You're staring at a coordinate plane with four different shaded regions. Each one looks plausible. Day to day, one has a solid line, another dashed. The shading goes up on one, down on another. And the question asks: which graph represents the system of inequalities?
Sound familiar? This is the moment where most students either guess or freeze.
Here's the thing — graphing systems of inequalities isn't about memorizing rules. In practice, it's about understanding what those lines and shaded regions actually mean*. Once you see the logic, the right graph jumps out.
What Is a System of Inequalities
A system of inequalities is just two or more inequalities that work together. Instead of finding a single point that satisfies one equation, you're looking for all the points that satisfy every* inequality in the system at the same time.
Think of it like this: each inequality draws a boundary line and says "the answer lives on this side." When you have multiple inequalities, the solution is where all those "sides" overlap.
The pieces you're working with
Every inequality in a system gives you three pieces of information:
The boundary line — this comes from replacing the inequality symbol with an equals sign. y > 2x + 1* becomes y = 2x + 1*. That line is the border between "works" and "doesn't work."
Solid or dashed — solid means the line itself is included (≤ or ≥). Dashed means it's not (< or >). This trips people up constantly. If the inequality says y ≥, the line is part of the solution. If it says y >, the line is just a boundary — points on the line don't count.
Which side to shade — this is the "test a point" part. Pick any point not on the line (0,0) is easiest if it's not on the line. Plug it into the original inequality. If it makes the statement true, shade that side. If false, shade the other side.
That's it. Still, three decisions per inequality. The system's solution is where the shading overlaps.
Why This Matters (And Where It Shows Up)
You might wonder when you'll ever use this outside of a math test. Fair question.
Linear programming — businesses use systems of inequalities to maximize profit or minimize cost under constraints. A factory can only produce so many units. Raw materials are limited. Labor hours are capped. Each constraint is an inequality. The feasible region — that overlapping shaded area — shows every possible production combination. The optimal answer lives at a corner of that region.
Budgeting with multiple constraints — you're planning a party. You need at least 50 drinks. You have $200. Soda costs $2, juice costs $3. You want at least twice as much soda as juice. That's a system of inequalities. The graph shows every combination that works.
Engineering tolerances — a part's dimensions must fall within certain ranges simultaneously. Length between 9.9 and 10.1 mm. Width between 4.8 and 5.2 mm. The acceptable region is a rectangle on a graph — a system of four inequalities.
The skill isn't "graphing for graphing's sake." It's visualizing constraints. That's a muscle you'll use whether you realize it or not.
How to Graph a System of Inequalities — Step by Step
Let's walk through a concrete example. Say the system is:
y ≥ 2x − 3*
y < −x + 4*
Step 1: Graph the first boundary line
Rewrite as y = 2x − 3*. Slope is 2, y-intercept is −3. Plot the intercept at (0, −3). Also, use the slope: up 2, right 1. Draw the line.
Solid or dashed? The inequality is ≥, so solid line. Points on the line count.
Step 2: Shade for the first inequality
Test (0,0) — it's not on the line. Plug into y ≥ 2x − 3*:
0 ≥ 2(0) − 3
0 ≥ −3
True. So shade the side containing (0,0). That's the region above* the line (since y is greater than the expression).
Step 3: Graph the second boundary line
y = −x + 4*. Slope −1, y-intercept 4. In real terms, plot (0,4). Practically speaking, down 1, right 1. Draw the line.
Solid or dashed? The inequality is <, so dashed line. Points on the line do not count.
Step 4: Shade for the second inequality
Test (0,0) again — not on this line either. Plug into y < −x + 4*:
0 < −(0) + 4
0 < 4
True. Shade the side containing (0,0). That's the region below* this line. No workaround needed.
Step 5: Find the overlap
The solution to the system is where the two shaded regions intersect. Not the union — the intersection*. Only points that satisfy both* inequalities.
If you did this on paper, you'd see a wedge-shaped region bounded by a solid line on one side, a dashed line on the other. That wedge — including the solid boundary, excluding the dashed one — is your answer.
What if the test point is on the line?
