One inequality, one half-plane
Replace the inequality sign by “$=$”. What you get is the boundary line. That line cuts the plane into two half-planes, and the solution set is one of them — always exactly one, never a strip and never a wedge.
- Boundary: solve $Ax+By=C$ and draw it.
- Style: $\le$ or $\ge$ gives a solid line (the boundary is included); $<$ or $>$ gives a dashed line (the boundary is excluded).
- Side: pick any test point not on the line — $(0,0)$ whenever the line misses the origin — and substitute. True means shade the side containing the test point; false means shade the other side.
Is $(3,-2)$ a solution of $\;4x-5y\ge 20$?
Substitute: $4(3)-5(-2)=12+10=22$. Since $22\ge 20$ is true, $(3,-2)$ is a solution. Note that the point does not have to lie on the line — it only has to sit on the correct side.
Một bất phương trình hai ẩn không có “nghiệm” theo nghĩa một cặp số. Nghiệm của nó là cả một miền chứa vô hạn điểm. Vì vậy mọi câu hỏi SAT về chủ đề này chỉ có ba kiểu: điểm này có nằm trong miền không, miền này ứng với bất phương trình nào, và hai miền cắt nhau ở đâu. Nhớ ba kiểu đó là đã đi được nửa đường.
Nét đứt: biên không thuộc miền. Nét liền: biên thuộc miền.
Rearranging, and the sign that flips
Test points work on any form, but reading a graph is far faster once the inequality is solved for $y$. Solving for $y$ means dividing by the coefficient of $y$, and dividing by a negative number reverses the inequality sign.
$y>mx+b$ shades above the line; $y<mx+b$ shades below it. There is no third possibility, so once the inequality is in this form the picture is decided.
Describe the graph of $\;5x-2y>8$.
Isolate $y$: $-2y>8-5x$. Divide by $-2$ and flip: $\;y<\dfrac{8-5x}{-2}=\dfrac52x-4$.
So the boundary is $y=\tfrac52x-4$, drawn dashed, and the shading is below it.
Check with the origin: $5(0)-2(0)=0$, and $0>8$ is false — the origin is not in the region. The origin does sit above the line $y=\tfrac52x-4$ (since $0>-4$), which agrees.
Chia cho hệ số âm của $y$ mà giữ nguyên chiều bất phương trình. Toàn bộ phần còn lại của bài làm sẽ đúng — và tô nhầm hẳn nửa mặt phẳng. Cách tự bảo hiểm rẻ nhất: giải xong, thử lại với $(0,0)$ trên dạng gốc chưa biến đổi.
Reading a region backwards
The graph is given, the inequality is not. This is the version the test likes, because the answer choices differ only by a sign or by a strict-versus-inclusive symbol.
- Find the boundary line from two visible points (intercepts are easiest).
- Solid line $\Rightarrow$ $\le$ or $\ge$; dashed line $\Rightarrow$ $<$ or $>$.
- Take any clearly shaded point, substitute into $Ax+By$, and compare with $C$. The direction you get is the direction you write.
A solid boundary line passes through $(0,3)$ and $(4,0)$, and the shaded region contains the origin. Write the inequality.
Slope $=\dfrac{0-3}{4-0}=-\dfrac34$, so the line is $y=-\tfrac34x+3$, i.e.\ $3x+4y=12$.
Test the origin in $3x+4y$: $3(0)+4(0)=0$, and $0<12$. The shaded side is therefore the “less than” side, and the line is solid, so the answer is $\boxed{3x+4y\le 12}$.
A solid boundary line passes through $(-2,0)$ and $(0,-5)$, and the origin is not in the shaded region. Write the inequality.
Slope $=\dfrac{-5-0}{0-(-2)}=-\dfrac52$, so $y=-\tfrac52x-5$, i.e.\ $5x+2y=-10$.
Check both given points: $5(-2)+2(0)=-10$ and $5(0)+2(-5)=-10$. Good.
Test the origin: $5(0)+2(0)=0$, and $0>-10$. Since the origin is excluded, the shaded side is the opposite one: $\boxed{5x+2y\le -10}$.
Bẫy ở dạng này nằm ở bước cuối. Nhiều học sinh tìm đúng đường thẳng, thử đúng gốc toạ độ, rồi viết dấu theo kết quả phép thử thay vì theo đề bài. Nếu đề nói gốc toạ độ không thuộc miền tô, dấu phải ngược với kết quả thử. Đọc lại câu đó một lần nữa trước khi khoanh.
Systems: the overlap and nothing else
For a system, a point must satisfy every inequality. Graphically the solution set is the intersection of the half-planes; a point in only one of them is not a solution.
Which of $(0,0)$, $(1,5)$, $(4,1)$, $(3,4)$ satisfies $\;y>-x+2\;$ and $\;y\le 3x-6$?
$(0,0)$: $0>2$ is false — out immediately.
$(1,5)$: $5>1$ is true, but $5\le 3(1)-6=-3$ is false — out.
$(4,1)$: $1>-4+2=-2$ true, and $1\le 3(4)-6=6$ true — in.
