SAT Systems of Inequalities | Graphs, Overlap & Practice | SATMath800
SAT Math · Algebra · 08

Systems of inequalities

Combine multiple constraints, identify the overlap, and reason about feasible regions on the coordinate plane.

Systems Overlap Feasible regions Boundary lines Test points

The big idea

A system of inequalities asks for points that satisfy every condition at the same time.

\(\text{solution set}=\text{region}_1\cap\text{region}_2\cap\cdots\)
Core idea: graph each inequality, then keep only the common region.

1. Each inequality creates a region

A single linear inequality divides the coordinate plane into two half-planes. A system combines those regions.

First condition
\(\displaystyle y\ge x-1\)

The boundary is solid, and the solution is above the line.

Second condition
\(\displaystyle y\le -x+5\)

The boundary is solid, and the solution is below the line.

2. Graph the boundaries first

Replace each inequality with equality to obtain its boundary line. Then decide whether the boundary is solid or dashed.

Inequality Boundary equation Boundary
\(y\ge x-1\) \(y=x-1\) Solid
\(y<-x+5\) \(y=-x+5\) Dashed

3. Shade each condition separately

Determine the solution side for each inequality before looking for the overlap. If the inequality is solved for \(y\), greater-than means above and less-than means below.

Above
\(\displaystyle y\ge x-1\)
Below
\(\displaystyle y\le -x+5\)

4. The overlap is the solution

A point belongs to the system only where both shaded conditions are true. The common region is the solution set.

System: \(y\ge x-1\) and \(y\le -x+5\)
(3, 2) y = x − 1 y = −x + 5 y x

The green region is the set of points satisfying both inequalities.

5. Find a boundary intersection

When the two boundary lines intersect, solve the boundary equations together.

\(\displaystyle y=x-1\)
\(\displaystyle y=-x+5\)
\(\displaystyle x-1=-x+5\)
\(\displaystyle 2x=6\)
\(\displaystyle x=3,\quad y=2\)

The intersection is \((3,2)\). Whether that point belongs to the system depends on whether each original inequality includes its boundary.

6. Test a point against the whole system

A point must pass every inequality.

1
Substitute the point into the first inequality.
2
Check the second inequality.
3
Accept the point only if all conditions are true.
One failure is enough. If a point violates even one condition, it is not a solution of the system.

7. Solid and dashed boundaries can occur together

Each inequality controls its own boundary. A system can therefore contain one solid line and one dashed line.

\(\displaystyle y\ge x-2\qquad\text{and}\qquad y<-x+4\)

The first boundary is included; the second is excluded.

8. Contexts create systems of constraints

SAT questions may describe several limits that must hold simultaneously.

\(\displaystyle 2x+3y\le60,\qquad x+y\ge18,\qquad x\ge0,\qquad y\ge0\)

Each inequality represents one restriction. The feasible region is the overlap of all four conditions.

9. Maximum and minimum questions

If a system creates a bounded feasible region, a linear quantity may reach its greatest or least value at a boundary point or corner.

For the SAT: first identify the feasible region. Then look for the feasible point that gives the requested maximum or minimum.

10. Parallel boundaries can eliminate the solution set

Parallel boundaries never intersect. If their inequality directions require opposite sides with no common region, the system has no solution.

\(\displaystyle y\le2x+1\qquad\text{and}\qquad y\ge2x+5\)

A point cannot be at or below \(2x+1\) and at or above \(2x+5\) at the same time.

11. Parameter questions

If the SAT gives a specific point and asks about a parameter, substitute the point into every inequality first.

\(\displaystyle y<-x+a\qquad\text{and}\qquad y>x+b\)

At \((0,0)\), these become \(0b\).

12. A reliable SAT workflow

1
Translate. Write each condition as an inequality.
2
Find the boundaries. Replace each inequality with equality.
3
Choose the regions. Use the inequality direction or a test point.
4
Find the overlap. Keep only points satisfying every condition.
5
Verify. Substitute a requested point or candidate answer into the original system.

Visual practice

Each visual question is self-contained: the equations needed to interpret the graph are shown directly on the graph.

Visual practice 1

The graph shows the system \(\displaystyle y\ge x-1\) and \(\displaystyle y\le -x+5\). Which point is in the shaded overlap?

Read the overlap
P = (2, 2) y = x − 1 y = −x + 5 y x
  • A) \((2,2)\)
  • B) \((0,-2)\)
  • C) \((4,4)\)
  • D) \((5,0)\)
Solution: For \((2,2)\), \(2\ge1\) and \(2\le3\), so both inequalities are true.
Visual practice 2

The graph shows the system \(\displaystyle y>x+2\) and \(\displaystyle y

Parallel boundaries
y = x + 2 y = x − 2 y x
  • A) Exactly one solution
  • B) Infinitely many solutions
  • C) No solution
  • D) A finite triangular region
Solution: The first condition requires points above \(x+2\), while the second requires points below \(x-2\). Those regions do not overlap.
Visual practice 3

The graph shows the system \(\displaystyle y\le x+4\) and \(\displaystyle y\ge -x+2\). Is \(P=(2,3)\) a solution?

