Algebra 1 · A.REI.D.12

The overlap of two shaded regions

One inequality keeps half of the plane. When two or more must hold at once, which points are left?

You need

Graphing a linear inequality in two variablesAlgebra 1 Compound inequalities: and, orAlgebra 1

Unlocks

Linear programmingAlgebra 2 Writing and interpreting constraintsAlgebra 1

Where you are 4 in Systems of linear inequalities

Algebra9

Systems of equations7

Systems of linear inequalities4

Graphing a linear inequality in two variables Linear programming

The overlap of two shaded regions

Writing and interpreting constraints

When you reach for it

One linear inequality in x and y, such as y < 2x + 1, is true on one side of a line and false on the other; Graphing a linear inequality in two variables draws it as a boundary line, solid or dashed, with the kept side shaded. A system of linear inequalities is two or more such inequalities, all required at once. A solution is an ordered pair that makes every one true, and the solution set is all of them.

This is the “and” of Compound inequalities: and, or moved from a line to a plane. There, x > -2 and x \le 3 each keep a ray and the answer is where the rays overlap. Here each inequality keeps a half-plane, and the solution set is where the half-planes overlap: where both shadings agree.

Unlike a system of equations (Solving systems by graphing), the answer is usually a whole region, and the crossing of two boundary lines is only a corner of it, which may or may not belong. Turning a situation into inequalities is the work of Writing and interpreting constraints; here they are given.

The procedure

Part 1 (the first day) puts two inequalities on one grid. Part 2 (the second day) adds more, which can close the region into a polygon, and finds its corners exactly.

Part 1: two inequalities on one grid

Graph both inequalities on the same axes. Each boundary is solid for \le or \ge (equality is allowed, so the line is kept) and dashed for < or >, and each kept side is shaded. A point solves the system exactly when it solves the first inequality and the second, so the solution set is exactly the part shaded twice.

One test point decides a side. For a line y = mx + b, the difference y - (mx + b) is zero on the line and cannot pass through zero without touching it, so it keeps one sign on each whole side. The origin is the easiest test point whenever the line misses it.

Crossing lines make four regions. Two boundaries with different slopes cut the plane into four regions. Inside one region you can move anywhere without crossing either line, so the whole region is in or out of each half-plane, and one test point classifies it. The four regions are the four combinations of sides, so exactly one is kept by both; a test point in it checks that neither shading went on the wrong side.

Which boundary points belong. A point on an edge of the region lies on one line and strictly inside the other half-plane, so it belongs exactly when that line is solid: on a dashed line the inequality would say a number is less than itself. The corner lies on both lines, so the corner belongs only when both lines are solid. A filled dot marks a corner that belongs, a hollow ring one that does not.

Parallel boundaries. Lines with the same slope (Parallel and perpendicular lines) never meet; they cut the plane into a strip and two half-planes. If the shadings face each other, the solution set is the strip. If they face away, nothing is shaded twice and the system has no solution. If they face the same way, the overlap is the smaller half-plane. The number line had the same three cases: x > 1 and x < 4, then x < 1 and x > 4, then x > 1 and x > 4.

Part 2: three or more inequalities, and their corners

Each new inequality cuts the region again. Two crossing boundaries leave a wedge that runs off the page. A third line can cut it into a triangle, a bounded region (one that fits inside some circle), and a fourth can leave a quadrilateral. Two inequalities alone never enclose a bounded region.

A corner solves two boundary equations. Every corner is where two boundary lines meet. Replace those two signs by = and the corner is the solution of a system of two linear equations, found by substitution when a variable is already alone or by elimination in standard form. A corner with fractional coordinates can only be guessed from a graph; the algebra gives it exactly.

Not every crossing is a corner. Four lines can cross in as many as six points, and some lie outside the region. A crossing is a corner only if it also satisfies every other inequality, so each candidate is tested before it is called a corner. It then belongs when every boundary line through it is solid.

x \ge 0 and y \ge 0 together keep the first quadrant, both axes included. They appear whenever x and y count or measure something that cannot be negative, and the corners they make are intercepts: set one variable to 0 in the other boundary equation.

Writing the system for a given region runs the method backwards. Find the equation of the line along each edge from two points on it. Substitute a point well inside the region: whichever of < and > is true there gives the direction, since the inside point is on the kept side. A solid edge takes \le or \ge, a dashed edge < or >.

