DeltaMath Guide

Algebra 1 · DeltaMath skill "Linear Inequality Systems Graphically"

Systems of Linear Inequalities Graphically

You are given two (sometimes three) linear inequalities and must either graph the solution region using DeltaMath's drawing tool or decide whether a given point is a solution.

A system of linear inequalities is solved by a region, not a single point. Every point inside the region makes all of the inequalities true at the same time. On DeltaMath the interactive version asks you to place each boundary line with two points, click to pick solid or dashed, and then click the side to shade.

The grading is strict about three things: the line must go through the right points, the line style must match the inequality symbol, and the shaded side must be correct for every inequality. Getting one of the three wrong marks the whole problem wrong, so it pays to work methodically.

How to solve it, step by step

  1. Rewrite each inequality in slope-intercept form

    Solve for y so you can read the slope and y-intercept. Remember to flip the inequality symbol if you divide or multiply by a negative number. For example −2y < 4x − 6 becomes y > −2x + 3.

  2. Graph the boundary line for each inequality

    Plot the y-intercept, use the slope to find a second point, and draw the line. On DeltaMath, drag or click the two points where the tool asks for them.

  3. Choose solid or dashed

    Use a solid line for ≤ or ≥ (points on the line are included). Use a dashed line for < or > (points on the line are not included).

  4. Shade the correct side of each line

    For y > or y ≥ shade above the line; for y < or y ≤ shade below. If you are unsure, test a point such as (0, 0): if it makes the inequality true, shade the side that contains it.

  5. Find the overlap

    The solution to the system is only the region where every shading overlaps. On the DeltaMath tool you shade one side per inequality and the platform highlights the intersection.

Worked example

Example 1

Graph the solution set of the system: y ≤ x + 1 and y > −2x + 4.

  1. Line 1: y = x + 1 → intercept (0, 1), slope 1 → second point (1, 2). Solid line (≤). Shade below.
  2. Line 2: y = −2x + 4 → intercept (0, 4), slope −2 → second point (1, 2). Dashed line (>). Shade above.
  3. Test (3, 0): 0 ≤ 3 + 1 ✓ and 0 > −6 + 4 ✓ → (3, 0) is in the solution.
  4. The solution region is to the right of where the lines cross at (1, 2), below the solid line and above the dashed line.

Answer The wedge opening to the right from (1, 2), including the solid boundary but not the dashed one.

Mistakes DeltaMath marks wrong

  • Forgetting to flip the inequality when dividing by a negative while solving for y.
  • Using a solid line for a strict inequality (< or >), or a dashed line for ≤ / ≥.
  • Shading above for y < because the line itself slopes upward. Direction of shading depends on the symbol, not the slope.
  • Shading the whole plane on the second inequality instead of just its own side, which hides the overlap.
  • Placing the second point using the slope in the wrong direction (for slope −2, move down 2 and right 1, not up).

How to type the answer

  • In the graphing tool, click two points to set each line, then use the toggle for solid or dashed.
  • Click once on the side you want shaded; click again to switch sides.
  • For "is the point a solution?" versions, substitute the point into every inequality and choose Yes only if all are true.

Systems of Linear Inequalities Graphically: questions

What if the two lines are parallel?
Parallel boundary lines produce either a strip between them or no solution at all, depending on the shading directions. Test a point between the lines to see whether it satisfies both inequalities.
Does the point where the lines cross count as a solution?
Only if both boundary lines are solid. If either line is dashed, the intersection point is excluded.
Why did DeltaMath mark my graph wrong when the shading looked right?
The most common cause is the wrong line style. Check that each line is solid for ≤ or ≥ and dashed for < or >, and that each line passes exactly through two correct lattice points.

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