Test the point against both boundaries
P = (2, 3) y = x + 4 y = −x + 2 y x
  • A) Yes, because it satisfies both inequalities.
  • B) No, because it is above both boundaries.
  • C) No, because it is below both boundaries.
  • D) Yes, because every point on either boundary is a solution.
Solution: \(3\le2+4\) and \(3\ge-2+2\), so \(P\) satisfies both inequalities.
Visual practice 4

The feasible region is defined by \(\displaystyle x\ge1\), \(\displaystyle y\ge2\), and \(\displaystyle x+y\le6\). What is the greatest possible value of \(x\)?

Three simultaneous constraints
(1, 2) (1, 5) (4, 2) x = 1 y = 2 x + y = 6 y x
  • A) 1
  • B) 2
  • C) 4
  • D) 6
Solution: The rightmost feasible point is \((4,2)\), so the greatest possible value of \(x\) is \(4\).
Visual practice 5

The graph shows the system \(\displaystyle y\ge x-2\) and \(\displaystyle y<-x+4\). Is the intersection point of the boundary lines part of the solution set?

One solid boundary and one dashed boundary
(3, 1) y = x − 2 y = −x + 4 y x
  • A) Yes, because both boundaries intersect there.
  • B) No, because one inequality is strict.
  • C) Yes, because the solid boundary is included.
  • D) No, because neither boundary is included.
Solution: The intersection is \((3,1)\). It lies on the dashed boundary \(y=-x+4\), which is excluded, so the point is not a solution.

Additional practice

Original SATMath800 practice. These questions vary the representation: algebraic systems, points, boundaries, contexts, and feasible-region reasoning.

Question 1

Which point satisfies both \(\displaystyle y\ge x-2\) and \(\displaystyle y\le-x+6\)?

  • A) \((0,-3)\)
  • B) \((1,0)\)
  • C) \((2,5)\)
  • D) \((4,1)\)
Solution: For \((1,0)\), \(0\ge-1\) and \(0\le5\), so both conditions hold.
Question 2

The system \(\displaystyle y\le2x+1\) and \(\displaystyle y\ge2x+5\) has which type of solution set?

  • A) Exactly one point
  • B) A finite segment
  • C) Infinitely many points
  • D) No points
Solution: The first condition requires \(y\) to be at most \(2x+1\), while the second requires \(y\) to be at least \(2x+5\). No point can satisfy both.
Question 3

For the system \(\displaystyle y\ge x+2\) and \(\displaystyle y\le-x+8\), the boundary lines intersect at which point?

  • A) \((2,4)\)
  • B) \((3,5)\)
  • C) \((4,6)\)
  • D) \((5,3)\)
Solution: \(x+2=-x+8\) gives \(x=3\), and then \(y=5\).
Question 4

Which system represents points that are above or on \(y=2x-1\) and below \(y=-x+8\)?

  • A) \(y\le2x-1\) and \(y\ge-x+8\)
  • B) \(y\ge2x-1\) and \(y\le-x+8\)
  • C) \(y>2x-1\) and \(y<-x+8\)
  • D) \(y\le2x-1\) and \(y\le-x+8\)
Solution: Above or on gives \(y\ge2x-1\), and below gives \(y\le-x+8\).
Question 5

In the system \(\displaystyle y<-x+a\) and \(\displaystyle y>x+b\), the point \((0,0)\) is a solution. Which relationship must be true?

  • A) \(a>b\)
  • B) \(b>a\)
  • C) \(a=0\)
  • D) \(b=0\)
Solution: Substitution gives \(0<a\) and \(0>b\), so \(a>b\).
Question 6

A school can use at most 26 total hours for two activities. If \(x\) is science-lab time and \(y\) is art-studio time, which system also requires at least 8 hours of science-lab time?

  • A) \(x+y\ge26,\ x\ge8\)
  • B) \(x+y\le26,\ x\ge8\)
  • C) \(x+y\le26,\ x\le8\)
  • D) \(x+y\ge26,\ x\le8\)
Solution: At most 26 gives \(x+y\le26\), and at least 8 gives \(x\ge8\).
Question 7

The system \(\displaystyle y\ge-2x+4\) and \(\displaystyle y\le x+1\) has a feasible region. Which point is on both boundary lines?

  • A) \((0,1)\)
  • B) \((1,2)\)
  • C) \((2,3)\)
  • D) \((3,4)\)
Solution: \(-2x+4=x+1\) gives \(x=1\), so \(y=2\).
Question 8

A company makes two products. Product A requires 2 machine-hours and product B requires 3 machine-hours. The factory has at most 60 machine-hours. Which inequality is one of the constraints?