Worked examples

Examples 1 to 3 belong to Part 1, Examples 4 to 7 to Part 2.

Example 1

Two crossing lines, one dashed

Graph the system

\begin{cases} y < 2x + 1 \\ y \ge -x + 4 \end{cases}

test one point in each of the four regions, and decide whether (1, 3), (4, 0) and (2, 5) are solutions.

Solution

Line A, y = 2x + 1, has slope 2 and y-intercept 1; the sign < makes it dashed. The origin gives 0 < 1, true, so the side below the line is shaded. Line B, y = -x + 4, has slope -1 and y-intercept 4; the sign \ge makes it solid. The origin gives 0 \ge 4, false, so the side above the line is shaded.

The lines meet where 2x + 1 = -x + 4, so 3x = 3, x = 1 and y = 3: the point (1, 3). The four points below lie one in each region.

Test point	y < 2x + 1	y \ge -x + 4	In both?
(0, 5)	5 < 1, false	5 \ge 4, true	no
(3, 5)	5 < 7, true	5 \ge 1, true	yes
(0, 0)	0 < 1, true	0 \ge 4, false	no
(-2, 0)	0 < -3, false	0 \ge 6, false	no

The solution set is the region holding (3, 5): the wedge to the right of (1, 3), below line A and on or above line B, shaded twice in Figure 1.

Figure 1Example 1. Each inequality shades its own side; the solution set is the darker wedge where both shadings agree. The corner (1, 3) is a hollow ring because line A is dashed.

(1, 3): 3 < 2(1) + 1 reads 3 < 3, false. Not a solution. It is the corner, and it lies on the dashed line A.
(4, 0): 0 < 9 and 0 \ge -4 + 4, both true. A solution, lying on the solid edge along B.
(2, 5): 5 < 5 is false, though 5 \ge 2 is true. Not a solution: it lies on the dashed edge.

Example 2

Both lines solid, from standard form

Graph the system

\begin{cases} x + 3y \le 9 \\ 2x - y \ge 4 \end{cases}

describe the solution set, and decide whether (6, 1) is a solution.

Solution

Intercepts draw standard-form lines fastest. Line A, x + 3y = 9, passes through (9, 0) and (0, 3), solid; the origin gives 0 \le 9, true, so the side below it is kept. Line B, 2x - y = 4, passes through (2, 0) and (0, -4), solid; the origin gives 0 \ge 4, false, so the side away from the origin is kept. Solved for y that reads y \le 2x - 4, since multiplying -y \ge -2x + 4 by -1 reverses the sign: the side below this steep line.

For the corner, line B gives y = 2x - 4, and substituting into line A,

x + 3(2x - 4) = 9, \qquad 7x = 21, \qquad x = 3

so y = 2. The inside point (4, 1) gives 7 \le 9 and 7 \ge 4. The solution set is the wedge below both lines, with its point at (3, 2) (Figure 2). Both lines are solid, so both edges belong, and so does the corner (3, 2).

Figure 2Example 2. Both boundaries are solid, so both edges and the corner (3, 2) belong to the solution set.

For (6, 1): 6 + 3 = 9 \le 9 and 12 - 1 = 11 \ge 4, so (6, 1) is a solution, on the solid edge along line A.

Example 3

Parallel boundaries

Describe the solution set of each system.

(a) \displaystyle y \ge -\frac{1}{2}x + 1 and \displaystyle y < -\frac{1}{2}x + 4 \qquad (b) x + 2y \le 2 and x + 2y > 8

Solution

(a) Both boundaries have slope \displaystyle -\frac{1}{2} and different intercepts, so they are parallel. The first is solid with the side above kept, the second dashed with the side below kept. The shadings face each other, so the solution set is the strip between the lines, lower edge included and upper edge not. A check: (0, 2) gives 2 \ge 1 and 2 < 4.

(b) Solved for y, the inequalities are \displaystyle y \le -\frac{1}{2}x + 1 and \displaystyle y > -\frac{1}{2}x + 4: the same two lines, but now the kept sides face away from each other (Figure 3). The system has no solution. The algebra says so directly: the one number x + 2y cannot be at most 2 and more than 8.

Figure 3Example 3. Parallel boundaries. In (a) the shadings face each other and overlap in a strip; in (b) they face away and never overlap.

