SAT Math – Algebra
Graphs of Linear Systems & Inequalities
Visual interpretation of solutions, intersections, and feasible regions
Graphing linear systems and inequalities transforms abstract algebra into visual problem-solving. Instead of manipulating symbols, you'll identify solution points by seeing where lines intersect or regions where shaded areas overlap. These questions test whether you can read graphs fluently and translate visual information into algebraic conclusions.
The SAT frequently presents graphs and asks you to determine solutions to systems, identify which inequality matches a shaded region, or find how many integer solutions exist in a feasible area. Success requires understanding how equations become lines, how inequalities create regions, and how multiple constraints combine to define solution spaces.
Understanding Graphical Representations
Systems of Linear Equations (Lines)
A system of linear equations appears as two or more lines on a graph. The solution is the point(s) where the lines intersect—coordinates that satisfy all equations simultaneously.
• One solution: Lines intersect at one point (different slopes)
• No solution: Lines are parallel (same slope, different intercepts)
• Infinite solutions: Lines are identical (same equation)
Linear Inequalities (Shaded Regions)
A linear inequality appears as a line plus a shaded region. The line is the boundary, and the shading shows all points that satisfy the inequality.
• Solid line: \(\leq\) or \(\geq\) (boundary included in solution)
• Dashed line: \(<\) or \(>\) (boundary NOT included)
Shading:
• Shade above line for \(y >\) or \(y \geq\)
• Shade below line for \(y <\) or \(y \leq\)
Systems of Inequalities (Overlapping Regions)
A system of inequalities shows multiple shaded regions. The solution is where all shadings overlap—the feasible region satisfying all constraints simultaneously.
How to Read These Graphs
Finding Solutions to Systems of Equations
1. Locate intersection point(s) where lines cross
2. Read coordinates \((x, y)\) at that point
3. Verify by substituting into both equations if needed
Determining Which Inequality Matches a Graph
1. Check the line type: Solid = \(\leq\) or \(\geq\), Dashed = \(<\) or \(>\)
2. Identify the boundary equation from the line
3. Test a point in the shaded region (often use origin if not on line)
4. Determine inequality direction based on which side is shaded
Finding Solutions in Feasible Regions
1. Identify the overlap region (usually darkest or double-shaded)
2. Test points to see if they're in the region
3. Count integer solutions if asked (lattice points with whole coordinates)
Common Pitfalls & Expert Tips
❌ Confusing solid and dashed lines
Solid lines include the boundary (\(\leq\), \(\geq\)); dashed lines exclude it (\(<\), \(>\)). Check carefully which type is shown!
❌ Misreading intersection coordinates
Always check the scale on both axes. If each square = 2 units, an intersection 3 squares right is at \(x = 6\), not \(x = 3\)!
❌ Identifying wrong overlap region
In systems of inequalities, the solution is where ALL shadings meet, not just any shaded area. Look for the darkest overlap.
❌ Forgetting to check if boundary is included
When counting integer solutions, points on a dashed boundary don't count, but points on a solid boundary do!
✓ Expert Tip: Use the origin as a test point
When determining which inequality matches a graph, test (0, 0) if it's not on the boundary line. Easy to substitute and reveals shading direction.
✓ Expert Tip: Look for lattice points systematically
To count integer solutions, go row by row (or column by column) within the shaded region. Don't estimate—count carefully!
✓ Expert Tip: Parallel lines = no solution
If two lines have the same slope but different y-intercepts, they'll never intersect. Recognize this pattern quickly on graphs.
Fully Worked SAT-Style Examples
The graph below shows the system of equations:
\(\begin{cases} y = 2x - 1 \\ y = -x + 5 \end{cases}\)
The lines intersect at point \(P\). What are the coordinates of point \(P\)?
Visual description: Two lines cross in the first quadrant. The first line has positive slope, crossing y-axis at -1. The second line has negative slope, crossing y-axis at 5.
Solution:
Method 1: Read from graph (if given precise coordinates)
Locate where the two lines intersect
Read the \(x\) and \(y\) coordinates at that point
Method 2: Solve algebraically (more reliable)
Set the equations equal since both equal \(y\):
\(2x - 1 = -x + 5\)
\(3x = 6\)
\(x = 2\)
Substitute \(x = 2\) into either equation:
\(y = 2(2) - 1 = 3\)
Verification:
Check \((2, 3)\) in both equations:
Equation 1: \(y = 2x - 1\) → \(3 = 2(2) - 1 = 3\) ✓
Equation 2: \(y = -x + 5\) → \(3 = -2 + 5 = 3\) ✓
Answer: Point \(P\) is at \((2, 3)\)
The graph shows a linear inequality. The boundary line passes through \((0, 3)\) and \((2, 0)\), is solid, and the region below the line is shaded. Which inequality represents this graph?
