2.8 Inequalities

Learning objectives

2.8A Inequality symbols and compound inequalitiesunderstand and use the symbols >, <, ≥ and ≤, including double-ended inequalities 2.8B Open and closed intervalsunderstand and use the convention for open and closed intervals on a number line 2.8C Solving linear inequalitiessolve simple linear inequalities in one variable and represent the solution set on a number line 2.8D Linear inequalities on Cartesian graphsRepresent simple linear inequalities on rectangular Cartesian graphs.2.8E Regions defined by linear inequalitiesidentify regions on rectangular Cartesian graphs defined by simple linear inequalities Conventions for the inclusion of boundaries are not required

An inequality describes an ordered set of possible values rather than one equality. The symbol points toward the smaller expression, while the bar in $\le$ or $\ge$ includes equality.

Symbol	Meaning	Example
$x>3$	greater than 3	3 excluded
$x\ge3$	at least 3	3 included
$x<7$	less than 7	7 excluded
$x\le7$	at most 7	7 included
$a<x\le b$	between $a$ and $b$	$a$ excluded, $b$ included

If only integer values are requested, list integers satisfying both ends. For $-3.4<n\le2$ , the values are $-3,-2,-1,0,1,2$ .

Equivalent statements can reverse order and symbol together: $x>3$ means $3<x$ . Read each relation from left to right to check meaning.

Do not infer that $<$ means ‘left’ without reading the variable position. Also distinguish ‘at most’ ( $\le$ ) from ‘less than’ ( $<$ ).

A number line shows a set by marking its boundary values and drawing the interval or ray containing all permitted values.

Algebra	Endpoint	Direction or segment
$x<a$	open at $a$	left
$x\le a$	closed at $a$	left
$x>a$	open at $a$	right
$x\ge a$	closed at $a$	right
$a<x\le b$	open at $a$ , closed at $b$	join between endpoints

To represent $-2<y\le5$ , mark an open endpoint at $-2$ , a closed endpoint at $5$ , and join the interval between them.

When reading a diagram, identify each boundary value, whether its marker is open or closed, and whether the set extends left, right or between endpoints before writing symbols.

An open marker excludes only its endpoint, not nearby values. A ray needs an arrow to show that the solution continues without bound.

Solve a linear inequality using the same balancing operations as an equation, except that multiplying or dividing by a negative number reverses the inequality symbol.

Operation on every part	Symbol action
add or subtract any value	keep direction
multiply or divide by a positive value	keep direction
multiply or divide by a negative value	reverse $<\leftrightarrow>$ and $\le\leftrightarrow\ge$
double-ended inequality	apply the operation to all three parts

Solve $7-3t<2t+15$ : $-5t<8$ . Dividing by $-5$ reverses the sign, so $t>-8/5$ .

For $-5\le2p+3<13$ , subtract 3 from every part and divide every part by 2: $-4\le p<5$ .

After solving, use an open or closed endpoint and shade the correct direction or interval. Test a simple value if the direction is uncertain.

A sign reverses because multiplying by a negative reverses order, not because a term merely ‘moves sides’. Never change the sign during addition or subtraction alone.

A linear inequality in $x$ and $y$ describes a half-plane. Its related equation is the boundary line; one side of that line satisfies the inequality.

Step	Graph action
boundary	replace the inequality sign by $=$
draw	plot the straight line using intercepts or gradient
choose side	test a point not on the line, often $(0,0)$
shade	shade the side whose test point satisfies the inequality
combine	retain only points satisfying every inequality

For $x\le6$ , draw the vertical line $x=6$ and select its left side. For $y\ge2$ , draw the horizontal line $y=2$ and select above it.

For $y\le x+1$ , the boundary is $y=x+1$ . Since $(0,0)$ satisfies $0\le1$ , shade the side containing the origin.

Never decide the side from the visual slope alone. A boundary test point must be substituted into the original inequality.

To identify a shaded region, write one inequality for each boundary line and choose the direction that includes an interior point of the region.

Boundary appearance	Equation form	Side test
vertical through $a$	$x=a$	compare interior $x$ with $a$
horizontal through $b$	$y=b$	compare interior $y$ with $b$
sloping line	derive $y=mx+c$ or $ax+by=c$	substitute one interior point

A region bounded by $x=-1$ , $x+y=4$ and $y=x/3-2$ that lies right of the vertical, below the descending line and above the rising line is $x\ge-1$ , $x+y\le4$ , $y\ge x/3-2$ .

Every point in the region must satisfy all inequalities simultaneously. Check one interior point against the complete list and verify that each boundary contributes an edge.

For this Foundation objective, syllabus conventions for inclusion of Cartesian boundaries are not required; the essential work is selecting the correct side of every line.

Do not infer the inequality direction from whether shading looks ‘above’ on a rotated or rearranged equation. Test a coordinate inside the labelled region.

A quadratic inequality asks where a quadratic expression is positive, negative or zero. First find its real roots, then determine the sign on the intervals separated by those roots.

Step	Action
standardise	move all terms to one side
roots	solve the associated equation $f(x)=0$
order	place critical values on a number line
sign	use factor signs, a test value, or parabola orientation
endpoints	include roots for $\le$ or $\ge$ ; exclude for $<$ or $>$

For $5y^2-17y\le40$ , factor $5y^2-17y-40=(5y+8)(y-5)$ . The upward-opening quadratic is non-positive between its roots, so $-8/5\le y\le5$ .

For an upward-opening quadratic, $f(x)>0$ usually lies outside two distinct roots and $f(x)<0$ between them; verify rather than memorise if the leading coefficient is negative.

In a contextual problem, intersect the algebraic solution with restrictions such as positive lengths or a non-zero denominator.

Solving only $f(x)=0$ gives boundary values, not the inequality solution. The required answer is one or more intervals with correct endpoint inclusion.

Harder Cartesian regions combine several oblique, vertical or horizontal constraints. Each boundary contributes a half-plane, and the required region is their intersection.

Stage	Control action
normalise	rewrite boundaries in a form that is easy to plot and compare
draw	use exact intercepts or two verified points per line
test	choose a point away from every boundary for each inequality
intersect	retain only the overlap of all allowed half-planes
audit	check every vertex and one interior point against every constraint

For $x\le4$ , $y\le2x+1$ and $5x+2y\le20$ , draw $x=4$ , $y=2x+1$ and $5x+2y=20$ , test the correct side of each, then keep only their common overlap.

Candidate vertices come from pairwise boundary intersections. A visible intersection is part of the feasible region only if it satisfies all remaining inequalities.

A feasible region may be unbounded or empty. Do not force a closed polygon when the half-planes do not create one.

Testing one point against only one line cannot establish the final region. The same point must satisfy the full system of constraints.

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—Pearson Edexcel International GCSE Mathematics (Specification A) (4MA1)Read

2.8 Inequalities

Learning objectives

Read and write inequalities precisely

Use open and closed number-line endpoints

Solve linear inequalities and graph the solution

Represent linear inequalities on Cartesian graphs

Identify inequalities defining a Cartesian region

Solve quadratic inequalities

Identify harder regions from linear inequalities

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