Inequalities: solving, number lines and graph regions

An inequality compares quantities that need not be equal. Its solution is usually a set of values rather than one value.

$$x<4,qquad y\geq -2,qquad 1<3x+4\leq 10$$

occur throughout A-level Mathematics in domains, ranges, optimisation, calculus, probability and hypothesis tests. The algebra often resembles solving an equation, but order must be preserved. In particular, multiplying or dividing by a negative number reverses the inequality sign.

Prerequisites

You should be able to:

• order positive and negative numbers;
• simplify algebraic expressions;
• solve linear and quadratic equations;
• plot straight lines and quadratic graphs;
• substitute a test value into an expression.

Review algebraic manipulation and factorisation, linear equations, quadratic equations or straight-line graphs if needed.

Reading inequality notation

Symbol	Meaning	Example
$<$	less than	$x<3$ excludes $3$
$>$	greater than	$x>3$ excludes $3$
$\leq$	less than or equal to	$x\leq3$ includes $3$
$\geq$	greater than or equal to	$x\geq3$ includes $3$
$\ne$	not equal to	$x\ne3$ excludes only $3$

The pointed end faces the smaller quantity. Thus $-5<-2$, even though $5>2$, because $-5$ lies further left on the number line.

The statements $x<7$ and $7>x$ mean exactly the same thing. If the sides are exchanged, the sign must turn around.

Number-line notation

Use an open circle for an excluded endpoint and a filled circle for an included endpoint. Shade every permitted value:

• $x<2$: open circle at $2$, shade left;
• $x\geq-1$: filled circle at $-1$, shade right;
• $-3<x\leq4$: open at $-3$, filled at $4$, shade between them.

An arrow means the set continues without bound. Infinity is not an endpoint that can be included.

Solving linear inequalities

You may add or subtract the same expression on both sides without changing the sign. You may also multiply or divide both sides by the same positive number.

Worked example 1: a one-sided inequality

Solve $5x-7<18$.

$$\begin{aligned} 5x-7&<18\\ 5x&<25\\ x&<5. \end{aligned}$$

The answer contains every real number below $5$, not merely the integers $4,3,2,\ldots$.

Why a negative multiplier reverses the sign

Start with a true statement:

$2<5.$

Multiplying by $-1$ gives $-2$ and $-5$. Their order is

$-2>-5.$

Reflection in zero swaps left and right on the number line. Therefore

$a<b\quad\Longrightarrow\quad -a>-b.$

Worked example 2: divide by a negative number

Solve $7-3x\geq19$.

$$\begin{aligned} 7-3x&\geq19\\ -3x&\geq12\\ x&\leq-4. \end{aligned}$$

The final sign reverses because both sides are divided by $-3$.

Check one value from each side. For $x=-5$,

$7-3(-5)=22\geq19,$

so an included value works. For $x=0$, $7\not\geq19$, so a value outside the solution does not work.

Worked example 3: variables on both sides

Solve $4(2x-1)>5x+11$.

$$\begin{aligned} 8x-4&>5x+11\\ 3x&>15\\ x&>5. \end{aligned}$$

Expanding and collecting first makes the final operation clear.

Misconception: turn the sign whenever a term moves

In

$-3x+4<10\quad\Longrightarrow\quad -3x<6,$

the sign does not reverse when $4$ is subtracted from both sides. It reverses only at the next step, when division by $-3$ gives $x>-2$.

Self-check 1

Solve:

1. $3x+8\leq20$;
2. $5-2x<11$;
3. $7x-3\geq4x+12$.

Answers

1. $3x\leq12$, so $x\leq4$.
2. $-2x<6$, so $x>-3$.
3. $3x\geq15$, so $x\geq5$.

Compound inequalities

A compound inequality such as

$-2<x\leq6$

means both $x>-2$ and $x\leq6$. It describes their overlap.

Perform the same operation on all three parts. If multiplying or dividing by a negative number, reverse both signs and then write the values in increasing order.

Worked example 4: operate on three parts

Solve $-5<2x+1\leq9$.

Subtract $1$ throughout:

$-6<2x\leq8.$

Divide throughout by $2$:

$\boxed{-3<x\leq4}.$

Worked example 5: negative coefficient

Solve $-8\leq2-5x<17$.

Subtract $2$ throughout:

$-10\leq-5x<15.$

Divide by $-5$ and reverse both signs:

$2\geq x>-3.$

Writing the interval from smaller to larger gives

$\boxed{-3<x\leq2}.$

Combining separate conditions

Suppose $x>-4$ and $x\leq3$. Both must hold, so

$-4<x\leq3.$

By contrast, $x<-2$ or $x\geq5$ describes two separate rays. Do not join these into one interval, because values between $-2$ and $5$ are excluded.

Integer solutions

If a question asks for integer solutions, solve over the real numbers first and then select integers.

Worked example 6: list the integers

Find the integer values satisfying $-4<3x+2\leq11$.

$$\begin{aligned} -4&<3x+2\leq11\\ -6&<3x\leq9\\ -2&<x\leq3. \end{aligned}$$

The integers in this interval are

$\boxed{x\in\{-1,0,1,2,3\}}.$

Notice that $-2$ is excluded and $3$ is included.

Quadratic inequalities

Solving $f(x)>0$ means finding where the graph $y=f(x)$ is above the $x$-axis. Solving $f(x)<0$ means finding where it is below. The roots divide the number line into intervals on which the sign cannot change.