Answer choices:
A) \(y > -\frac{3}{2}x + 3\)
B) \(y \leq -\frac{3}{2}x + 3\)
C) \(y < -\frac{3}{2}x + 3\)
D) \(y \geq -\frac{3}{2}x + 3\)
Solution:
Step 1: Determine the boundary line equation
Find slope using points \((0, 3)\) and \((2, 0)\):
\(m = \frac{0 - 3}{2 - 0} = \frac{-3}{2} = -\frac{3}{2}\)
Y-intercept is 3 (line crosses y-axis at \((0, 3)\))
Equation of boundary line: \(y = -\frac{3}{2}x + 3\)
Step 2: Determine inequality symbol from line type
Line is solid → includes boundary → use \(\leq\) or \(\geq\)
Eliminates choices A and C (which use \(>\) and \(<\))
Step 3: Determine direction from shading
Shading is below the line
"Below" means \(y \leq\) (y-values less than or equal to the line)
This eliminates choice D (which is \(\geq\), meaning above)
Verification:
Test a point in the shaded region, say \((0, 0)\):
\(0 \leq -\frac{3}{2}(0) + 3\) → \(0 \leq 3\) ✓ (True!)
Test a point NOT in shaded region, say \((0, 4)\):
\(4 \leq -\frac{3}{2}(0) + 3\) → \(4 \leq 3\) ✗ (False, as expected)
Answer: B) \(y \leq -\frac{3}{2}x + 3\)
A graph shows two parallel lines that never intersect. Line 1 has equation \(y = 3x + 2\) and Line 2 has equation \(y = 3x - 4\). How many solutions does this system have?
Solution:
Step 1: Analyze the slopes
Line 1: \(y = 3x + 2\) → slope = 3
Line 2: \(y = 3x - 4\) → slope = 3
Same slope!
Step 2: Analyze the y-intercepts
Line 1: y-intercept = 2
Line 2: y-intercept = -4
Different y-intercepts!
Step 3: Determine solution type
Same slope + Different y-intercepts = Parallel lines
Parallel lines never intersect
Therefore: NO SOLUTION
Key Pattern Recognition:
One solution: Different slopes → lines intersect
No solution: Same slope, different intercepts → parallel
Infinite solutions: Same slope, same intercept → identical lines
Answer: Zero solutions (no solution)
A graph shows the system:
\(\begin{cases} y \leq x + 3 \\ y > 2x - 1 \end{cases}\)
Which of the following points is in the solution region?
Test points:
A) \((0, 5)\)
B) \((2, 4)\)
C) \((1, 2)\)
D) \((3, 1)\)
Solution:
Strategy: Test each point in BOTH inequalities
A point is in the solution region only if it satisfies BOTH constraints
Testing point A: \((0, 5)\)
Inequality 1: \(5 \leq 0 + 3\) → \(5 \leq 3\) ✗ False
Fails first inequality, so not in solution region
Testing point B: \((2, 4)\)
Inequality 1: \(4 \leq 2 + 3\) → \(4 \leq 5\) ✓ True
Inequality 2: \(4 > 2(2) - 1\) → \(4 > 3\) ✓ True
Satisfies both! This is in the solution region
Testing point C: \((1, 2)\)
Inequality 1: \(2 \leq 1 + 3\) → \(2 \leq 4\) ✓ True
Inequality 2: \(2 > 2(1) - 1\) → \(2 > 1\) ✓ True
Also satisfies both!
Testing point D: \((3, 1)\)
Inequality 1: \(1 \leq 3 + 3\) → \(1 \leq 6\) ✓ True
Inequality 2: \(1 > 2(3) - 1\) → \(1 > 5\) ✗ False
Fails second inequality
Answer: B) \((2, 4)\) and C) \((1, 2)\) are both in the solution region
If only one answer is allowed, both B and C work. Check question carefully!
The system of inequalities below defines a feasible region:
\(\begin{cases} x \geq 0 \\ y \geq 0 \\ x + y \leq 4 \end{cases}\)
How many points with integer coordinates \((x, y)\) are in this region?
Solution:
Step 1: Understand the region
\(x \geq 0\) and \(y \geq 0\) → First quadrant (including axes)
\(x + y \leq 4\) → Below/on the line \(x + y = 4\)
This forms a triangle with vertices at \((0, 0)\), \((4, 0)\), and \((0, 4)\)
Step 2: List integer points systematically
Go through each possible \(x\) value:
When \(x = 0\): \(y \leq 4\) → \(y\) can be 0, 1, 2, 3, 4
Points: \((0, 0), (0, 1), (0, 2), (0, 3), (0, 4)\) → 5 points
When \(x = 1\): \(1 + y \leq 4\) → \(y \leq 3\) → \(y\) can be 0, 1, 2, 3
Points: \((1, 0), (1, 1), (1, 2), (1, 3)\) → 4 points
When \(x = 2\): \(2 + y \leq 4\) → \(y \leq 2\) → \(y\) can be 0, 1, 2
Points: \((2, 0), (2, 1), (2, 2)\) → 3 points
When \(x = 3\): \(3 + y \leq 4\) → \(y \leq 1\) → \(y\) can be 0, 1
Points: \((3, 0), (3, 1)\) → 2 points
When \(x = 4\): \(4 + y \leq 4\) → \(y \leq 0\) → \(y\) can be 0
Points: \((4, 0)\) → 1 point
Step 3: Count total points
\(5 + 4 + 3 + 2 + 1 = 15\) points
Quick Formula:
For \(x + y \leq n\) in first quadrant: \(\frac{(n+1)(n+2)}{2}\) points
Here: \(\frac{(4+1)(4+2)}{2} = \frac{5 \times 6}{2} = 15\) ✓
Answer: 15 integer points
Which inequality would be graphed with a dashed boundary line?
