Graphing Inequalities

A solved inequality is a boundary together with a direction, and drawing it means being exact about both, which side of the boundary the solutions lie on and whether the boundary itself is one of them. On a number line those are a direction and a single circle, and in the plane they are a side of a line together with how that line is drawn.

Problem

The solutions of

4 x + 7 \geq 83

fill a ray on the number line, running right from a single point. Enter the number where the ray starts.

Show a hint

The point at the end of the ray is the value where the two sides come out exactly equal, so start by solving.
Subtracting $7$ from both sides leaves $4 x \geq 76$ , and $4$ is positive, so dividing by it leaves the direction alone.

Show the full solution

Subtract

7

from both sides for

4 x \geq 76

, then divide by the positive

4

for

x \geq 19

. At

x = 19

the left side is exactly

83

, and

83 \geq 83

is true, so the ray starts at

19

and includes that point.

Problem

On a number line the endpoint of a solution ray is drawn as an open circle when that endpoint is not a solution and as a filled circle when it is. Solve

6 x + 13 > 55

, put the boundary number you get back into the inequality, and enter the word, open or filled, for the circle at that endpoint.

Show a hint

The circle at the endpoint answers one yes or no question, whether the boundary number itself satisfies $6 x + 13 > 55$ .
Isolating $x$ gives $x > 7$ , so the endpoint is $7$ . Put $7$ into the left side and compare what you get against $55$ under a strict $>$ .

Show the full solution

Subtracting

13

gives

6 x > 42

, and dividing by

6

gives

x > 7

. Putting

7

back in gives

55 > 55

, which is false, so

7

is not a solution and its circle is

open

. Everything to the right of

7

is shaded and

7

itself is not. With

\geq

in place of

>

, that one point would be a solution and the circle would be filled.

Problem

The chain

- 5 < 2 x + 3 \leq 17

says both

- 5 < 2 x + 3

and

2 x + 3 \leq 17

at once, so its solutions graph as a single segment of the number line. Enter how many integers satisfy the chain.

Show a hint

Work on all three parts at each step, the same way 8.3 solved a chain, and keep track of which end carries the strict symbol.
Subtracting $3$ from all three parts gives $- 8 < 2 x \leq 14$ . Divide all three by the positive $2$ , then count the integers between the two ends, remembering that the end with the strict symbol is drawn open and is left out.

Show the full solution

Subtracting

3

from all three parts gives

- 8 < 2 x \leq 14

, and dividing all three by the positive

2

gives

- 4 < x \leq 7

. The integers on the segment run from

- 3

7

, and

7 - (- 3) + 1 = 11

. The end at

- 4

is drawn open, so

- 4

is left out, while the end at

7

is drawn filled, so

7

is counted.

Problem

The point

(9, 7)

lies on the line

y = 3 x - 20

. Enter yes if

(9, 7)

is a solution of

y < 3 x - 20

, and no if it is not.

Show a hint

Substitute the point, putting $9$ in for $x$ and $7$ in for $y$ , then check whether the resulting statement is true.
The point is on the line, so the two sides come out equal. The question is whether $<$ is true when the two sides are equal.

Show the full solution

Substituting gives

7 < 3 (9) - 20

, which is

7 < 7

, false, so the answer is

no

. A strict inequality is false wherever the two sides are equal, so no point of the line

y = 3 x - 20

is a solution, just as an open circle on a number line means the endpoint is not included.

The two planes have the same boundary and the same side is shaded, so the dash is the only difference between them. The left boundary is dashed because its own points are not solutions and the right one is solid because they are. Below the rule the same distinction appears one dimension down, an open circle where the endpoint is left out and a filled circle where it is kept.

With one letter the boundary is a single point. With two letters it is the set of points where the two sides come out equal, the line $y = m x + b$ that chapter 7 drew. Every other point of the plane sits on one side of it or the other, and two things are undecided, which side is the graph and whether the line belongs to it.

