Two regions, and the only answers are the points that lie in both of them.
Which of the following ordered pairs (x, y) is a solution to the given system of inequalities?
y ≤ −2x + 6
y ≥ x − 3
At (1, 3): −2(1) + 6 = 4 and 3 ≤ 4 ✓; 1 − 3 = −2 and 3 ≥ −2 ✓. Both hold, so it is a solution. A satisfies only the first: −4 ≥ −3 is false. C satisfies only the second: at x = 4 the first needs y ≤ −2, and 2 is not. D fails both. A system needs every inequality at once, so one failure is enough to eliminate a choice — which is why you should test the cheaper inequality first.
1
The solution is the part that is shaded twice.
Each inequality on its own owns half the plane. The system owns only what the two halves have in common.
Where a point sits
What it means
shaded twice
satisfies both — a solution to the system
shaded once
satisfies one and fails the other — not a solution
not shaded
fails both
Both boundaries here are solid, so both lines belong to their own region — and the
corner where they cross, (3, 0), belongs to the solution too. On the SAT the
shading is usually drawn for you and the only thing you have to decide is which
patch is the doubly covered one.
2
"Which of these four points is a solution?" — substitute, and stop at the first failure.
This is how the SAT asks about a system nine times out of ten. You never have to draw anything.
Point
y ≤ −2x + 6
y ≥ x − 3
Verdict
(0, −4)
−4 ≤ 6 ✓
−4 ≥ −3 ✗
one only
(1, 3)
3 ≤ 4 ✓
3 ≥ −2 ✓
solution
(4, 2)
2 ≤ −2 ✗
2 ≥ 1 ✓
one only
(5, 0)
0 ≤ −4 ✗
0 ≥ 2 ✗
neither
Two of the four choices always satisfy exactly one
(0, −4) and (4, 2) are each in one hatching and each in the choices for a reason:
a student who checks one inequality, sees a tick and moves on will pick one of
them. Check both, every time — and start with whichever inequality is quicker to
evaluate, because a single ✗ ends the work on that choice.
3
A budget and a capacity are just two inequalities.
A food pantry's van holds at most 40 cases. Rice costs 12 dollars a case, beans cost 8 dollars a case, and the pantry may spend at most 384 dollars.
x = cases of rice, y = cases of beans
name the variables first
x + y ≤ 40
the capacity: how many cases
12x + 8y ≤ 384
the budget: how many dollars
x ≥ 0 and y ≥ 0
you cannot buy −3 cases
The two extra inequalities x ≥ 0 and y ≥ 0 close the region into a four-cornered
shape, with vertices at (0, 0), (32, 0), (16, 24) and (0, 40). Notice which
constraint stops you on each axis: 40 cases of rice would cost 480 dollars, so the
budget bites first there; 48 cases of beans would fit the budget, but not the van.
Both signs are ≤ here, so both boundaries are solid and every corner counts as a
real plan the pantry could buy. A word like "more than 40 cases is impossible"
would keep the ≤; only wording such as "fewer than 40" would make that edge
dashed and push the answer inside.
4
A corner is where two boundaries meet — so solve them as equations.
Turn both ≤ signs into =, solve the pair, and you have the corner's coordinates exactly.
x + y = 40 and 12x + 8y = 384
the two boundaries that cross
y = 40 − x
substitute into the second
12x + 8(40 − x) = 384
12x + 320 − 8x = 384
4x = 64, so x = 16
then y = 40 − 16 = 24
check: 12(16) + 8(24) = 192 + 192 = 384
and 16 + 24 = 40 ✓ both tight at once
the corner is (16, 24)
In words: 16 cases of rice and 24 of beans fill the van exactly and spend the budget
exactly. It is the only plan that uses up both limits, which is why "the greatest
number of cases the pantry can carry while spending all 384 dollars" lands there.
The greatest or least value always sits at a corner
When an item asks for the greatest possible value of y in a region, look at where
the boundaries meet, not where the shading looks tallest. Read the corner off the
picture if the grid gives it to you, and solve the two equations when it does not
— an intersection at (73,53) is never going to be readable from a graph.
5
On the boundary: solid keeps it, dashed throws it out.
A point may satisfy one inequality with room to spare and the other exactly. Whether that counts depends on one line style.
y ≤ −2x + 6 at (3, 0)0 ≤ 0exactly on it, and ≤ allows it ✓
y ≥ x − 3 at (3, 0)0 ≥ 0the corner is in the solution ✓
y < −2x + 6 at (3, 0)0 < 0 is falseone strict sign and the corner is gone ✗
y ≥ x − 3 at (0, −4)−4 ≥ −3 is falsemissing by 1 is missing ✗
Desmos takes a system as two lines
Type y ≤ −2x + 6 on one line and y ≥ x − 3 on
the next, and Desmos shades each region; the solution is the patch where the two
shadings sit on top of each other and look darkest. Add the four answer choices as
points, such as (1, 3), and the right one is visibly the only dot
inside the dark patch. For a corner, type both boundaries as equations and click
the intersection — it reports (3, 0) exactly.
✓
Check: every inequality, every time.
Run the winner through both, and confirm that each rejected choice really does fail one.
(1, 3) in both3 ≤ 4 ✓ · 3 ≥ −2 ✓the answer
(16, 24) in both40 ≤ 40 ✓ · 384 ≤ 384 ✓the corner of the pantry's region
(20, 20)40 ≤ 40 ✓ · 400 ≤ 384 ✗fits the van, breaks the budget
(0, 48)48 ≤ 40 ✗ · 384 ≤ 384 ✓fits the budget, breaks the van
Remember the flow
Both regions · Keep the overlap · Corners from the equations · Test every inequality
Satisfying one of the two is exactly what the wrong answers do.
Now you try
Practice problem
In the xy-plane, a point (x, y) satisfies all three of the given inequalities. What is the greatest possible value of y?
y ≤ −x + 9
y ≤ 3x + 1
y ≥ 0
The region lies below both slanted lines, so the highest point is where they cross: −x + 9 = 3x + 1 gives 4x = 8, x = 2, and then y = −2 + 9 = 7. Check the other line: 3(2) + 1 = 7 ✓. B is 2, the x of that corner rather than the y — the quantity the question did not ask for. D is 9, the highest point of the first line alone, at (0, 9), which fails y ≤ 3x + 1 because 9 ≤ 1 is false. A is 1, the y-intercept of the second line.
Hint: the tallest point of a region bounded above by two lines is
the point where those two lines meet. Set the two right-hand sides equal, solve for
x — and then remember that the question asked for y.