Happens more than you'd think. Because of that, if (0,0) lies on the boundary, pick any other point. (1,0), (0,1), (−1,0) — whatever's easy. In real terms, just don't pick a point on the line. It'll give you 0 = 0 or something equally useless.
What if the inequalities are in standard form?
2x + 3y ≤ 12
x − y > 2*
Same process. Find intercepts for the boundary lines. For 2x + 3y = 12, x-intercept is (6,0), y-intercept is (0,4). Connect them. Solid line (≤). Test (0,0): 0 ≤ 12, true. Shade toward origin.
For x − y = 2*, intercepts (2,0) and (0,−2). Dashed line (>). Think about it: test (0,0): 0 > 2? Also, false. Shade away* from origin. Most people skip this — try not to.
Overlap is the solution.
Common Mistakes (And How to Avoid Them)
I've graded hundreds of these. The same errors show up every time.
Mistake 1: Confusing "greater than" with "shade up"
y > mx + b* means shade above* the line. y < mx + b* means shade below*. But this only works when the inequality is solved for y and the coefficient of y is positive.
If you have −y > 2x − 3, you can't just shade up. Which means divide by −1 (flipping the inequality): y < −2x + 3*. Now shade down.
Fix: Always solve for y first. Or just test a point. Testing a point never lies. Practical, not theoretical.
Mistake 2: Forgetting to flip the inequality when multiplying/dividing by a negative
−2x + 4y ≤ 8
Subtract 2x: 4y ≤ 2x + 8
Divide by 4: y ≤ ½x + 2* — wait, that's fine, 4 is positive.
Continue exploring with our guides on books to read for ap lit and passive transport goes against the gradient. true or false.
But *−4y ≤ 2x
…−4y ≤ 2x
Divide both sides by −4 (remember to flip the inequality sign):
[ y \ge -\frac{1}{2}x . ]
Now the inequality is in the familiar y ≥ mx + b* form with a positive coefficient on y, so you can shade above the line y = −½x*.
A Worked‑out Example in Standard Form
Consider the system
[ \begin{cases} -3x + 2y \ge 6 \[4pt] 5x - y < 10 \end{cases} ]
First inequality
Solve for y:
[ 2y \ge 3x + 6 ;\Longrightarrow; y \ge \tfrac{3}{2}x + 3 . ]
The boundary line y = ½·3x + 3* has slope 3/2 and y‑intercept 3. Because the inequality is ≥, draw a solid line. Test the origin (0,0):
[ 0 \ge \tfrac{3}{2}(0)+3 ;\Longrightarrow; 0 \ge 3 \quad\text{false}. ]
Since the test point fails, shade the side away from the origin – that is, the region above the line.
Second inequality
[ 5x - y < 10 ;\Longrightarrow; -y < -5x + 10 ;\Longrightarrow; y > 5x - 10 . ]
(We multiplied by −1, flipping the sign.Here's the thing — )
The boundary y = 5x − 10* has slope 5 and y‑intercept −10. The strict > means a dashed line.
[ 0 > 5(0)-10 ;\Longrightarrow; 0 > -10 \quad\text{true}. ]
Thus shade the side containing the origin – the region above this dashed line.
Intersection
Overlay the two shadings. The feasible region is the wedge that lies above both lines: it starts where the solid line y = ½·3x + 3* meets the dashed line y = 5x − 10* and extends upward indefinitely. Because the solid line is part of the solution, points on it are included; points on the dashed line are excluded.