Problem

The line

y = 5 x - 14

splits the plane into two sides, and

y > 5 x - 14

is true at every point of one side and false at every point of the other. The origin is not on that line, so substituting

(0, 0)

settles which side is which. Enter the word above or below for the side that holds the solutions.

Show a hint

Substituting the origin puts $0$ in for both letters, which leaves only the constant term on the right.
A true test means the origin is one of the solutions, so the side you want is the side the origin sits on. Put $x = 0$ into $5 x - 14$ for the line's height there, then compare it with the origin's height of $0$ .

Show the full solution

Substituting

(0, 0)

gives

0 > - 14

, which is true, so the origin is one of the solutions. At

x = 0

the line is at

y = - 14

while the origin is at

y = 0

, so the origin's side is the one

above

. Testing one point settles the whole side, since the inequality is true at every point of one side and false at every point of the other.

Problem

The point

(12, 16)

lies on the line

y = \frac{5}{3} x - 4

, which is the boundary of

y \geq \frac{5}{3} x - 4

. A boundary is drawn dashed when its own points are not solutions and solid when they are. Decide whether

(12, 16)

is a solution, then enter the word, dashed or solid.

Show a hint

The word depends on one yes or no question, whether the points of the boundary line are solutions. Substitute $(12, 16)$ , a point of that line, and see whether the inequality holds.
The right side at $x = 12$ is $\frac{5}{3} (12) - 4$ , which is $20 - 4$ . Compare that with the point's $y$ -coordinate of $16$ .

Show the full solution

The right side at

x = 12

\frac{5}{3} (12) - 4 = 16

, the point's own

y

-coordinate, and

16 \geq 16

is true, so

(12, 16)

is a solution and the boundary is drawn

solid

. Every point of the line makes the two sides equal, so one check settles all of them. Under

>

and

<

equality is not allowed and the boundary is dashed, the number line's open and filled circles one dimension up.

The statement is true at every point of one side of the boundary and false at every point of the other, so substituting one point that is off the boundary settles the shaded side. The origin is the cheapest test, since substituting it puts zero in for both letters, and it works whenever the boundary line does not pass through it.

Problem

The boundary of

y > \frac{3}{2} x

is a line through the origin, so

(0, 0)

is not available as a test point. Enter the smallest integer

y

for which

(6, y)

is a solution.

Show a hint

Put $x = 6$ into the inequality. The right side becomes a single number, so what is left is an inequality in $y$ alone.
That number is $\frac{3}{2} (6)$ , and the symbol is strict, so the boundary value itself is not a solution. The answer is the first integer above it.

Show the full solution

x = 6

the inequality reads

y > \frac{3}{2} (6)

, which is

y > 9

, so the smallest integer is

10

. The origin lies on the boundary and settles nothing, while

(6, 10)

lies off it and satisfies the inequality, so the shaded side is the one above the line.

Problem

The inequality

3 x - 4 y > 20

is not solved for

y

. Enter the largest integer

y

for which the point

(8, y)

satisfies this inequality.

Show a hint

Substituting $x = 8$ leaves an inequality in $y$ alone. Check the sign of the coefficient of $y$ before dividing, since 8.1's rule about dividing by a negative applies.
That inequality is $24 - 4 y > 20$ , so $- 4 y > - 4$ . Dividing both sides by $- 4$ reverses the direction.

Show the full solution

Substituting

x = 8

gives

24 - 4 y > 20

, so

- 4 y > - 4

, and dividing by

- 4

reverses the direction to

y < 1

. The largest integer below

1

0

. Checking,

3 (8) - 4 (0) = 24 > 20

, while

3 (8) - 4 (1) = 20

is not above

20

. Keeping

>

through that division would give

y > 1

, which has no largest integer at all.

Several inequalities can hold at once, and a point is a solution only when it satisfies them all. Each has its own half-plane, and some have only one letter. The inequality $x \geq k$ says nothing about $y$ , so its graph is every point on or right of the vertical line $x = k$ , and $y < k$ graphs as everything below the horizontal line $y = k$ .

