Solve the inequality (x - 4)/3 < (2x - 1)/2.

Step-by-step solution

1. 1

   Multiply by the LCD 6 rather than cross-multiplying blindly. With a single fraction on each side, cross-multiplication looks tempting, but it is only safe when you know the signs of the denominators. Here both denominators are the positive constants $3$ and $2$, so multiplying both sides by $6$ is legitimate and keeps the $<$ direction.

2. 2

   Clear the denominators.

   $6\cdot\frac{x-4}{3}<6\cdot\frac{2x-1}{2}\ \Longrightarrow\ 2(x-4)<3(2x-1)$

   Each numerator must stay bracketed: $2x-4$ instead of $2(x-4)$ would change the constant term.

3. 3

   Expand both sides.

   $2x-8<6x-3$

4. 4

   Collect terms, keeping the $x$ coefficient positive. Subtract $2x$ from both sides, then add $3$:

   $-8<4x-3\ \Longrightarrow\ -5<4x$

   Had you instead subtracted $6x$, you would get $-4x<5$ and then have to divide by $-4$ and flip the sign — same answer, more risk.

5. 5

   Divide by 4 and keep the answer exact.

   $-\frac{5}{4}<x\ \Longrightarrow\ x>-\frac54,\qquad\text{i.e. }\left(-\tfrac54,\infty\right)$

   Leave it as $-\tfrac54$; writing $-1.25$ is fine numerically but exact fractions are what a solution set should contain.

6. 6

   Verify the endpoint. At $x=-\tfrac54$: left $=\dfrac{-\frac54-4}{3}=\dfrac{-\frac{21}{4}}{3}=-\dfrac74$ and right $=\dfrac{2(-\frac54)-1}{2}=\dfrac{-\frac72}{2}=-\dfrac74$. The sides are equal, so $-\tfrac54$ is the exact boundary and is correctly excluded.