Multiply by a Negative and the Inequality Turns Around

Because an inequation describes a range, not a point.

The equation $x+3=8$ pins the variable down to exactly one value, $x=5$. The inequation $x+3<8$ only says $x$ is somewhere below $5$ — which leaves $4$, $3$, $2$, $1$ and, if negative numbers are allowed, endlessly many more.

So the answer to an inequation is not a number but a set of numbers, and two things must be settled before you can write it down: what the inequation reduces to, and which numbers you were allowed to choose from in the first place.

There is also one rule that behaves unlike anything in the equations chapter — multiplying or dividing by a negative number flips the sign round. That rule is what this page is really about. It covers the ICSE Class 8 Mathematics chapter on linear inequations.

**An equation uses $=$ and claims two things are equal; an inequation uses an inequality symbol and claims one is bigger or smaller.**

There are four symbols, and each has a precise reading:

- $x<5$ — $x$ is less than $5$. The value $5$ itself is excluded.
- $x>5$ — $x$ is greater than $5$, again excluding $5$.
- $x\le 5$ — $x$ is less than or equal to $5$. Now $5$ is included.
- $x\ge 5$ — $x$ is greater than or equal to $5$, including $5$.

Translating words into symbols. At most means $\le$, at least means $\ge$, under or fewer than means $<$, and more than or exceeds means $>$. The pair that trips students is at most and at least, because both sound restrictive while pointing in opposite directions.

Worked example — everyday wording. A lift may carry at most 8 people. If $p$ is the number of people, this is $p\le 8$ — eight people is allowed, nine is not. Written with $<$ it would wrongly forbid a full lift.

**Why $\le$ and $\ge$ exist at all.** Many real conditions include their boundary. A square whose perimeter must not exceed $40$ cm may have a perimeter of exactly $40$ cm, and the difference between $<$ and $\le$ is precisely whether that one borderline case counts.

Reading an inequation backwards. The statement $5>x$ says the same thing as $x<5$; the wide end of the symbol always faces the larger quantity. Rewriting with the variable on the left, as $x<5$, makes the solution set far easier to read off, and it is worth doing every time.

The replacement set is the pool of numbers you are allowed to choose from, and the solution set is the part of that pool which satisfies the inequation.

The three pools used in this chapter are:

- Natural numbers $N=\{1,2,3,\dots \}$
- Whole numbers $W=\{0,1,2,3,\dots \}$
- Integers $Z=\{\dots ,-2,-1,0,1,2,\dots \}$

Worked example 1 — one inequation, three answers. Solve $x+3<8$, which reduces to $x<5$.

- If $x\in N$, the solution set is $\{1,2,3,4\}$ — four values.
- If $x\in W$, it is $\{0,1,2,3,4\}$ — five values, because zero is now allowed.
- If $x\in Z$, it is every integer below $5$, which never ends.

The same inequation, three different answers. This is why solve over the given replacement set is written into the question, and why an answer that ignores it cannot be marked right.

Worked example 2. Solve $3x+2\le 14$ where $x\in N$.

$$3x\le 12\Rightarrow x\le 4$$

The solution set is $\{1,2,3,4\}$. Note $4$ is included, because the symbol was $\le$.

Worked example 3. Solve $2x-1\ge 7$ where $x\in W$.

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

The solution set is $\{4,5,6,\dots \}$, which is infinite even though the pool started at zero.

Worked example 4 — a double inequation. Solve $-3\le x<4$ where $x\in Z$.

Read it as two conditions at once: $x$ is at least $-3$ and less than $4$. The solution set is

$$\{-3,-2,-1,0,1,2,3\}$$

Seven values. The left end is included and the right end is not, because the two symbols differ — and copying both symbols correctly is the whole difficulty of a double inequation.

Worked example 5 — the largest or smallest member. Find the largest integer satisfying $3x-5<10$.

$$3x<15\Rightarrow x<5$$

The largest integer below $5$ is $4$, not $5$. And for the smallest natural number satisfying $2x+1>9$: $2x>8$ gives $x>4$, so the answer is $5$.