A reliable method is:

1. move everything to one side;
2. factorise or find the roots;
3. mark the roots in order;
4. determine the sign in each interval;
5. include roots only for $\leq$ or $\geq$.

Worked example 7: an upward quadratic

Solve $x^2-x-6<0$.

Factorise:

$x^2-x-6=(x-3)(x+2).$

The critical values are $x=-2$ and $x=3$. Test one value in each interval:

Interval	Test value	Sign of $(x-3)(x+2)$
$x<-2$	$x=-3$	positive
$-2<x<3$	$x=0$	negative
$x>3$	$x=4$	positive

We need the negative interval, so

$\boxed{-2<x<3}.$

The roots are excluded because the inequality is strict.

Worked example 8: two outside intervals

Solve $2x^2+x\geq6$.

Move all terms to the left and factorise:

$$\begin{aligned} 2x^2+x-6&\geq0\\ (2x-3)(x+2)&\geq0. \end{aligned}$$

The roots are $x=-2$ and $x=\frac32$. The product is non-negative outside the roots, including the roots:

$\boxed{x\leq-2\quad\text{or}\quad x\geq\frac32}.$

Do not write $-2\leq x\geq\frac32$. That notation does not express two separate intervals.

Worked example 9: a downward quadratic

Solve $-x^2+5x+6>0$.

Factorise:

$-x^2+5x+6=-(x-6)(x+1).$

Its roots are $-1$ and $6$. Testing $x=0$ gives $6>0$, so the required interval is between the roots:

$\boxed{-1<x<6}.$

Do not memorise that positive always means outside or inside. That depends on the sign of the $x^2$ coefficient. A sign table or sketch settles it.

Self-check 2

Solve:

1. $x^2+3x-10\leq0$;
2. $x^2-9>0$;
3. $x^2+4x+5<0$.

Answers

1. $(x+5)(x-2)\leq0$, so $-5\leq x\leq2$.
2. $(x-3)(x+3)>0$, so $x<-3$ or $x>3$.
3. $x^2+4x+5=(x+2)^2+1$, which is always positive. There are no real solutions.

Linear inequalities in two variables

An inequality such as

$y<2x+1$

describes a region of the coordinate plane. The line $y=2x+1$ is its boundary.

• Draw a solid boundary for $\leq$ or $\geq$, because points on the line are included.
• Draw a dashed boundary for $<$ or $>$, because points on the line are excluded.
• Test a point not on the boundary to decide which side satisfies the inequality.

Worked example 10: choose the correct half-plane

Represent $y\geq2x-3$ graphically.

First draw the solid line $y=2x-3$. Test $(0,0)$:

$0\geq2(0)-3,$

which is true. Therefore shade the side containing the origin. Equivalently, shade on or above the line.

The phrase “above the line” is safe when $y$ has been isolated. For an inequality such as $2x+3y<6$, testing a point is less error-prone than relying on appearance.

Worked example 11: a vertical boundary

Represent $x<-2$.

The boundary is the dashed vertical line $x=-2$. Shade to its left, where all points have $x$-coordinate below $-2$. The $y$-coordinate is unrestricted.

Regions satisfying several inequalities

For simultaneous inequalities, draw every boundary and retain only the overlap of all the required half-planes.

Worked example 12: describe a finite feasible region

Consider

$$x\geq0,qquad y\geq0,qquad x+y\leq6,qquad y\leq2x.$$

All boundaries are solid. The first two inequalities restrict the region to the first quadrant. The condition $x+y\leq6$ selects the side of $x+y=6$ containing $(0,0)$. The condition $y\leq2x$ selects points on or below $y=2x$.

Find vertices by intersecting boundary lines:

• $y=0$ and $y=2x$ meet at $(0,0)$;
• $y=0$ and $x+y=6$ meet at $(6,0)$;
• $y=2x$ and $x+y=6$ give $3x=6$, so $(x,y)=(2,4)$.

The feasible region is the triangle with vertices

$\boxed{(0,0),\ (6,0),\ (2,4)}.$

At A-level, objective functions in linear programming attain extrema at vertices of such regions, so accurate boundaries matter.

Misconception: shade the labelled side without testing

For $x+y>4$, the origin gives $0>4$, which is false. The solution is therefore the side not containing the origin. If the origin lies on a boundary, choose another simple point such as $(1,0)$.

Self-check 3

1. Should the boundary for $3x-y<7$ be solid or dashed?
2. Does $(2,1)$ satisfy $x+2y\leq5$?
3. Which side of $x+y=3$ represents $x+y\geq3$: the side containing $(0,0)$ or the opposite side?

Answers

1. Dashed, because equality is excluded.
2. Yes, because $2+2(1)=4\leq5$.
3. The opposite side, because $0+0\geq3$ is false. The boundary itself is included.

Final checklist

Before accepting an answer, ask:

• Did I reverse the sign only when multiplying or dividing by a negative number?
• Did I preserve both signs in a compound inequality?
• Are strict endpoints excluded and non-strict endpoints included?
• For a quadratic, did I select intervals rather than just finding roots?
• For a graph region, is the boundary solid or dashed as required?
• Have I tested a value or point to confirm the chosen interval or side?

Next, connect these skills to graphs of common functions, function notation and coordinate geometry. Inequalities also become essential when determining domains and ranges and when interpreting solutions in calculus.