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Writing Systems of Inequalities from Shaded Graphs: Reverse Reasoning Practice

Learn to derive two inequalities from a pre-shaded region by reading boundary lines, testing points, and checking boundary type and shading direction.

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Four exercises
Worked answers

Algebra and Function Practice

Most graphing lessons start with inequalities and ask you to shade. This reverse task starts with a shaded region and asks you to write the system that produces it. You will identify each boundary line from its slope and intercept, decide whether the line is solid or dashed, and use a test point to choose the correct inequality symbol. The key habit is verification: after writing both inequalities, test a point inside the shaded region and a point outside it to confirm your system matches the graph.

Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.

Reading boundary lines from a graph

Each boundary is a straight line. Find two clear points on the line, compute slope as rise over run, and write the equation in slope-intercept form y = mx + b. If the line is vertical, write x = c instead. Record whether the line is solid or dashed before choosing the inequality symbol.

A solid boundary means points on the line are included, so use ≤ or ≥. A dashed boundary means points on the line are excluded, so use < or >. This decision is separate from shading direction and must be checked independently.

  1. Locate two points where the boundary crosses grid intersections.
  2. Compute slope m = (change in y) / (change in x).
  3. Identify the y-intercept b and write y = mx + b.
  4. Note solid or dashed to select ≤, ≥, <, or >.
  5. Choose a test point not on the line and substitute to decide the symbol direction.

Choosing the inequality symbol with test points

After writing the boundary equation, pick a test point that is clearly inside the shaded region. Substitute its coordinates into the equation form. If the resulting statement is true for the intended symbol, keep that symbol; if false, reverse it. Repeat with a point outside the region to confirm the reverse is false.

For example, suppose a boundary is y = 2x + 1 and the shaded region lies below a solid line. Test (0,0): 0 ≤ 2(0)+1 is true, so y ≤ 2x + 1 is consistent. Test a point above the line such as (0,3): 3 ≤ 1 is false, confirming the outside point is excluded.

  1. Select an inside test point with integer coordinates when possible.
  2. Substitute into the boundary equation using the candidate symbol.
  3. Keep the symbol if the statement is true; reverse it if false.
  4. Test an outside point to confirm it makes the inequality false.
  5. Write the final pair and verify both inequalities together at one inside point.

Combining two inequalities into a system

A system of inequalities is the pair of statements joined by 'and' because the shaded region is the overlap of two half-planes. Write both inequalities with the same variable isolated when possible, such as y ≤ 2x + 1 and y > -x + 4. If one boundary is vertical, write x ≥ c or x ≤ c.

Check that the overlap of the two half-planes matches the shaded region. A common error is reversing one symbol while keeping the other correct, which produces a region on the wrong side of one boundary. Testing one inside point against both inequalities catches this quickly.

  1. Write each boundary equation separately.
  2. Decide solid or dashed for each line.
  3. Use test points to fix each symbol.
  4. Join the two inequalities as a system.
  5. Verify one inside point satisfies both and one outside point fails at least one.

Checking boundary type and shading direction

Boundary type and shading direction are independent decisions. A solid line can shade above or below; a dashed line can also shade above or below. Do not assume that solid means 'above' or dashed means 'below'. Read each feature from the graph.

When a line is vertical, shading to the right means x ≥ c or x > c depending on solid or dashed. Shading to the left means x ≤ c or x < c. For horizontal lines, shading above means y ≥ c or y > c, and shading below means y ≤ c or y < c.

  1. Mark each boundary as solid or dashed.
  2. Mark the shaded side of each boundary.
  3. Match solid to ≤ or ≥ and dashed to < or >.
  4. Match the shaded side using a test point.
  5. Recheck both decisions after writing the system.

Worked example

A graph shows two solid boundary lines. Line A passes through (0,1) and (2,5). Line B passes through (0,4) and (2,0). The shaded region is below Line A and below Line B. Write the system of inequalities.

Worked answer

y ≤ 2x + 1 and y ≤ -2x + 4

Why: Line A slope is (5-1)/(2-0)=2, intercept 1, so y=2x+1. Solid and shaded below gives y ≤ 2x+1. Test (0,0): 0 ≤ 1 is true. Line B slope is (0-4)/(2-0)=-2, intercept 4, so y=-2x+4. Solid and shaded below gives y ≤ -2x+4. Test (0,0): 0 ≤ 4 is true. The system is y ≤ 2x+1 and y ≤ -2x+4.