Additional Pitfalls to Watch For
| Pitfall | Why it Happens | Quick Fix |
|---|---|---|
| Misreading the direction after multiplying/dividing by a negative | Forgetting to reverse the inequality sign leads to shading the wrong half‑plane. | Always write the step “multiply/divide by − … → flip sign*” explicitly, or simply test a point after rearranging. In real terms, |
| Using the wrong intercept when the line is not in slope‑intercept form | Plotting only the y‑intercept can give a skewed line if the x‑intercept is mis‑calculated. | Find both intercepts (set x=0 for y‑int, y=0 for x‑int) and draw through them; this works regardless of the form. |
| Assuming “greater than” always means “shade up” | This rule holds only when y is isolated with a positive coefficient. | If y has a negative coefficient, isolate y first (which flips the inequality) or rely on a test point. |
| Overlooking that a dashed line excludes its points | Including the boundary can turn a correct region into an incorrect one, especially when the inequality is strict. Even so, | After shading, lightly erase or omit the dashed line; remember that any point exactly on it does not satisfy the system. |
| Testing a point that lies exactly on a boundary | The test yields a true/false statement that gives no information about which side to shade. | Choose a point with coordinates that are clearly off the line (e.Here's the thing — g. , (0,0), (1,0), (0,1), (−1,−1)). If (0,0) happens to be on the line, pick another simple point. |
A Final Checklist
- Isolate y (if needed) and note any sign flips.
- Draw each boundary: solid for ≤ or ≥, dashed for < or >.
- Select a test point not on any line (commonly the origin).
- Shade the half‑
4. Shade the half‑plane – Once a test point (usually the origin) satisfies an inequality, shade the entire side of the corresponding boundary that contains that point. Use a hatch, a colour, or a pattern that is distinct from the other inequality’s shading; this visual cue makes it easy to see where the two regions overlap.
5. Identify the feasible region – The feasible region is the intersection of all shaded half‑planes. Its boundary consists of pieces of the original lines, with solid segments included and dashed segments excluded. Locate the vertices where the boundaries meet; these corner points are the only places where the constraints change from one active inequality to another. In the present system the only intersection of the two lines occurs when
[ \frac32x+3 ;=;5x-10 ]
which gives
[ x=\frac{26}{7},\qquad y=\frac32!\left(\frac{26}{7}\right)+3=\frac{60}{7}. ]
Thus the unique vertex is (\bigl(\frac{26}{7},,\frac{60}{7}\bigr)). Because the solid line (y=\frac32x+3) is part of the solution, the vertex lies inside the feasible region, whereas any point on the dashed line (y=5x-10) is excluded.
6. Verify the solution – Pick a few points from the shaded overlap (for example ((0,0)), ((4,10)), and a point just above the vertex such as ((\frac{26
…/7), (\frac{26}{7}+1,\frac{60}{7}+2) to confirm that each satisfies both inequalities.
For (0,0):
- (0 \le \ge 2(0)+3 → 0)? (0\le\frac32(0)+3=3) → true.
- (0>5(0)-10=-10) → true.
For (4,10):
- (10\le\frac32(4)+3=6+3=9)? A better test point is (5,12):
- (12\le\frac32(5)+3=7.No, 8 < 10, so (4,8) fails the second. 5)? 5+3=10.No, 10 > 9, so (4,10) fails the first inequality; instead pick (4,8):
- (8\le\frac32(4)+3=9) → true.
- (8>5(4)-10=20-10=10)? No.
Let’s choose a point clearly inside the overlap, e.Plus, 5) → true. 5+3=7.And g. , (3,7):
- (7\le\frac32(3)+3=4.- (7>5(3)-10=15-10=5) → true.
Thus (3,7) lies in the feasible region, confirming that the shading is correct.
Any point on the dashed line, such as (2,0) because (0=5·2-10), violates the strict inequality (y>5x-10) and must be excluded, while points on the solid line, like (0,3), satisfy (y\le\frac32x+3) and are included.
Conclusion
Graphing a system of linear inequalities reduces to three reliable actions: (1) draw each boundary with the correct line style, (2) use a single test point not lying on any boundary to decide which side to shade, and (3) take the intersection of all shaded half‑planes as the feasible region. By following the checklist—isolating (y) when necessary, distinguishing solid from dashed lines, selecting an unambiguous test point, and verifying a few sample points—one can avoid the common pitfalls of mis‑shading, incorrect boundary inclusion, and erroneous test‑point selection. Applying this method to the given system yields the feasible region whose only vertex is (\bigl(\frac{26}{7},\frac{60}{7}\bigr)); the region includes the solid line (y=\frac32x+3) and excludes the dashed line (y=5x-10). The final shaded overlap represents all ((x,y)) that satisfy both inequalities simultaneously.
