⚙️ Function Kingdom · Functions

Systems of Inequalities

Graph a linear inequality in two variables as a half-plane with a solid or dashed boundary, test whether a point satisfies a system of inequalities, and write a system from the constraints of a situation.

In short

A linear inequality in two variables has a whole half-plane of solutions, bounded by the line you get when you replace the sign with $=$ .
The sign alone decides the boundary style: $<=$ and $>=$ give a solid line whose points count, while $<$ and $>$ give a dashed line whose points do not.
To find which side to shade, substitute one test point that is not on the boundary; the origin is the easiest choice unless the boundary runs through it.
A point is a solution of a system only when it makes every inequality in the system true, so each candidate has to be tested against all of them.
The corners of a bounded region are found by solving pairs of boundary equations, and a linear quantity always takes its greatest and least values at a corner.

An inequality in two variables shades half the plane

$y = 2x + 1$ is a line: a thin thread of points, each one making the two sides exactly equal.

Change the $=$ to $a <$ and something much bigger happens. $y < 2x + 1$ asks for every point whose y-value is smaller than the height of the line above that $x. (0, 0)$ qualifies, because $0 < 1$ . So does $(5, 3)$ , because $3 < 11$ . So do infinitely many others, and together they fill in a whole half-plane — everything on one side of the line.

That is the one idea this topic is built on. A linear equation in two variables draws a line; a linear inequality in two variables shades one side of that line.

The line itself is called the boundary. Find it by replacing the inequality sign with $=$ .
The shaded part is the solution region. Every point in it is a solution; every point outside it is not.
The region is unbounded: it runs off the edge of any grid you draw it on.

Testing a single point is all it takes to check membership. Put its x and its y into the inequality and ask whether the statement that comes out is true.

The boundary: solid or dashed, and which side

Graphing one inequality is three decisions, and none of them is hard on its own.

1. Draw the boundary. Swap the sign for $=$ and graph the line, exactly as you always have.

2. Solid or dashed? This depends only on the sign.

$<=$ and $>=$ include the boundary, so the line is drawn solid — its points are solutions.
$<$ and $>$ leave the boundary out, so the line is drawn dashed — its points are not solutions.

3. Which side? Pick any point that is not on the line, substitute it, and see what happens. If the statement comes out true, shade the side that point is on. If it comes out false, shade the other side.

The origin $(0, 0)$ is the test point to reach for first: putting zeros in is almost no work. There is one case where it will not do: if the boundary passes through the origin, then $(0, 0)$ is on the line and it cannot tell you anything about either side. Pick something else — $(1, 0)$ or $(0, 1)$ usually settles it.

In the form $y <$ mx $+ b$ you can read the side straight off: less than means below the line, greater than above it. In the form ax $+$ by $<= c$ that shortcut is not safe, because dividing by a negative b reverses the sign. Test a point instead.

A system is the overlap

A system of inequalities is two or more inequalities that have to hold at the same time. Its solution region is where all the shaded half-planes overlap — and only there.

Take $y >= 1$ together with $y <$ -x $+ 6$ . The first shades everything on or above the horizontal line $y = 1$ ; the second shades everything strictly below the slanted line $y =$ -x $+ 6$ . The overlap is the wedge caught between them.

Checking a point is now two checks, not one, and it has to pass both.

$(2, 3)$ : is $3 >= 1$ ? Yes. Is $3 < -2 + 6 = 4$ ? Yes. So $(2, 3)$ is a solution.
$(5, 4)$ : is $4 >= 1$ ? Yes. Is $4 < -5 + 6 = 1$ ? No. So $(5, 4)$ is not a solution, even though it passed the first test.

That second line is where most marks are lost. Passing one inequality feels like success, and it is only half the work. Test every inequality, every time, before you call a point a solution.

Overlaps come in shapes. Two half-planes whose boundaries cross always overlap in a wedge. Add more inequalities and the wedge can close up into a triangle or a quadrilateral — a bounded region with corners.

Writing a system from constraints

Real constraints turn into inequalities almost word for word, once you have named the unknowns.

Say a pack holds at most 12 things, an arrow costs 2 gold, a lantern costs 3 gold, the purse holds 30 gold, and you want at least 4 arrows. Let x be the number of arrows and y the number of lanterns:

"at most 12 things altogether" becomes $x + y <= 12$
"30 gold to spend" becomes $2x + 3y <= 30$
"at least 4 arrows" becomes $x >= 4$

Three phrases worth memorising: at most and no more than mean $<=$ ; at least and no fewer than mean $>=$ ; more than and under are the strict signs $>$ and $<$ .

Two traps sit in the coefficients. The number in front of x has to be the price of the thing x counts, not the other one — swapping 2 and 3 gives a different system with different solutions. And the two totals do different jobs: 12 counts objects, 30 counts gold, and they cannot change places.

Counting problems bring one more condition that is easy to leave out: you cannot pack a negative number of arrows, so $x >= 0$ and $y >= 0$ belong in the system too.

Corners, and where this is going

When a region is bounded, its corners are where two boundary lines cross. Finding one is a job you already know: replace both inequality signs with $=$ , and solve the pair of equations by substitution or elimination. The signs play no part in that step — they only decide which side of each line the region lies on.

Corners matter far more than they look. If you are trying to make something as large or as small as possible — the most gold, the least travel time — and the quantity you are measuring is linear, then its best value always turns up at a corner of the region. So instead of testing infinitely many points you test a handful.

