The solution is a region, not a point.
Every point in the shaded half makes the inequality true.
An equation in two variables gives a line. An inequality gives everything on one side of that line — infinitely many solutions, all at once.
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Linear Inequalities in Two Variables
Draw the boundary line, then shade the half that works.
Test the point
Is (1, 2) a solution of 3x − y < 5?
Substitute: 3(1) − 2 = 3 − 2 = 1, and 1 < 5 is true, so (1, 2) is a solution. B and C both compute 3x − y as 5, which comes from adding the y instead of subtracting it. The whole method rests on substituting carefully — the rest is just shading.
Solid or broken depends on the symbol.
| Symbol | Boundary line | Because |
|---|---|---|
| < or > | broken (dashed) | points on the line do not count |
| ≤ or ≥ | solid | points on the line do count |
It is the same idea as the open and closed circles on a one-variable number line, stretched into two dimensions.
The method, in four steps.
y ≥ 2x − 1
the inequality
draw y = 2x − 1
solid, because of the ≥
test (0, 0): 0 ≥ −1 ✓
true
shade the side containing (0, 0)
done
Use (0, 0) as the test point whenever the line does not pass through it — the arithmetic is as easy as it gets. If the line does go through the origin, pick (1, 0) or (0, 1) instead.
Guessing the side by the symbol is risky.
y > 2x − 1
y isolated, > means above
shade up ✓
−y > 2x − 1
y is negative here
"above" is wrong ✗
Either way
test a point
always right ✓
The shortcut only works with y alone on the left
"Greater than means shade above" is true for y > mx + b and false the moment a negative coefficient appears. Testing a point costs ten seconds and never lies.
Where these come from: constraints.
Pens cost ₱20, notebooks ₱50
you have ₱500
20p + 50n ≤ 500
what you can afford
p ≥ 0 and n ≥ 0
you cannot buy negative pens
The shaded region is every affordable combination. Only whole-number points inside it make sense here, which is a good reminder that the maths describes the situation rather than replacing it.
Variations: a point on the line, a vertical boundary, a line through the origin.
The three cases where the four-step method needs a second thought.
Is (4, 3) a solution of x + 2y ≥ 10? Of x + 2y > 10?
4 + 2(3) = 10
the point is ON the boundary
10 ≥ 10 is true
yes for ≥: solid line, it counts
10 > 10 is false
no for >: broken line, it does not
A point that makes both sides equal sits on the line, so the verdict depends only on whether the symbol includes "equal".
Graph x < −2.
boundary x = −2
a vertical line, broken
test (0, 0): 0 < −2 is false
the origin is out
shade left of the line
every x smaller than −2
With only one letter, the boundary is vertical (x = …) or horizontal (y = …). The test point still decides the side.
Graph y ≤ −3x.
boundary y = −3x, solid
it passes through (0, 0)
test (1, 0): 0 ≤ −3 is false
(1, 0) is out
shade the side away from (1, 0)
below and to the left of the line
(0, 0) sits on this line, so it cannot tell the sides apart. Any point not on the line will do; (1, 0) keeps the arithmetic easy.
On the exam
Items ask whether a point is a solution, whether the boundary is solid or broken, which side is shaded, or which inequality matches a budget or a capacity ("at most" is ≤, "at least" is ≥). The fastest route: substitute carefully, let the symbol decide the line and a test point decide the side. For a 10-second check, test (0, 0) — or any point clearly inside your shading — in the original inequality.
Now you try
Practice problem
Describe the graph
How should the graph of 2x + y > 6 be drawn?
The symbol is >, not ≥, so the boundary is broken — points on the line are not solutions. Testing (0, 0): 2(0) + 0 = 0, and 0 > 6 is false, so the origin is not in the region and the shading goes on the other side. That gives a broken line shaded away from the origin.
Hint: two separate decisions. The symbol decides solid or broken; the test point decides which side.