One line, two sides, and a single test point that decides which side you keep.
The shaded region in the xy-plane shown represents all the solutions to which inequality?
The line crosses the y-axis at −3 and climbs 2 for every 1 across, so the boundary is y = 2x − 3. It is dashed, so the sign is strict, and the shading covers the origin, where 0 > 2(0) − 3 = −3 is true. The answer is y > 2x − 3. C reads the shading correctly but ignores the dashed line — a solid line would be needed for ≥. A and B shade the wrong side: (0, 0) is in the picture but fails y < −3.
1
Swap the sign for an equals sign and draw that line first.
The line style carries the sign: solid for ≤ and ≥, dashed for < and >.
≤ or ≥
solid line
the points on the line are solutions
< or >
dashed line
the points on the line are not
y > 2x − 3 → draw y = 2x − 3
slope 2, y-intercept −3, dashed
2x + 4y ≤ 8 → draw 2x + 4y = 8
intercepts (4, 0) and (0, 2), solid
For a line written in standard form, the two intercepts are the fastest plot: set
x = 0 and read y, then set y = 0 and read x. Two points and a ruler — no
rearranging into slope form at all.
2
Test one point. Never trust "above" or "below".
Pick any point that is not on the line — (0, 0) whenever the line misses the origin — and substitute. True means you shade that side; false means you shade the other one.
3x − y > 3, test (0, 0)
3(0) − 0 = 0
0 > 3 is false
so the origin is not a solution
shade the other side
check it: (3, 0) gives 9 > 3 ✓
A > sign does not mean "shade above"
Here the sign is > and the shading is below the line. Rearranged,
3x − y > 3 becomes −y > 3 − 3x, and dividing by −1 flips it to
y < 3x − 3. The "above for >" shortcut only works once y is alone on the
left with a positive coefficient — and the SAT writes standard form precisely
because that shortcut then fails. One test point never fails.
3
A solid boundary belongs to the region.
Same two moves: draw the line from its intercepts, then test the origin.
2x + 4y ≤ 8
≤, so the line is solid
x = 0 → 4y = 8 → y = 2
the point (0, 2)
y = 0 → 2x = 8 → x = 4
the point (4, 0)
test (0, 0): 0 ≤ 8 ✓
shade the origin's side
(4, 0) is a solution here: 2(4) + 4(0) = 8, and 8 ≤ 8 is true. On a dashed boundary
that same point would fail. The line style is the only thing that decides it.
4
Reading a drawn region backwards: line, then style, then side.
Three questions in a fixed order — what is the line, is it solid or dashed, and does a point you can test lie in the shading?
the line hits the y-axis at −3
so b = −3
from (0, −3) to (1, −1): up 2, across 1
so the slope is 2 and the line is y = 2x − 3
the line is drawn dashed
so the sign is < or >, never ≤ or ≥
(0, 0) is shaded: 0 and 2(0) − 3 = −3
0 > −3, so the sign is >
y > 2x − 3
Use a lattice point, not the picture's "feel"
Read the slope from two points the line passes through exactly, where the grid
crosses. Sketching the slope by eye is how 2 turns into 3, and a region drawn
with slope 3 instead of 2 covers a noticeably different part of the plane.
5
"Is (a, b) a solution?" is one substitution.
Put the numbers in and read whether the sentence is true. No graph required.
y > 2x − 3 at (4, 6)6 > 2(4) − 3 = 5true — a solution ✓
y > 2x − 3 at (4, 5)5 > 5 is falseon the dashed line, so it is out ✗
2x + 4y ≤ 8 at (4, 0)8 ≤ 8on the solid line, so it is in ✓
2x + 4y ≤ 8 at (2, 2)4 + 8 = 12, not ≤ 8the wrong side ✗
The point that lands exactly on the boundary is the SAT's favourite distractor.
Dashed excludes it; solid keeps it.
Desmos shades it for you
Type y > 2x − 3 into Desmos exactly as written — it accepts
inequalities — and the region appears with a dashed edge. Type
(4, 5) on the next line and you can see at a glance whether the
dot sits inside the shading or on its edge. For a "which of these four points"
item, entering all four points and one inequality answers the question in about
fifteen seconds.
✓
Check: one point inside, one point outside.
Two substitutions confirm both the side and the line style.
y > 2x − 3 at (0, 0)0 > −3shaded in the picture ✓
y > 2x − 3 at (3, 0)0 > 3 is falseunshaded in the picture ✓
3x − y > 3 at (3, 0)9 > 3the region below the line ✓
2x + 4y ≤ 8 at (−6, 4)−12 + 16 = 4 ≤ 8the far top-left corner is shaded ✓
Remember the flow
Line · Solid or dashed · Test a point · Shade that side
Going backwards, ask the same three questions in the same order.
Now you try
Practice problem
Which of the following ordered pairs (x, y) is a solution to the given inequality in the xy-plane?
4x − 3y < 12
At (1, −2): 4(1) − 3(−2) = 4 + 6 = 10, and 10 < 12 ✓. A and B are the two intercepts of the boundary line 4x − 3y = 12 — each gives exactly 12, and 12 < 12 is false, so a dashed boundary shuts them out. D gives 16 − 3 = 13, on the wrong side altogether. Watch the double negative in C: subtracting −2 adds 6.
Hint: substitute rather than graph. Two of these choices land
exactly on the boundary line, and the sign is strict — so "exactly 12" is not
good enough.
