The Direction That Flips Linear Inequalities

Ravi has Rs.200 to spend on notebooks priced at Rs.30 each, and wants to know how many he can afford -- the honest condition here isn't 30x = 200 (an exact equation), it's 30x < 200 or 30x <= 200, since spending LESS than the full budget is perfectly fine too. Reshma, buying registers at Rs.40 and pens at Rs.20 with a Rs.120 budget, faces the same kind of condition: 40x + 20y <= 120. These everyday budget constraints, expressed with < or <=, > or >=, rather than =, are exactly what a linear inequality captures.

An inequality using < or > is called strict (the boundary value itself is excluded), while one using <= or >= is called slack (the boundary value is included). A double, or compound, inequality squeezes a variable between two boundaries at once, like 2 < y <= 4, meaning y must be strictly greater than 2 but can be equal to (or less than) 4.

Solving an inequality follows two rules. The first: adding or subtracting the same number on both sides never changes the direction of the inequality, exactly like solving an equation. The second, and the one genuinely worth remembering carefully: multiplying or dividing both sides by the same POSITIVE number keeps the direction unchanged, but multiplying or dividing by a NEGATIVE number flips it completely. This is easy to verify directly: 3 is greater than 2, a true statement -- but multiplying both sides by -1 gives -3 and -2, and now -3 is actually LESS than -2, so the direction had to flip to keep the statement true. Forgetting this single flip is the most common mistake in solving inequalities.

Once an inequality is solved down to something like x < 3 or x >= 1, its solution set can be drawn directly on a number line: an open circle at the boundary marks a strict inequality (the boundary point itself is excluded), while a filled-in circle marks a slack inequality (the boundary point is included), with a ray extending in the direction the inequality points.

A student scoring 70 and 75 in the first two of three exams, needing an average of at least 80 across all three, faces a genuine inequality: with the third score called x, the condition is (70+75+x)/3 >= 80, which solves to x >= 95 -- meaning at least 95 is required on the final exam, a real, checkable consequence of the averaging requirement.