Questions of this kind exist purely to test whether you noticed the difference between $<$ and $\le$.

The inequality sign reverses. Adding and subtracting are safe; multiplying and dividing are safe only by a positive number.

See it with plain numbers first. Start with a statement that is obviously true:

$$4>-2$$

Multiply both sides by $-1$. The left becomes $-4$, the right becomes $2$ — and $-4$ is smaller than $2$. So the true statement is now

$$-4<2$$

The sign had to turn round. Multiplying by a negative number reflects both values across zero, and reflection swaps which one is larger.

Worked example 1. Solve $-2x>6$.

Dividing both sides by $-2$ reverses the sign:

$$x<-3$$

Verify. Take $x=-4$, which satisfies $x<-3$: then $-2(-4)=8$, and $8>6$ is true. Now take $x=-2$, which does not: $-2(-2)=4$, and $4>6$ is false. The solution set is exactly right.

Worked example 2. Solve $5-2x\le 11$.

$$-2x\le 6\Rightarrow x\ge -3$$

Verify. At $x=-3$: $5+6=11$, and $11\le 11$ is true, so the boundary belongs in the set. At $x=-4$: $5+8=13$, and $13\le 11$ is false. Correct.

Worked example 3 — avoiding the flip altogether. Solve $7-3x<1$.

Instead of dividing by $-3$, move the variable to the side where its coefficient is positive:

$$7-1<3x\Rightarrow 6<3x\Rightarrow 2<x\Rightarrow x>2$$

Same answer, no reversal to remember. Collecting the variable on the side that makes its coefficient positive removes the commonest error in the chapter before it can happen.

Worked example 4 — with a fraction. Solve $\frac{x}{3}-1>2$.

$$\frac{x}{3}>3\Rightarrow x>9$$

Multiplying by $3$ is multiplying by a positive number, so the sign stays as it is.

The rules, gathered. You may add or subtract the same quantity on both sides freely. You may multiply or divide both sides by the same positive number freely. Multiply or divide by a negative number and you must reverse the sign. And you may never multiply by zero, which destroys the statement entirely.

Mark the boundary, choose a hollow or filled dot, then shade the side that satisfies the inequation.

The convention is fixed and worth stating exactly:

- A hollow (unfilled) circle at the boundary means the value is excluded — use it for $<$ and $>$.
- A filled (solid) circle means the value is included — use it for $\le$ and $\ge$.
- A thick line with an arrow shows the direction that continues without end.

Worked example 1. Show $x>2$ for $x\in Z$.

Put a hollow circle at $2$, then thicken the line to the right through $3$, $4$, $5$ with an arrow. The hollow circle is the entire message here: $2$ itself is not a solution.

Worked example 2. Show $x\ge -3$ for $x\in Z$.

A filled circle at $-3$, and the line thickened to the right with an arrow.

Worked example 3 — both ends. Show $-3\le x<4$ for $x\in Z$.

A filled circle at $-3$, a hollow circle at $4$, and the segment between them thickened. No arrows — the set is finite, running from $-3$ to $3$.

**When the replacement set is $N$ or $W$, draw dots and not a solid line.** For $x<5$ over natural numbers, only the four points $1$, $2$, $3$, $4$ are solutions, so four dots is the honest picture. A continuous shaded line would claim that $2.5$ is a solution too, and over $N$ it is not.

Worked example 4 — a word problem on the line. The perimeter of a square must be less than 40 cm. What can its side be?

With side $s$ cm, the perimeter is $4s$:

$$4s<40\Rightarrow s<10$$

But a side length must also be positive, so the full answer is $0<s<10$, shown with hollow circles at both $0$ and $10$ and the segment between them thickened.

That extra condition is the point of the example. The algebra gave $s<10$ and said nothing about the lower end, because algebra does not know that a square cannot have a side of $-3$ cm. A real-world inequation almost always carries a hidden condition of this kind — lengths and counts are positive, and a number of people must be a whole number as well.