That single fact is the whole of linear programming, a method used to plan deliveries, timetables, menus and factory schedules. A working method looks like this:

write the constraints as a system of inequalities
shade the region and find every corner
work out the quantity you care about at each corner
pick the best one

Next you will meet regions bounded by curves rather than lines, and quadratic inequalities, where the boundary is a parabola and the same three questions — boundary, strict or inclusive, which side — still apply.

Worked examples

Example 1

Is $(3, -2) a$ solution of $4x + 5y > 6$ ?

Substitute $x = 3$ and $y = -2$ into the left side: $4(3) + 5(-2)$ .
$4(3) = 12$ and $5(-2) = -10$ , so the left side is 12 - $10 = 2$ .
The inequality now reads $2 > 6$ , which is not true.
So $(3, -2)$ is not a solution. It lies on the other side of the boundary $4x + 5y = 6$ .

Example 2

Describe the graph of 3x - $2y <= 12$ : is the boundary solid or dashed, and is the origin shaded?

Boundary first: replace the sign with $=$ to get the line 3x - $2y = 12$ .
The sign is $<=$ , so the boundary is included and the line is drawn solid.
Test the origin, which is not on that line: 3(0) - $2(0) = 0$ , and $0 <= 12$ is true.
So the origin is inside the region, and the side of the line containing $(0, 0)$ is the side to shade.

Example 3

The region of the system $y <= 2x + 1$ and $y <=$ -x $+ 7$ has a corner. Find it.

A corner sits on both boundary lines, so replace both signs with $=: y = 2x + 1$ and $y =$ -x $+ 7$ .
Both equations give y, so set the right sides equal: $2x + 1 =$ -x $+ 7$ .
Add x to both sides and subtract $1: 3x = 6$ , so $x = 2$ .
Substitute back: $y = 2(2) + 1 = 5$ . Check in the other line: $-2 + 7 = 5$ as well.
The corner is $(2, 5)$ .

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Which of these points is a solution of $y > 2x$ - 5?

$(-5, 4)$
$(4, 3)$
$(2, -3)$
$(7, 3)$

Answer: $A. (-5, 4)$

At $(-5, 4)$ the right side is 2x - $5 = 2(-5)$ - $5 = -15$ , and $4 > -15$ is true.
So $(-5, 4)$ is a solution.
At $(4, 3)$ the right side is 2x - $5 = 2(4)$ - $5 = 3$ , and $3 > 3$ is not true.
The solution set is every point above the boundary $y = 2x$ - 5, with the boundary itself left out.

Problem 2

Difficulty 3 of 5

The boundary line $y = -3x + 4$ is drawn below. Everything below that line is shaded, and the points on the line itself are part of the shaded region. Which inequality does the picture show?

$y <= -3x + 4$
$y < -3x + 4$
$y <= 3x + 4$
$y >= -3x + 4$

Answer: $A. y <= -3x + 4$

The boundary is $y = -3x + 4$ .
The shading runs below the line, so every solution has a y-value less than the height of the line at that x.
The boundary is included, so the sign carries the line with it $(<=$ or $>=)$ .
Check with the shaded point: At $(3, -6)$ the right side is $-3x + 4 = -3(3) + 4 = -5$ , and $-6 <= -5$ is true.
The picture shows $y <= -3x + 4$ .

Problem 3

Difficulty 4 of 5

The region of 3x - $y < 1$ is shaded on a grid. Is the boundary line solid or dashed, and is the point $(0, 0)$ inside the shaded region?

solid, and $(0, 0)$ is in the shaded region
dashed, and $(0, 0)$ is not in the shaded region
solid, and $(0, 0)$ is not in the shaded region
dashed, and $(0, 0)$ is in the shaded region

Answer: D. dashed, and $(0, 0)$ is in the shaded region

The sign is $<$ , so the boundary 3x - $y = 1$ is left out, and the line is drawn dashed.
At $(0, 0)$ the left side is 3(0) - $(0) = 0$ , and $0 < 1$ is true.
So the line is dashed and $(0, 0)$ is in the shaded region.

Common mistakes

Shading the wrong side: reading $y > 2x + 1$ as the region below the line. Substituting one test point settles it in seconds.
Treating a strict inequality as inclusive: counting a point that sits exactly on a dashed boundary as a solution of $y < 2x + 1$ .
Testing a point with the sign reversed: working out both sides correctly and then checking $2 > 6$ as though it read $6 > 2$ .
Checking only one inequality of a system, and calling a point a solution because it passed the first line.
Using the origin as the test point when the boundary passes through it, so the test gives $0 <= 0$ and settles nothing.
Flipping "at most" into $>=$ when writing a system, or swapping the coefficients so the price of one item ends up in front of the wrong variable.

What you should be able to do

Decide whether a point is a solution of a linear inequality in two variables, or of a system.
Match a shaded graph to its inequality, choosing the boundary style and the side to shade.
Describe the solution region of a system of two linear inequalities.
Write a system of inequalities from the constraints of a word problem.

Where this fits in the curriculum

Common Core

HSA-REI.D.12
High school — Graph the solutions to a linear inequality in two variables as a half-plane, and the solution set of a system of linear inequalities as the intersection of half-planes.
HSA-CED.A.3
High school — Represent constraints by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modelling context.

Ontario

MTH1W.C4.2
Grade 9 de-streamed — Graph relations of the forms x = k, y = k, x + y = k, ax + by = k and their associated inequalities, and identify the points and regions they define.
MTH1W graphs the region of a single inequality; the intersection of two regions as a system is Grade 11 and beyond, so the tag covers the half-plane half of this skill.

SAT

Heart of Algebra
Systems of linear inequalities in two variables and their solution regions.