College Algebra / Unit 3 Alg 163 worked example, 60 to 90 seconds
Solving a Linear Inequality
The idea
Solving a linear inequality runs the equation moves — clear, collect, isolate — with one extra law: a negative multiplier or divisor reverses the symbol, and the answer lands in interval notation.
In practice
Solve 5 − 3x ≥ −7. Subtract 5 from both sides: −3x ≥ −12. Divide by −3 and flip: x ≤ 4. In interval notation: (−∞, 4]. Test the claim twice: x = 0 gives 5 ≥ −7, true and inside the interval; x = 5 gives −10 ≥ −7, false and outside. The two tests bracket the boundary, and the boundary itself, x = 4, gives −7 ≥ −7 — true, which is why the bracket is square. Every solved inequality earns a two-sided test like this one.
Where people slip
One flip per negative multiplier — count them, since two flips restore the original direction.
Your place on the trail
Every lesson rests on earlier ones and opens later ones. Tap any to jump.
Builds on
- Nothing. This is a starting point.
Unlocks
- Nothing yet depends on this.
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