Linear Equations and Inequalities: Question 3

Worked solution

Step 1: Treat it as one inequality acting on all three parts

A double (compound) inequality such as $-5 \le 3x - 2 < 7$ means both of these hold at the same time: $-5 \le 3x-2 \qquad \text{and} \qquad 3x-2 < 7.$ The quickest method is to keep the three parts together and do the same operation to every part.

Step 2: Add $2$ to all three parts

To undo the "$-2$", add $2$ everywhere: $-5 + 2 \;\le\; 3x - 2 + 2 \;<\; 7 + 2$ $-3 \;\le\; 3x \;<\; 9.$

Step 3: Divide all three parts by $3$

Dividing by a positive number does not reverse the inequality signs: $\frac{-3}{3} \;\le\; \frac{3x}{3} \;<\; \frac{9}{3}$ $-1 \;\le\; x \;<\; 3.$

So $x$ lies in the range $-1 \le x < 3$.

Step 4: Pick out the integers

On a number line this is a solid dot at $-1$ (because of $\le$, so $-1$ is included) and an open dot at $3$ (because of $<$, so $3$ is excluded):

$\bullet\!\!=\!-1 \quad\longrightarrow\quad 3\!=\!\!\circ$

The integers between $-1$ (included) and $3$ (not included) are $-1,\; 0,\; 1,\; 2.$

Notice that $3$ is not in the set, because the right-hand sign is strict ($<$), but $-1$ is in the set, because the left-hand sign allows equality ($\le$).

Alternative check: substitute the end values

• $x=-1:\;\; 3(-1)-2 = -5$, and $-5 \le -5$ is true, while $-5 < 7$ is true. ✓ include
• $x=3:\;\; 3(3)-2 = 7$, but $7 < 7$ is false. ✗ exclude

This confirms the boundary behaviour.

Final answer

$\boxed{\{-1,\,0,\,1,\,2\}}$

which is option B.