Example 4

A triangle with one dashed side

Graph the system

\begin{cases} y \le x + 2 \\ y \ge -\frac{1}{2}x - 1 \\ 2x + y < 8 \end{cases}

find every corner of the region, and say which corners belong to the solution set.

Solution

The boundaries are y = x + 2 and \displaystyle y = -\frac{1}{2}x - 1, both solid, and 2x + y = 8, dashed. The origin satisfies all three (0 \le 2, 0 \ge -1, 0 < 8), so the region is the triangle around the origin (Figure 4). Its corners come from the three pairs of boundary equations.

First and second: \displaystyle x + 2 = -\frac{1}{2}x - 1, so \displaystyle \frac{3}{2}x = -3, giving P(-2, 0).
First and third: y = x + 2 into 2x + y = 8 gives 3x + 2 = 8, so Q(2, 4).
Second and third: \displaystyle 2x - \frac{1}{2}x - 1 = 8 gives \displaystyle \frac{3}{2}x = 9, so R(6, -4).

Each passes the inequality that did not make it: 2(-2) + 0 = -4 < 8, 4 \ge -2 and -4 \le 8.

Figure 4Example 4. Three inequalities close the region into a triangle. Only the corner P, where two solid lines meet, belongs.

P lies on two solid lines, so it belongs. Q and R lie on the dashed line, so neither belongs, nor does any point of the edge QR.

Example 5

A quadrilateral in the first quadrant

Find the corners of the region given by

x \ge 0, \qquad y \ge 0, \qquad x + y \le 7, \qquad 2x + y \le 10

Solution

The first two keep the first quadrant. The line x + y = 7 meets the axes at (7, 0) and (0, 7), the line 2x + y = 10 at (5, 0) and (0, 10), and the origin satisfies both, so each kept side is the origin’s. The slanted lines meet where, by elimination, (2x + y) - (x + y) = 10 - 7, so x = 3 and y = 4. Each crossing is tested in the inequalities that did not make it.

Crossing	Made by	Test	Corner?
(0, 0)	the two axes	0 \le 7 and 0 \le 10	yes
(5, 0)	2x + y = 10 and y = 0	5 + 0 \le 7	yes
(7, 0)	x + y = 7 and y = 0	2(7) + 0 = 14 \le 10 fails	no
(0, 7)	x + y = 7 and x = 0	0 + 7 \le 10	yes
(0, 10)	2x + y = 10 and x = 0	0 + 10 \le 7 fails	no
(3, 4)	the two slanted lines	3 \ge 0 and 4 \ge 0	yes

The region is the quadrilateral with corners (0, 0), (5, 0), (3, 4) and (0, 7), every edge and corner included (Figure 5). The crossings (7, 0) and (0, 10) are not corners because each breaks the third inequality, the one whose line does not pass through it.

Figure 5Example 5. The axes, x + y = 7 and 2x + y = 10 bound a quadrilateral. The points (7, 0) and (0, 10), where the lines meet the axes, lie outside it.

Example 6

Writing the system for a shaded region

Figure 6 shows a shaded triangle with corners (-2, -1), (4, -1) and (1, 5). Its bottom and right sides are solid and its left side is dashed. Write a system of inequalities whose solution set is this region.

Figure 6Example 6. Read the region and write the system that gives it.

Solution

The bottom side is the line y = -1. The left side joins (-2, -1) and (1, 5), slope \displaystyle \frac{6}{3} = 2, and 5 = 2(1) + b gives y = 2x + 3. The right side joins (4, -1) and (1, 5), slope \displaystyle \frac{6}{-3} = -2, and 5 = -2(1) + b gives y = -2x + 7.

The inside point (1, 1) has 1 > -1, so the region is above the bottom line. Both slanted lines give 5 at x = 1, and 1 < 5, so the region is below each. The dashed side takes a strict sign:

\begin{cases} y \ge -1 \\ y < 2x + 3 \\ y \le -2x + 7 \end{cases}

This matches the corners drawn: (4, -1) lies on two solid lines and is filled, while (-2, -1) and (1, 5) lie on the dashed line and are hollow.

Example 7

Delfina’s dog walks

Delfina walks dogs. A short walk takes half an hour and a long walk one hour. She walks at most 15 hours a week and has promised four regular clients a long walk each week. Let s and l be her numbers of short and long walks in a week. Write a system and find the corners of its region. Can she do 20 short and 5 long walks? What about 10 short and 12 long?

