Solve the system of inequalities x − 3(x − 2) ≥ 4 and (2x − 1)/5 < (x + 1)/2.

Step-by-step solution

1. 1

   Solve the first inequality. Expand, remembering that $-3$ multiplies both terms:

   $x-3x+6\ge 4\quad\Longrightarrow\quad -2x+6\ge 4\quad\Longrightarrow\quad -2x\ge -2$

2. 2

   Divide by the negative coefficient and flip the sign. Dividing both sides by $-2$ reverses the inequality:

   $x\le 1$

   This sign flip is the single most common error in the whole problem.

3. 3

   Solve the second inequality by clearing denominators. The LCM of $5$ and $2$ is $10$, which is positive, so multiplying does not change the direction:

   $2(2x-1)<5(x+1)\quad\Longrightarrow\quad 4x-2<5x+5$

4. 4

   Isolate x. Subtract $4x$ and $5$ from both sides:

   $-7<x$

5. 5

   Intersect the two solution sets. Both conditions must hold simultaneously:

   $-7<x\le 1$

   $\boxed{(-7,\,1]}$

6. 6

   Describe the number-line picture and test the endpoints. Draw an open circle at $-7$ (excluded, strict inequality) and a closed circle at $1$ (included), shading between them. Checking: at $x=1$ the first gives $1-3(-1)=4\ge 4$ ✓ and the second gives $\tfrac15<1$ ✓; at $x=-7$ the second gives $\tfrac{-15}{5}=-3<\tfrac{-6}{2}=-3$, which is false, confirming $-7$ must be excluded; at $x=0$ (inside) both hold ✓.