Simultaneous linear inequalities

A chained inequality requires both comparisons

For −1 < 2x + 3 ≤ 7, subtract 3 from all three parts to obtain −4 < 2x ≤ 4, then divide by positive 2: −2 < x ≤ 2. The left endpoint is excluded and the right included.

For −6 ≤ −2x < 4, dividing by −2 reverses both inequalities: 3 ≥ x > −2. Reorder it as −2 < x ≤ 3. Reversing only one relation changes the solution set.

“And” means intersection

If 2x − 1 < 5 and 3 − x ≤ 5, the first gives x < 3 and the second x ≥ −2. Both hold exactly on −2 ≤ x < 3. Draw one number line and shade only values meeting every condition.

For x < 0 or x > 2, either condition suffices, so the set is the union of two disjoint intervals. Do not replace an “or” statement with the empty overlap between them.

Check your understanding 1 Keep both restrictions

Solve 3x + 1 ≥ -5 and 2-x > 0 simultaneously.

Your answer or drawing on paper

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The first gives x ≥ −2, the second x < 2. Their intersection is −2 ≤ x < 2. A filled point at −2 and open point at 2 match the original endpoint conditions.

Check a boundary in the original conditions

Substitute each proposed endpoint into both original inequalities. A value allowed by one condition may still fail the other. A complete solution describes the interval, not only the boundary equation roots.