$(3,4)$: $4>-1$ true, but $4\le 3(3)-6=3$ is false — out.
The answer is $\boxed{(4,1)}$.
Test the cheapest inequality first. One false statement kills the point, so a single substitution often eliminates an answer choice — there is no need to check the second inequality at all.
The point $(a,2a)$ satisfies both $\;y>x+3\;$ and $\;y\le 3x-1$. Find all possible values of $a$.
First inequality: $2a>a+3$, so $a>3$.
Second inequality: $2a\le 3a-1$, so $1\le a$, that is $a\ge 1$.
Both must hold, so the intersection is $\boxed{a>3}$. The weaker condition $a\ge1$ contributes nothing; only the stronger one survives.
For what values of $k$ does the system $\;y\ge 2x+k\;$ and $\;y\le 2x-3\;$ have at least one solution?
The two boundary lines are parallel (slope $2$ each), so the region is a strip — and the strip is nonempty exactly when the lower boundary is not above the upper one. A point exists when $2x+k\le y\le 2x-3$ is possible, which needs $2x+k\le 2x-3$, i.e.\ $\boxed{k\le -3}$.
At $k=-3$ the strip collapses to the single line $y=2x-3$, which still counts. For $k>-3$ there is no solution at all.
Hai đường thẳng song song thì hệ có thể vô nghiệm — nhưng không phải lúc nào cũng vô nghiệm. Phải xem hai nửa mặt phẳng quay vào nhau hay quay lưng nhau. Quay vào nhau thì miền chung là một dải, quay lưng nhau thì rỗng. Vẽ nhanh hai mũi tên chỉ hướng tô là thấy ngay.
Miền chung của $y\le -x+8$, $y\le 3x$, $y\ge 0$ là tam giác với ba đỉnh $(0,0)$, $(8,0)$, $(2,6)$.
Vertices, areas, and extreme values
When the overlap is a bounded polygon, its corners are intersections of pairs of boundary lines. Every “largest” or “smallest” question about a linear expression on such a region is answered at a corner, so the whole task reduces to solving small systems.
The system $\;y\le -x+8$, $\;y\le 3x$, $\;y\ge 0\;$ has a triangular solution region. Find its area.
Corner 1 — $y=0$ with $y=3x$: gives $(0,0)$.
Corner 2 — $y=0$ with $y=-x+8$: gives $(8,0)$.
Corner 3 — $y=3x$ with $y=-x+8$: $3x=-x+8$, so $4x=8$, $x=2$, $y=6$, giving $(2,6)$.
The base along the $x$-axis runs from $x=0$ to $x=8$, length $8$; the height is the $y$-coordinate of the top vertex, $6$. Area $=\tfrac12(8)(6)=\boxed{24}$.
For points satisfying $\;y\le -2x+12$, $\;y\le x+3$, $\;x\ge 0$, $\;y\ge 0$, what is the greatest possible value of $x+y$?
Corners: $(0,0)$; $(0,3)$ from $x=0$ with $y=x+3$; $(6,0)$ from $y=0$ with $y=-2x+12$; and $x+3=-2x+12$ gives $x=3$, $y=6$, so $(3,6)$.
Evaluate $x+y$ at each: $0$, $3$, $6$, $9$. The greatest value is $\boxed{9}$, at $(3,6)$.
Đừng thử “điểm bất kỳ trông có vẻ xa” rồi kết luận. Giá trị lớn nhất của một biểu thức bậc nhất trên miền đa giác luôn rơi vào một đỉnh. Vì vậy quy trình là: tìm hết đỉnh, thay vào, so sánh. Ba phép giải hệ hai ẩn nhanh hơn mọi cách mò.
Turning words into constraints
Context problems give two limits and ask for the system, for an interpretation, or for a largest feasible quantity. Two habits keep this clean: name the variables with units, and match each English phrase to exactly one symbol.
“no more than”, “at most”, “a maximum of” $\rightarrow \le$. “at least”, “no fewer than”, “a minimum of” $\rightarrow \ge$. “more than”, “exceeds” $\rightarrow >$. “fewer than” $\rightarrow <$.
A caterer buys $x$ trays of chicken at $\$18$ each and $y$ trays of pasta at $\$12$ each. She spends no more than $\$300$ and needs at least $20$ trays in total. Write the system, then find the greatest number of chicken trays she could buy.
Cost: $18x+12y\le 300$. Count: $x+y\ge 20$.
To push $x$ as high as possible, make $y$ as small as the count allows: $y=20-x$. Then $18x+12(20-x)\le 300$ gives $18x+240-12x\le 300$, so $6x\le 60$ and $x\le 10$.
The greatest number of chicken trays is $\boxed{10}$, with $y=10$: cost $180+120=300$, exactly the budget, and $20$ trays exactly.
Đề cho hai ràng buộc thì đáp án gần như luôn nằm ở chỗ cả hai cùng chặt. Bỏ quên ràng buộc thứ hai — ví dụ chỉ dùng $18x\le 300$ — sẽ ra $x\le 16$, một số có mặt trong đáp án nhiễu. Viết đủ hệ trước, tối ưu sau.