Linear Inequalities and Written Region Descriptions
Interpret boundary styles and solution sides from linear inequalities and written region descriptions. Check coordinate pairs against several constraints, count integer-coordinate solutions, and write inequalities for described regions.
What a learner can do afterwards
- Use inequality signs, substitution and rearrangement to explain boundary inclusion and which side of a line satisfies a linear inequality
- Check whether coordinate pairs satisfy every inequality in a system and count its integer-coordinate solutions
- Translate given boundary equations and written descriptions of sides and boundary inclusion into inequalities
1 · Read
You have plotted lines and solved one-variable inequalities. Here you will use equations and written region descriptions. A solution is a pair (x, y). For Ax + By < C, A and B are not both zero; Ax + By = C is the boundary. Check a point by substituting its coordinates: (1, 0) in y < 2x + 1 gives 0 < 3, which is true.
Substitution checks whether this particular point belongs.
A boundary point makes the two sides equal. For < or > it is excluded, so draw the line dashed. For <= or >= it is included, so draw the line solid. For y < 2x + 1, equality on y = 2x + 1 does not pass the strict test. A solid line includes boundary points for that one inequality.
Rearrange x - 2y >= 8. Subtract x: -2y >= 8 - x. Divide by -2 and reverse the sign: y <= x/2 - 4. Draw the boundary solid and shade below it. The original >= sign does not tell you to shade above. First isolate y or test a point.
The rearranged inequality identifies the shaded side.
Choose a test point off the boundary. If it passes, shade its side; if it fails, shade the opposite side. When the origin lies on the line, choose another point. With y isolated, < shades below and > above. The line x = a is vertical: x < a is left and x > a is right. The line y = b is horizontal: y < b is below and y > b is above. Equality changes line style, not which side is shaded.
A system requires every inequality to hold. Substitute a point into each one; keep it only if all pass. The overlap may be bounded, unbounded or empty. To count integer-coordinate solutions, fix each allowed integer x and find all allowed integer y-values. For x + y <= k, use y <= k - x. List the pairs column by column, then add the counts. Include a boundary point only when every constraint permits it.
Translate a written region description using its given boundary equation and allowed side. For y = mx + c, below gives y < mx + c and above gives y > mx + c. Add equality when the boundary is included. For x = a, left gives x < a and right gives x > a. Solid boundaries include equality; dashed ones exclude it. For example, on or below y = 5 - 2x gives y <= 5 - 2x. Apply this to each described boundary.
For a rectangle with independent x and y limits, count the integer choices for each coordinate and multiply. If -1 <= x <= 1 and 0 <= y <= 1, there are 3 x-values and 2 y-values. Every combination works. Each row below represents one y choice and each column one x choice, giving 2 times 3 = 6 coordinate pairs.
This array counts combinations. It is not a coordinate grid.
Strict limits leave their boundary points out. For integer y, y > 0 starts at 1. Consider x >= 0, y > 0 and x + y < 3. Rearrange the last condition to y < 3 - x. At x = 0, y can be 1 or 2. At x = 1, only y = 1 works. At x >= 2, no positive integer y works. Total: 3 points.
The strict horizontal and sloping boundaries contribute no points.
When two sloping limits bound y, compare their lower and upper values in each column. For 0 <= x <= 2, y >= x and y <= 3 - x: at x = 0, y is 0 to 3, giving 4 points; at x = 1, y is 1 to 2, giving 2. At x = 2, no y can be both at least 2 and at most 1. Total: 6 points.
If the lower limit exceeds the upper limit, the column is empty.
Use dashed boundaries for < and >, solid for <= and >=. Test a point off each line. Keep only the overlap, and check every inequality when listing points.
The lesson, practice questions, worked solutions and flashcards were written for this dataset.
Readings, extra questions and flashcards written for this dataset. Video references: Modeling with systems of inequalities and Writing systems of inequalities word problem, Khan Academy.
More to read
Using a point to choose the shaded side
Draw y < 6 - 2x on paper. Begin with its boundary equation, y = 6 - 2x. Two points on this line are (0, 6) and (3, 0). Use a dashed line through them because the inequality excludes equality.
The point (1, 1) gives 1 < 4, a true comparison. Shade the side containing that point. At each fixed x, the allowed y-values are below the line's y-value.
Now check (2, 2). It gives 2 < 2, which is false. This point belongs to the boundary line but not to the inequality's solution. Changing < to <= would include the boundary and require a solid line.
To choose a side, use a point off the boundary. The origin is convenient only when it is off the line. For a boundary such as y = -x, the origin lies on it; (0, 1) is a possible alternative.
Direction depends on the rearranged comparison
Consider 3x - 2y < 10. Subtracting 3x gives -2y < 10 - 3x. Divide by -2 and reverse the sign: y > (3/2)x - 5. The solution is above a dashed boundary, even though the original comparison used <.
A substitution checks the direction. At (0, 0), the original comparison is 0 < 10, which is true. At x = 0, the boundary has y-value -5, so the origin is above it.
A vertical boundary needs left or right rather than above or below. The condition x < -3 selects the left side of x = -3 and excludes the line. A horizontal condition such as y >= 5 selects the line y = 5 and the region above it.
When reading a shaded region, work through its boundaries separately. Find each line's equation, choose the comparison matching the shaded side, then include equality only for a solid boundary. The resulting inequalities must all hold.
Listing points in the overlap
Use all the constraints when deciding whether a point belongs. For x >= 0, y >= 0 and x + y <= 1, the point (1, 1) satisfies the coordinate limits but fails the sum limit. The point (0, 1) satisfies all three.
List integer-coordinate solutions by columns. At x = 0, y can be 0 or 1. At x = 1, only y = 0 works. No larger integer x can satisfy the sum limit with y nonnegative. The complete list is (0, 0), (0, 1) and (1, 0).
Check whether each boundary is allowed before including its points. A strict condition can remove an entire edge from the list. When lower and upper y limits both vary with x, calculate both limits in each column. If the lower limit exceeds the upper limit, there are no points in that column.
An overlap need not be a bounded polygon. The conditions x >= 0 and y >= 0 extend without an upper limit. The conditions y >= 3 and y < 1 have no common solutions.
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