  • A) \(2x+3y\ge60\)
  • B) \(2x+3y\le60\)
  • C) \(2x+3y<0\)
  • D) \(2x+3y=60\)
Solution: The total machine time is \(2x+3y\), and at most 60 means \(2x+3y\le60\).
Question 9

The boundary lines of a feasible system intersect at \((150,750)\). If every feasible point satisfies \(y\le-15x+3000\) and \(y\le5x\), what is the greatest possible value of \(y\)?

  • A) 150
  • B) 600
  • C) 750
  • D) 3000
Solution: The two upper boundaries meet at \((150,750)\), so the stated feasible maximum is \(750\).
Question 10

Which point satisfies both \(\displaystyle 2x+y\le10\) and \(\displaystyle x+2y\le8\)?

  • A) \((2,2)\)
  • B) \((4,3)\)
  • C) \((5,2)\)
  • D) \((1,5)\)
Solution: For \((2,2)\), \(2(2)+2=6\le10\) and \(2+2(2)=6\le8\).
Question 11

Which system has a solution set that includes the boundary \(y=x+3\) but excludes the boundary \(y=-x+7\)?

  • A) \(y>x+3,\ y\le-x+7\)
  • B) \(y\ge x+3,\ y<-x+7\)
  • C) \(y\le x+3,\ y\ge-x+7\)
  • D) \(y>x+3,\ y>-x+7\)
Solution: Including \(y=x+3\) requires \(\ge\); excluding \(y=-x+7\) requires \(<\).
Question 12

For \(\displaystyle y\ge2x-4\) and \(\displaystyle y\le-2x+8\), what is the \(y\)-coordinate of the intersection of the boundaries?

  • A) 2
  • B) 4
  • C) 6
  • D) 8
Solution: \(2x-4=-2x+8\) gives \(x=3\), and then \(y=2\).
Question 13

A point is a solution of a system of three inequalities. What must be true?

  • A) It satisfies at least one inequality.
  • B) It satisfies exactly two inequalities.
  • C) It satisfies all three inequalities.
  • D) It lies on all three boundary lines.
Solution: A system requires every condition to be true.
Question 14

A feasible region is bounded by \(x=1\), \(y=2\), and \(x+y=6\), with \(x\ge1\), \(y\ge2\), and \(x+y\le6\). What is the greatest possible value of \(x\)?

  • A) 1
  • B) 2
  • C) 4
  • D) 6
Solution: The greatest \(x\) occurs when \(y=2\): \(x+2=6\), so \(x=4\).
Question 15

Which system represents \(x\) and \(y\) both nonnegative and their sum at least 12?

  • A) \(x\le0,\ y\le0,\ x+y\le12\)
  • B) \(x\ge0,\ y\ge0,\ x+y\ge12\)
  • C) \(x\ge0,\ y\le0,\ x+y\ge12\)
  • D) \(x\le0,\ y\ge0,\ x+y\le12\)
Solution: Nonnegative means \(x\ge0\) and \(y\ge0\), while at least 12 means \(x+y\ge12\).
Question 16

For the system \(y\le3x+2\) and \(y\ge3x-4\), which statement is true?

  • A) The system has no solution.
  • B) The boundaries intersect at one point.
  • C) The system has infinitely many solutions.
  • D) The solution is only the origin.
Solution: The lines are parallel, and the region between them is nonempty, so infinitely many points satisfy both conditions.
Question 17

For the system \(y\ge x+1\) and \(y\le4\), which \(x\)-values can occur in a solution?

  • A) \(x\le3\)
  • B) \(x\ge3\)
  • C) \(x<4\)
  • D) \(x>4\)
Solution: A feasible \(y\) must satisfy \(x+1\le4\), so \(x\le3\).
Question 18

A point \((a,b)\) lies on both boundaries \(y=2x+1\) and \(y=-x+7\). What is \(a+b\)?

  • A) 6
  • B) 7
  • C) 8
  • D) 9
Solution: \(2a+1=-a+7\) gives \(a=2\), then \(b=5\), so \(a+b=7\).
Question 19

Which point lies on the boundary \(x+y=10\) and satisfies \(x\ge4\) and \(y\ge3\)?

  • A) \((3,7)\)
  • B) \((4,6)\)
  • C) \((6,5)\)
  • D) \((7,4)\)
Solution: \((4,6)\) is on \(x+y=10\) and satisfies both lower bounds.
Question 20

The feasible region of a system is the overlap of two half-planes. Which operation describes that overlap?

  • A) Union
  • B) Intersection
  • C) Reflection
  • D) Translation
Solution: Points satisfying both conditions belong to the intersection of the two regions.
SATMath800.com · Content prepared by Dr. Aytekin Vargün · Original SAT Math instruction and practice created for SATMath800.

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