Problem

In the plane the graph of

x \geq - 6

is a half-plane. Enter how many of the points

(- 6, 25)

(- \frac{13}{2}, 0)

(3, - 41)

(- 7, - 7)

and

(- 5, 12)

satisfy

x \geq - 6

Show a hint

The inequality is a condition on $x$ alone, so compare each first coordinate with $- 6$ and ignore the second one entirely.
Two of the five are close calls. One turns on whether $\geq$ is true when the two sides are equal, and the other on how $- \frac{13}{2}$ compares with $- 6$ .

Show the full solution

Only the first coordinate matters.

- 6 \geq - 6

is true,

- \frac{13}{2} \geq - 6

is false,

3 \geq - 6

is true,

- 7 \geq - 6

is false, and

- 5 \geq - 6

is true, so the count is

3

. The graph is the vertical line

x = - 6

drawn solid together with everything to its right, and the second coordinate is free because

y

never appears in it.

Problem

A point is on the graph of two inequalities at once exactly when both substitutions come out true. Enter how many of

(3, 4)

(4, 7)

(5, 4)

(6, 5)

(2, 1)

and

(7, 6)

satisfy both

y < 2 x - 1

and

x + y \geq 7

Show a hint

Each inequality has its own half-plane, so every point needs two substitutions, and one false result is enough to leave it out.
Two of the points sit on a boundary. $(4, 7)$ makes the first read $7 < 7$ and $(3, 4)$ makes the second read $7 \geq 7$ , and only one of those two symbols is true when the sides are equal.

Show the full solution

(3, 4)

(5, 4)

(6, 5)

and

(7, 6)

satisfy both, so the count is

4

(4, 7)

gives

7 < 7

, false, and

(2, 1)

gives

3 \geq 7

, false. Of the two points on a boundary,

(3, 4)

is included and

(4, 7)

is not, since

x + y = 7

is drawn solid under

\geq

while

y = 2 x - 1

is drawn dashed under

<

The work is two moves per inequality. Draw the boundary, dashed for a strict symbol and solid otherwise, then test one point off it to decide which side is shaded. A region closed on every side contains finitely many points with integer coordinates, and those are counted one column at a time, fixing $x$ and reading which $y$ are allowed.

Problem

The points that satisfy all three of

x \geq 1

y \geq 2

and

3 x + 2 y \leq 20

form a triangular region, and the edges are part of the region. Enter how many points with integer coordinates lie in that region.

Show a hint

Count one vertical line at a time. Since $y \geq 2$ , every point in the region has $3 x \leq 20 - 4 = 16$ , so $x$ is at most $5$ .
Set $x = 1$ in the third inequality to get $2 y \leq 17$ , so $y$ is an integer from $2$ to $8$ . Do the same at each $x$ and add the counts.

Show the full solution

x = 1

the third inequality becomes

2 y \leq 17

, so

y

is an integer from

2

8

and that column has

7

points, and the same count at

x = 2, 3, 4, 5

6

4

3

and

1

, with nothing left at

x = 6

, so the total is

7 + 6 + 4 + 3 + 1 = 21

. At

x = 1

x = 3

and

x = 5

the bound on

y

is not a whole number, so the largest integer

y

is below it.

Problem

The points that satisfy

3 x + 4 y \leq 26

x - 2 y \geq - 3

and

y \geq 0

all at once form a triangular region. Two of its corners sit on the

x

-axis, and the third is where the boundary lines of the first two inequalities cross. Enter the

y

-coordinate of that third corner.

Show a hint

The two boundary lines are $3 x + 4 y = 26$ and $x - 2 y = - 3$ , and the corner is the one point on both of them, so finding it is a chapter 5 system.
The second equation gives $x = 2 y - 3$ . Substituting that into the first leaves an equation in $y$ alone.