Four practice exercises

Exercise 1

A graph shows a solid line through (0,2) and (3,5). The shaded region is above this line. A second solid line passes through (0,6) and (3,0), and the shaded region is below this line. Write the system.

Show answer and check

y ≥ x + 2 and y ≤ -2x + 6

Why: First line slope (5-2)/(3-0)=1, intercept 2, solid above gives y ≥ x+2. Test (0,3): 3 ≥ 2 true. Second line slope (0-6)/(3-0)=-2, intercept 6, solid below gives y ≤ -2x+6. Test (0,3): 3 ≤ 6 true. System: y ≥ x+2 and y ≤ -2x+6.

Check: Test (1,4): 4 ≥ 3 true and 4 ≤ 4 true, so inside. Test (0,0): 0 ≥ 2 false, so outside fails first inequality.

Exercise 2

A graph shows a dashed line through (0,1) and (2,3). The shaded region is below this line. A solid vertical line is at x = 1, and the shaded region is to the right. Write the system.

Show answer and check

y < x + 1 and x ≥ 1

Why: Slanted line slope (3-1)/(2-0)=1, intercept 1, dashed below gives y < x+1. Test (2,0): 0 < 3 true. Vertical line solid at x=1, shaded right gives x ≥ 1. Test (2,0): 2 ≥ 1 true. System: y < x+1 and x ≥ 1.

Check: Test (2,1): 1 < 3 true and 2 ≥ 1 true, so inside. Test (0,0): 0 < 1 true but 0 ≥ 1 false, so outside fails second inequality.

Exercise 3

A graph shows a solid line through (0,5) and (5,0). The shaded region is below this line. A dashed line passes through (0,0) and (2,4), and the shaded region is above this line. Write the system.

Show answer and check

y ≤ -x + 5 and y > 2x

Why: First line slope (0-5)/(5-0)=-1, intercept 5, solid below gives y ≤ -x+5. Test (1,1): 1 ≤ 4 is true, confirming the shaded side is below this boundary. Second line slope (4-0)/(2-0)=2, intercept 0, dashed above gives y > 2x. Test (1,3): 3 > 2(1) is true, confirming the shaded side is above this boundary. Test an outside point below the line, such as (1,0): 0 > 2 is false. System: y ≤ -x+5 and y > 2x.

Check: Test (1,3): 3 ≤ 4 true and 3 > 2 true, so inside. Test (1,0): 0 ≤ 4 true but 0 > 2 false, so outside fails the second inequality.

Exercise 4

A graph shows a dashed horizontal line at y = 2 with shading above it. A solid line passes through (0,0) and (2,2), and the shaded region is below this line. Write the system.

Show answer and check

y > 2 and y ≤ x

Why: Horizontal dashed line at y=2, shaded above gives y > 2. Test (3,3): 3 > 2 true. Slanted solid line slope (2-0)/(2-0)=1, intercept 0, shaded below gives y ≤ x. Test (3,3): 3 ≤ 3 true. System: y > 2 and y ≤ x.

Check: Test (3,3): 3 > 2 true and 3 ≤ 3 true, so inside. Test (1,1): 1 > 2 false, so outside fails first inequality.

Common mistakes

Self-check rubric

Questions and answers

How do I know whether to use ≤ or < from a graph?

Look at the boundary line. A solid line includes the boundary, so use ≤ or ≥. A dashed line excludes the boundary, so use < or >. Then use a test point to choose the correct side.

Can a system have one vertical boundary and one slanted boundary?

Yes. Write the vertical boundary as x ≥ c or x ≤ c, using solid or dashed to choose the symbol. Write the slanted boundary in slope-intercept form and test a point to fix its symbol.

What if my test point satisfies one inequality but not the other?

An inside point must satisfy both inequalities. If it fails one, recheck that boundary's symbol by testing a point on the other side. The error is usually in the symbol for the failed inequality.

Do I need to test two points for every boundary?

Testing one inside point is usually enough to choose the symbol, but testing one outside point confirms the reverse is false. For accuracy, test one inside and one outside point for each boundary.

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