Two half-planes, and the answer is the part they share.
Given the system
y ≥ 2x − 3
x + y < 4
Is (0, 0) a solution?
Test it in each line. y ≥ 2x − 3 becomes 0 ≥ −3 ✓, and x + y < 4 becomes 0 < 4 ✓. Both hold, so the origin is in the region. D drops the minus: 0 ≥ 3 would fail, but the right side is 2(0) − 3 = −3. Testing a point beats reading a shaded graph — and (0, 0) is the cheapest point to test.
Replace the inequality sign with = and plot that line first.
solid line
the boundary is included
dashed line
the boundary is not
y = 2x − 3
slope 2, y-intercept −3 — solid
x + y = 4
intercepts (4, 0) and (0, 4) — dashed
For a line like x + y = 4, the two intercepts are the fastest plot: set x = 0, then set y = 0. Two points and a ruler.
Pick any point off the line — (0, 0) if the line misses the origin — and see if it works.
It works when y is alone on the left, but 3x − y ≥ 6 catches people out. Testing a point never does — one substitution and the shading is settled.
| Region | Meaning |
|---|---|
| shaded twice | satisfies both — the solution |
| shaded once | satisfies one, fails the other |
| not shaded | fails both |
Shade lightly, or with strokes leaning different ways. A solution region you cannot see through is no use to you at the end.
(3, 3): is 3 ≥ 2(3) − 3?
3 ≥ 3 — true, just
(3, 3): is 3 + 3 < 4?
6 < 4 is false
so (3, 3) is not a solution
one failure is enough
A system means every inequality at once. (3, 3) sits exactly on the solid boundary, so the first one holds — and it still fails, because the second does not.
2x − 3 = 4 − x
where the boundaries meet
3x = 7, x = 73
then y = 4 − 73 = 53
corner at (73, 53)
not included — one boundary is dashed
You cannot make −4 chairs. Those two extra inequalities close the region into a polygon, and the best answer to a “maximum profit” question always sits at one of its corners — which is why the corners are worth finding.
Over the region x ≥ 0, y ≥ 0, x + y ≤ 5, y ≤ 3, what is the greatest value of 2x + 3y? List the corners, then evaluate at each one.
x + y = 5 meets y = 3 at x = 2
the corner you have to solve for
corners: (0, 0), (5, 0), (2, 3), (0, 3)
the others lie on the axes
| Corner | 2x + 3y |
|---|---|
| (0, 0) | 0 |
| (5, 0) | 10 |
| (2, 3) | 4 + 9 = 13 — the greatest |
| (0, 3) | 9 |
(0, 5) is where x + y = 5 meets the y-axis, but it breaks y ≤ 3, so it is outside the region. Every corner must satisfy every inequality — test it before you evaluate.
Shade 3x − y ≥ 6.
boundary 3x − y = 6: (2, 0) and (0, −6), solid
intercepts; ≥ means solid
(0, 0): 0 ≥ 6 is false
test point
shade the side away from the origin — below the line
same as y ≤ 3x − 6
Different: the y term is negative, so “≥” ends up shading below — solving for y flips the sign.
Shading above 3x − y = 6 puts (0, 0) in the answer, and 0 ≥ 6 is false. Test a point instead of reading the sign.
Shade y < 2x.
dashed line through (0, 0) and (1, 2)
< means dashed
(1, 0): 0 < 2 is true
(0, 0) is on the line — use another point
shade the side containing (1, 0), below the line
—
Different: when the boundary passes through the origin, (0, 0) tells you nothing; test a point clearly off the line.
Describe the region x ≤ 2 and y > −1.
x = 2: a vertical solid line
shade left of it
y = −1: a horizontal dashed line
shade above it
overlap: left of x = 2 and above y = −1, corner (2, −1) excluded
(0, 0) ✓ inside
Different: an inequality with only x or only y gives a vertical or horizontal boundary; the corner sits on the dashed line, so it is not in the region.
Whole-number points: how many points with whole-number coordinates satisfy x ≥ 0, y ≥ 0 and x + 2y ≤ 4? Go row by row in y.
y = 0: x ≤ 4 → x = 0, 1, 2, 3, 4
5 points
y = 1: x ≤ 2 → x = 0, 1, 2
3 points
y = 2: x ≤ 0 → x = 0
1 point
5 + 3 + 1 = 9
y = 3 would need x ≤ −2: none
Area: x ≥ 0, y ≥ 0 and 3x + 4y ≤ 24 is a right triangle cut off by the axes. Its legs are the intercepts: x = 8 when y = 0, and y = 6 when x = 0. Area = ½ × 8 × 6 = 24 square units.
No solution: y ≥ 2x + 3 and y ≤ 2x − 1 have parallel boundaries (both slope 2). A point would have to be on or above the upper line and on or below the lower one at once — impossible, so the system has no solution.
Expect “which point satisfies the system” (substitute — never trust the picture), “which system matches the graph” (solid or dashed, then test (0, 0)), the greatest value at a corner, a count of whole-number points, or the area of a triangular region from its intercepts.
Given the system
y < −x + 5
y ≥ 2x − 1
Is (1, 1) a solution?
1 < −1 + 5 = 4 ✓, and 1 ≥ 2(1) − 1 = 1 ✓. The second is an equality, and ≥ allows it, so the point sits exactly on that boundary and still counts. A rejects the boundary — a strict > would, but ≥ does not.
