left(-\infty,\tfrac23\right)$.

5. 5

   Verify on both sides of the boundary. At $x=0$: left $9$, right $7$, and $9>7$ ✓ (inside). At $x=1$: left $12$, right $13$, and $12>13$ ✗ (outside). At the boundary $x=\tfrac23$: left $11$, right $11$ — equal, so the strict inequality fails and the endpoint is correctly excluded. Comparing the raw inequality with $x<\tfrac23$ at $120$ exact rational points agrees everywhere ✓.