Solution

The hours give \displaystyle \frac{1}{2}s + l \le 15, and doubling both sides gives s + 2l \le 30. The promise gives l \ge 4, and a count cannot be negative, so s \ge 0 (l \ge 4 already covers l).

\begin{cases} s + 2l \le 30 \\ l \ge 4 \\ s \ge 0 \end{cases}

With s across and l up, the region lies below s + 2l = 30, on or above l = 4 and to the right of the l-axis: a triangle (Figure 7). Its corners are (0, 4); (22, 4), from s + 8 = 30; and (0, 15), from 2l = 30. All three lines are solid, so all three are possible plans: (22, 4) is the most short walks that fit beside the four long ones, (0, 15) a week of long walks only.

Figure 7Example 7. Every plan Delfina can keep lies in the triangle. The plan of 20 short and 5 long walks sits on the edge, using all 15 hours.

(20, 5): 20 + 10 = 30 \le 30 and 5 \ge 4. Yes, using exactly \displaystyle \frac{1}{2}(20) + 5 = 15 hours, on the edge. (10, 12): 10 + 24 = 34 > 30. No: it would take \displaystyle \frac{1}{2}(10) + 12 = 17 hours. Walks are counted, so only the whole-number points of the triangle are real plans.

Boundaries

Straight boundaries. Corners come from pairs of linear equations only when every boundary is a line. With a curved boundary such as y \ge x^2 the overlap and the test points still work, but corners need a system that is not linear.

A test point must miss every boundary. A point on the line says nothing about which side is kept: the origin cannot test y > 3x, while (1, 0) can.

Corners need both conditions. A crossing is a corner only if it satisfies every other inequality, and it belongs only if every boundary through it is solid.

The solution set can be empty. Parallel boundaries facing away have no overlap, and three inequalities can have none even when every pair of boundaries crosses: x \ge 0, y \ge 0, x + y \le -1.

A region need not be bounded. x \ge 0, y \ge 0 and x + y \ge 4 leave a region that runs off the page.

In context, only sensible points count, such as whole numbers of walks. The region says which plans are allowed, never which is best.

Where this goes next

Linear programming looks for the point of such a region that makes a profit or a cost as large or as small as possible, and the answer sits at a corner, so finding corners exactly is its first step. Writing and interpreting constraints starts from a situation, writes each limit as an inequality, and asks which points of the region make sense. Together with Graphing a linear inequality in two variables, these make up Systems of linear inequalities.

Exercises

Part 1: is the point a solution?

Is (2, 1) a solution of the system y < 3x - 2 and y \ge -x + 2?
Is (1, 1) a solution of the same system, y < 3x - 2 and y \ge -x + 2? Where does the point lie on the graph?

Part 1: graph the system

For each system, graph both inequalities on one grid, find the corner where the boundaries meet, say whether that corner belongs to the solution set, and give one point that is a solution.

y > x - 4 and y \le -2x + 2
\displaystyle y \ge \frac{1}{2}x + 1 and y \ge -x + 7
x + y < 5 and x - 2y \le 2
y \le 2 and x > -3
3x - 4y \ge 8 and y > -3

Part 1: parallel boundaries

Describe the solution set of each system.

y \ge 3x - 2 and y \le 3x + 4
y > -x + 2 and y < -x - 1
2x - y \le 3 and 2x - y < 7
For which values of c does the system y \ge 2x + c and y \le 2x + 3 have at least one solution? What is the solution set when c = 3?

Part 2: corners of a region

Graph each system, find every corner of the region, and say which corners belong to the solution set.

y \le x + 1, y \ge -x + 1 and x < 4
y \ge 1, y \le 2x - 3 and x + y \le 7
x \ge 0, y \ge 0, x + 2y \le 10 and 3x + y \le 15
x \ge -2, x \le 3, y \ge -1 and \displaystyle y \le \frac{1}{2}x + 3

Part 2: write the system

Write a system of inequalities whose solution set is the region described.

The triangle with corners (0, 0), (4, 0) and (0, 6), every side solid.
The strip between the line y = x + 1, which is solid, and the line y = x - 2, which is dashed.
The triangle with corners (0, 0), (6, 0) and (6, 3), where the side from (0, 0) to (6, 3) is dashed and the other two sides are solid.

