Why Multiplying an Inequality by a Negative Number Flips It

Real limits are rarely exact: a lift carries at most a certain load, a medicine dose must stay between two values, and a budget cannot be exceeded. Inequalities describe these ranges, and their solutions are intervals rather than single numbers.

This part covers linear inequalities, quadratic inequalities by the method of intervals, and rational inequalities.

A linear inequality is solved like a linear equation, except that multiplying or dividing both sides by a negative number reverses the inequality sign; the solution is an interval, drawn on the number line with a filled dot for an included end point and a hollow dot for an excluded one.

Rules:

- Adding or subtracting the same number keeps the sign
- Multiplying or dividing by a positive number keeps the sign
- Multiplying or dividing by a negative number reverses the sign

Worked example. Solve $3-2x\le 11$.

$$-2x\le 8\Rightarrow x\ge -4$$

The solution is $[-4,\infty )$: a filled dot at $-4$ with an arrow to the right. Check with $x=0$: $3\le 11$, true.

Worked example 2. Solve $-1<\frac{2x+3}{5}\le 3$. Multiplying by 5 gives $-5<2x+3\le 15$, so $-8<2x\le 12$ and $-4<x\le 6$, the interval $(-4,6]$.

An everyday example. A student needs an average of at least 60 over two tests and scored 52 in the first, so $\frac{52+x}{2}\ge 60$ gives $x\ge 68$ in the second test.

The substance. The sign flips because a negative multiplier reverses order — $2<5$, but $-2>-5$.

To solve a quadratic inequality, bring every term to one side, factorise to find the critical points, mark them on the number line, and read the sign in each interval — with a positive leading coefficient the expression is positive outside the roots and negative between them.

Method of intervals:

- Write the inequality with $a>0$, such as $a{x}^2+bx+c<0$
- Find the roots and mark them on the number line
- Signs alternate $+$, $-$, $+$ from right to left across distinct roots
- Include the roots only for $\le$ or $\ge$

Worked example. Solve $x^2-x-6<0$. Factorising, $(x-3)(x+2)<0$. The critical points are $-2$ and 3, and the expression is negative between them, so $-2<x<3$. Check $x=0$: $-6<0$, true.

Perfect-square cases:

- $(x-4{)}^2>0$ — every real x except 4
- $(x-4{)}^2\ge 0$ — every real x
- $(x-4{)}^2<0$ — no solution
- $(x-4{)}^2\le 0$ — only $x=4$

An everyday example. A rectangular stall with a perimeter of 20 m must have an area above 21 square metres, so $x(10-x)>21$ gives $x^2-10x+21<0$ and a side between 3 m and 7 m.

The substance. Never divide both sides by an expression in x — its sign is unknown, so the direction of the inequality is unknown too.

A rational inequality is solved by moving everything to one side as a single fraction, marking the zeros of the numerator and the denominator as critical points, testing the sign in each interval, and always excluding the points where the denominator is zero.

Steps:

- Bring all terms to one side — never cross-multiply, because the denominator may be negative
- Combine into $\frac{P(x)}{Q(x)}$ and factorise
- Mark the zeros of $P(x)$ and $Q(x)$; zeros of $Q(x)$ are always excluded
- Test one point in each interval

Worked example. Solve $\frac{x+1}{x-2}\ge 2$.

$$\frac{x+1}{x-2}-2\ge 0\Rightarrow \frac{x+1-2x+4}{x-2}\ge 0\Rightarrow \frac{5-x}{x-2}\ge 0$$

The critical points are 2 and 5. At $x=3$ the fraction is $\frac{2}{1}>0$, so it is positive between 2 and 5. Including 5, where the numerator is zero, but excluding 2, the solution is $2<x\le 5$.

Check. At $x=5$: $\frac{6}{3}=2$, which satisfies the inequality. At $x=6$: $\frac{7}{4}=1.75$, which does not.

An everyday example. A caterer charging a fixed ₹3000 plus ₹80 per plate has an average cost of $\frac{3000+80n}{n}$ rupees per plate, which is at most ₹100 only when $3000\le 20n$, that is $n\ge 150$ plates.

The substance. **Cross-multiplying by $x-2$ would be wrong for $x<2$** — the denominator is then negative and the inequality would have to reverse.