Show the full solution

The boundaries are

3 x + 4 y = 26

and

x - 2 y = - 3

. The second gives

x = 2 y - 3

, and substituting gives

3 (2 y - 3) + 4 y = 26

, so

10 y - 9 = 26

and

y = \frac{7}{2}

. The corner is

(4, \frac{7}{2})

, and

y \geq 0

holds there, so it really is a corner of the region.

Every region in this lesson was drawn and then read, for a count of the points inside it or for one corner. Nothing here asked which point of a region is the best one. Lesson 8.5, Optimization, asks exactly that, and the corners found here are the points to check.

Practice these ideas

Practice

The solutions of

x - 14 > 9

fill a ray on the number line, and the ray does not include the number it starts at. Enter that number.

Show the solution

Adding

14

to both sides leaves the direction alone and gives

x > 23

. At

x = 23

the statement reads

9 > 9

, which is false, so the ray starts at

23

without including it.

Practice

Solve

5 x + 7 \leq - 13

. The solutions form a ray on the number line with a single circle at its endpoint. Enter one word, open or filled, for that circle.

Show the solution

Subtracting

7

gives

5 x \leq - 20

, and dividing both sides by the positive

5

leaves the direction alone, so

x \leq - 4

. At

x = - 4

the left side is exactly

- 13

, and

\leq

allows equality, so the circle is

filled

. Everything to the left of

- 4

is shaded, and

- 4

is included in the solution set.

Practice

The solutions of

- 7 < 3 x + 2 \leq 14

form a segment with one end included and the other excluded. Enter the number at the excluded end.

Show the solution

Subtracting

2

gives

- 9 < 3 x \leq 12

, and dividing by

3

gives

- 3 < x \leq 4

, so the excluded end is

- 3

. The strict symbol is at the lower end, so the open circle is at

- 3

and the filled one is at

4

Practice

Enter yes if

(6, - 1)

satisfies

y > \frac{1}{3} x - 5

, and no if it does not.

Show the solution

The right side is

\frac{1}{3} (6) - 5 = - 3

, and

- 1 > - 3

is true, so the answer is

yes

. On the number line

- 1

sits to the right of

- 3

, which is what makes it the larger of the two.

Practice

The graph of

y < 7 x + 2

is a half-plane. Enter one word, dashed or solid, for how its boundary line is drawn.

Show the solution

At a point of

y = 7 x + 2

the two sides are equal, and

<

is strict, so those points are not solutions and the line is drawn

dashed

. A dashed boundary means its points are excluded, the same convention as an open circle on a number line.

Practice

Test the origin in

y \leq - 4 x + 9

to settle which side of the boundary line

y = - 4 x + 9

holds the solutions. Enter the word above or below.

Show the solution

Substituting

(0, 0)

gives

0 \leq 9

, which is true, so the origin is a solution. At

x = 0

the origin is at

y = 0

while the line is at

y = 9

, so the solutions lie

below

. Checking the other side,

(0, 20)

gives

20 \leq 9

, which is false.

Practice

The graph of

y < 6

in the plane is a half-plane. Enter how many of the seven points

(11, 6)

(- 2, 5)

(0, - 30)

(4, \frac{13}{2})

(7, - 1)

(- 9, 0)

and

(15, 5)

satisfy

y < 6

Show the solution

Only the second coordinate matters, and the points with second coordinate below

6

are

(- 2, 5)

(0, - 30)

(7, - 1)

(- 9, 0)

and

(15, 5)

, so the count is

5

. The other two fail because

\frac{13}{2} = 6.5

is more than

6

, and

(11, 6)

is on the boundary line, where the strict

6 < 6

is false.

Practice

In the plane the graph of

3 x + 4 < 43

is a half-plane whose boundary is a vertical line. Enter the

x

-coordinate that every point of that boundary has.

Show the solution

Subtracting

4

gives

3 x < 39

, and dividing by the positive

3

gives

x < 13

. The shaded region is everything to the left of the line

x = 13

, and that line is drawn dashed, since the bound is strict.

Practice

The graph of

6 x + y \geq 5

is a half-plane, and its boundary line is part of it. Enter the smallest integer

y

for which

(- 2, y)

lies in that half-plane.