Part 2: Delfina’s walks again

Delfina from Example 7 changes her plan. She drops the promise of long walks but wants at least 10 walks a week in total, still within 15 hours, with a short walk taking half an hour and a long walk one hour. Using s \ge 0 and l \ge 0, write the system, find the corners of the region, and decide whether 8 short and 3 long walks fit, and whether 25 short and 4 long walks fit.

Reasoning

Two boundary lines cross. Explain why testing one point in each of the four regions decides every point of the plane, not just the four points tested.
Two boundary lines cross at a point P. Explain why P belongs to the solution set when both lines are solid, and why it does not when either line is dashed.
Explain why a system of two linear inequalities whose boundary lines cross always has solutions. Then explain why the system x \ge 0, y \ge 0, x + y \le -1 has none.

Answers

Yes. 1 < 3(2) - 2 = 4 and 1 \ge -2 + 2 = 0.
No. 1 < 3(1) - 2 reads 1 < 1, false. The point lies on both lines, so it is the corner; the first line is dashed, so the corner is left out.
Corner (2, -2), from x - 4 = -2x + 2. It does not belong: y = x - 4 is dashed. The origin is a solution.
Corner (4, 3), from \displaystyle \frac{3}{2}x = 6. It belongs, since both lines are solid. The region lies on or above both lines; (4, 6) is a solution.
Corner (4, 1): subtracting, 3y = 3. It does not belong, because x + y = 5 is dashed. The origin is a solution.
Corner (-3, 2). It does not belong, because x = -3 is dashed. The origin is a solution.
Corner \displaystyle \left(-\frac{4}{3}, -3\right): with y = -3, 3x + 12 = 8. It does not belong, because y = -3 is dashed. Dividing -4y \ge -3x + 8 by -4 reverses the sign: \displaystyle y \le \frac{3}{4}x - 2. (4, 0) is a solution.
The strip between y = 3x - 2 and y = 3x + 4, both lines included.
No solution: -x - 1 is less than -x + 2 for every x, so no y is above the second line and below the first.
The half-plane 2x - y \le 3, on or above y = 2x - 3. Every such point also has 2x - y < 7, so the second inequality removes nothing.
c \le 3, since a solution needs 2x + c \le y \le 2x + 3. When c = 3 the solution set is the line y = 2x + 3 itself.
Corners (0, 1), (4, 5) and (4, -3). Only (0, 1) belongs; the other two lie on the dashed line x = 4.
Corners (2, 1), (6, 1) and \displaystyle \left(\frac{10}{3}, \frac{11}{3}\right), all belonging. The last comes from y = 2x - 3 in x + y = 7: 3x = 10.
Corners (0, 0), (5, 0), (4, 3) and (0, 5), all belonging. With x = 10 - 2y in 3x + y = 15: 5y = 15. The intercepts (10, 0) and (0, 15) fail the other slanted inequality, so they are not corners.
Corners (-2, -1), (3, -1), \displaystyle \left(3, \frac{9}{2}\right) and (-2, 2), all belonging. The slanted line meets y = -1 at x = -8, which fails x \ge -2.
x \ge 0, y \ge 0, 3x + 2y \le 12. The slanted side is \displaystyle y = -\frac{3}{2}x + 6, and (1, 1) inside gives 5 \le 12.
y \le x + 1 and y > x - 2, checked with (0, 0).
y \ge 0, x \le 6, \displaystyle y < \frac{1}{2}x, checked with (5, 1). Only the corner (6, 0) belongs; the other two lie on the dashed side.
s + l \ge 10, s + 2l \le 30, s \ge 0, l \ge 0. Corners (10, 0), (30, 0), (0, 15) and (0, 10), all belonging; s + l = 10 and s + 2l = 30 meet at (-10, 20), which fails s \ge 0. (8, 3) fits: 11 \ge 10 and 14 \le 30. (25, 4) does not: 33 > 30, which would be \displaystyle \frac{33}{2} = 16.5 hours.
Within one region a point can move anywhere without crossing either line, and an inequality changes from true to false only on its boundary, where its two sides are equal. So every point of a region answers each inequality as its test point does. Points on the lines are decided by solid or dashed.
P lies on both lines, so in each inequality its two sides are equal. Equality satisfies \le and \ge (solid) and fails < and > (dashed). P must satisfy both, so both lines must be solid.
Crossing lines make four regions, one for each choice of side of each line, so one region is on the kept side of both and holds solutions. In the second system, x \ge 0 and y \ge 0 force x + y \ge 0, which is never at most -1.