Answer choices:
A) \(y \leq 2x + 1\)
B) \(y \geq -x + 3\)
C) \(y < \frac{1}{2}x - 2\)
D) \(3x - 2y \leq 6\)
Solution:
Key Rule: Boundary line type depends on inequality symbol
Solid line: \(\leq\) or \(\geq\) (boundary included)
Dashed line: \(<\) or \(>\) (boundary NOT included)
Analyzing each choice:
A) \(y \leq 2x + 1\) → has \(\leq\) → solid line ✗
B) \(y \geq -x + 3\) → has \(\geq\) → solid line ✗
C) \(y < \frac{1}{2}x - 2\) → has \(<\) → DASHED line ✓
D) \(3x - 2y \leq 6\) → has \(\leq\) → solid line ✗
Why this matters:
Dashed line means points exactly ON the line are NOT solutions
For \(y < \frac{1}{2}x - 2\), a point like \((4, 0)\) on the line wouldn't satisfy the inequality
Only points strictly below the line work
Answer: C) \(y < \frac{1}{2}x - 2\)
The feasible region for a system of inequalities is bounded by the vertices \((0, 0)\), \((5, 0)\), \((3, 4)\), and \((0, 6)\). What is the maximum value of \(P = 2x + 3y\) in this region?
Solution:
Key Principle: Linear optimization theorem
For a linear objective function on a bounded feasible region:
Maximum and minimum values occur at vertices (corner points)
Step 1: Evaluate \(P = 2x + 3y\) at each vertex
At \((0, 0)\): \(P = 2(0) + 3(0) = 0\)
At \((5, 0)\): \(P = 2(5) + 3(0) = 10\)
At \((3, 4)\): \(P = 2(3) + 3(4) = 6 + 12 = 18\)
At \((0, 6)\): \(P = 2(0) + 3(6) = 18\)
Step 2: Identify maximum
Values: 0, 10, 18, 18
Maximum value = 18
Occurs at both \((3, 4)\) and \((0, 6)\)
Important Note:
When the maximum occurs at two adjacent vertices, all points on the line segment between them also give the maximum value
But the maximum VALUE is still 18
Answer: Maximum value is 18
The graph shows the solution to \(|x - 2| \leq 3\). What interval represents all solutions?
Visual: A shaded region on the number line from -1 to 5, with solid dots at both endpoints.
Solution:
Step 1: Understand absolute value inequality
\(|x - 2| \leq 3\) means "distance from 2 is at most 3"
This translates to: \(-3 \leq x - 2 \leq 3\)
Step 2: Solve the compound inequality
\(-3 \leq x - 2 \leq 3\)
Add 2 to all parts:
\(-1 \leq x \leq 5\)
Step 3: Verify with graph
Left endpoint: \(x = -1\) (solid dot, so included)
Right endpoint: \(x = 5\) (solid dot, so included)
Shaded between them → all values from -1 to 5
Test values:
At \(x = 2\): \(|2 - 2| = 0 \leq 3\) ✓
At \(x = -1\): \(|-1 - 2| = 3 \leq 3\) ✓
At \(x = 5\): \(|5 - 2| = 3 \leq 3\) ✓
At \(x = 6\): \(|6 - 2| = 4 \leq 3\) ✗
Answer: \([-1, 5]\) or \(-1 \leq x \leq 5\)
Quick Reference: Systems Solution Types
SAT Graph Reading Checklist
For Systems of Equations
- Check axis scales carefully
- Find intersection point(s)
- Read coordinates precisely
- Verify slopes if checking solution type
For Inequalities
- Check if line is solid or dashed
- Identify which side is shaded
- Test a point to verify direction
- Find overlap for systems
For Integer Solutions
- Count systematically (row by row)
- Check if boundaries are included
- Don't estimate—count carefully
- Verify a few points if unsure
For Optimization
- Identify all corner points
- Evaluate function at each vertex
- Compare values
- Report max/min as asked
Graphs: Where Algebra Becomes Visual
Graph interpretation transforms abstract algebraic relationships into visual patterns you can see and analyze at a glance. While solving systems algebraically requires manipulation and calculation, reading graphs lets you identify solutions, understand constraints, and verify answers instantly. This visual literacy is invaluable not just for the SAT, but for any field involving data analysis, optimization, or constraint satisfaction—from engineering and economics to logistics and computer science. Master the art of reading these graphs, understanding what solid vs. dashed lines mean, recognizing where regions overlap, and counting solutions systematically. The ability to move fluently between algebraic and graphical representations is a hallmark of mathematical maturity that serves you throughout quantitative disciplines.