Show the solution

Substituting

x = - 2

gives

- 12 + y \geq 5

, so

y \geq 17

and the smallest integer available is

17

. The inequality is not strict, so

17

itself counts. Checking,

6 (- 2) + 17 = 5

, and

5 \geq 5

is true.

Practice

The graph of

2 x + 3 y \leq 31

is a half-plane. Enter the largest integer

y

for which the point

(5, y)

lies in it.

Show the solution

Substituting gives

10 + 3 y \leq 31

, so

3 y \leq 21

and

y \leq 7

. Checking,

2 (5) + 3 (7) = 31

, and

31 \leq 31

is true, so

7

is allowed.

Practice

Enter the smallest integer

y

for which the point

(3, y)

satisfies

4 x - 5 y \leq - 28

Show the solution

Substituting

x = 3

gives

12 - 5 y \leq - 28

, so

- 5 y \leq - 40

, and dividing by

- 5

reverses the direction to

y \geq 8

. At

y = 8

the left side is

12 - 40 = - 28

, and a

\leq

statement is true when the two sides are equal, so

8

itself works, while

y = 7

gives

- 23

, which is greater than

- 28

Practice

The origin lies on the boundary line of

y > \frac{5}{4} x

, so a test point has to come from somewhere off that line. Enter the smallest integer

y

for which

(20, y)

is a solution.

Show the solution

x = 20

the inequality reads

y > 25

, so the smallest integer is

26

. The point

(20, 26)

is then a usable test point, and it sits above the line, so the graph is the region above.

Practice

Enter how many of the six points

(0, 0)

(1, 6)

(5, 1)

(3, 5)

(2, 1)

and

(4, 0)

satisfy both

y \leq x + 2

and

x + y > 4

Show the solution

(5, 1)

gives

1 \leq 7

and

6 > 4

, and

(3, 5)

gives

5 \leq 5

and

8 > 4

, so both pass.

(1, 6)

fails the first, and

(0, 0)

(2, 1)

and

(4, 0)

fail the second, so the count is

2

. The two boundary cases come out differently because

\leq

includes its boundary line, drawn solid, while

>

excludes its own, drawn dashed.

Practice

The boundary lines of

x + y \leq 20

and

x - y \geq 4

cross at one point. Enter the

x

-coordinate of that point.

Show the solution

Adding

x + y = 20

and

x - y = 4

gives

2 x = 24

, so

x = 12

. The crossing is

(12, 8)

, and both inequalities hold there with equality, so it is a corner of the region they share.

Practice

Enter how many points with integer coordinates satisfy all three of

x \geq 0

y \geq 0

and

2 x + y \leq 6

Show the solution

x = 0

the allowed

y

run from

0

6

, which is

7

points, then

5

x = 1

3

x = 2

and

1

x = 3

, so the count is

7 + 5 + 3 + 1 = 16

. All three inequalities are non-strict, so the points on all three edges are counted.

Practice

The inequality

6 x - 3 y < 12

is graphed in the plane. Enter the word, above or below, for the side of its boundary line where the solutions lie.

Show the solution

Subtracting

6 x

gives

- 3 y < 12 - 6 x

, and dividing by

- 3

reverses the direction to

y > 2 x - 4

, so the solutions lie

above

. Testing the origin gives the true statement

0 < 12

, and the origin sits above

y = 2 x - 4

, which crosses the

y

-axis at

- 4

Practice

The boundary lines of

3 x - 2 y \geq - 4

and

x + 2 y \leq 16

cross at a corner of the region where both hold. Enter the

y

-coordinate of that corner.

Show the solution

Adding

3 x - 2 y = - 4

and

x + 2 y = 16

gives

4 x = 12

, so

x = 3

, and

3 + 2 y = 16

gives

y = \frac{13}{2}

. Both

\geq

and

\leq

allow equality, so their boundary lines are solid and the corner

(3, \frac{13}{2})

is itself in the region.