Algebra 2 Systems Of Inequalities

3.seven: Solving Systems of Inequalities with Two Variables

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Learning Objectives

Bank check solutions to systems of inequalities with two variables.
Graph solution sets of systems of inequalities.

Solutions to Systems of Inequalities

A system of inequalities ³³ consists of a gear up of two or more inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For case,

$\left\{ \brainstorm{array} { l } { y > 10 - 2 } \\ { y \leq 2 x + 2 } \finish{assortment} \right.$

We know that each inequality in the set contains infinitely many ordered pair solutions defined past a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets will define the set up of simultaneous ordered pair solutions. When we graph each of the above inequalities separately we have:

And when graphed on the same set of axes, the intersection can be determined.

The intersection is shaded darker and the final graph of the solution gear up will exist presented every bit follows:

The graph suggests that $(3, ii)$ is a solution considering it is in the intersection. To verify this, nosotros can show that it solves both of the original inequalities equally follows:

Table $\PageIndex{i}$
$\colour{Cerulean}{Check :}\:\:\color{YellowOrange}{(three,2)}$
Inequality $one$: $\begin{assortment} { l } { y > ten - 2 } \\ { \color{Cerulean}{2}\color{blackness}{ >}\colour{Cerulean}{ 3}\colour{black}{ -} 2 } \\ { 2 > one }\:\: \color{Cerulean}{✓} \stop{array}$	Inequality $2$: $\begin{assortment} { fifty } { y \leq 2 x + ii } \\ { \colour{Cerulean}{ii}\color{black}{ \leq} 2 (\color{Cerulean}{ iii}\colour{blackness}{ )} + 2 } \\ { ii \leq eight } \:\:\color{Cerulean}{✓} \finish{assortment}$

Points on the solid boundary are included in the set of simultaneous solutions and points on the dashed purlieus are non. Consider the signal $(−1, 0)$ on the solid purlieus defined by $y = 2x + 2$ and verify that it solves the original system:

Table $\PageIndex{2}$
$\color{Cerulean}{Cheque :}\:\:\color{YellowOrange}{(-one,0)}$
Inequality $one$: $\begin{assortment} { l } { y > x - ii } \\ { \color{Cerulean}{0}\color{black}{ >}\color{Cerulean}{ -i}\color{black}{ -} 2 } \\ { 0 > -iii }\:\: \colour{Cerulean}{✓} \end{array}$	Inequality $ii$: $\begin{array} { fifty } { y \leq 2 x + 2 } \\ { \color{Cerulean}{0}\color{blackness}{ \leq} 2 (\color{Cerulean}{ -1}\color{black}{ )} + ii } \\ { 0 \leq 0 } \:\:\color{Cerulean}{✓} \end{array}$

Find that this betoken satisfies both inequalities and thus is included in the solution set. Now consider the point $(ii, 0)$ on the dashed purlieus divers by $y = x − two$ and verify that it does not solve the original system:

Table $\PageIndex{3}$
$\color{Cerulean}{Bank check :}\:\:\colour{YellowOrange}{(two,0)}$
Inequality $one$: $\brainstorm{assortment} { l } { y > x - 2 } \\ { \color{Cerulean}{0}\color{black}{ >}\color{Cerulean}{ two}\color{black}{ -} ii } \\ { 0 > 0 } \:\:\color{red}{✗}\end{array}$	Inequality $2$: $\begin{array} { l } { y \leq ii x + 2 } \\ {\color{Cerulean}{ 0}\color{blackness}{ \leq} ii (\color{Cerulean}{ 2}\color{black}{ )} + 2 } \\ { 0 \leq vi } \:\:\colour{Cerulean}{✓}\end{array}$

This signal does not satisfy both inequalities and thus is not included in the solution set.

Instance $\PageIndex{1}$

Determine whether or not $(-3,iii)$ is a solution to the following organisation:

$\left\{ \begin{aligned} ii x + 6 y \leq 6 \\ - \frac { 1 } { 3 } 10 - y \leq three \end{aligned} \right.$

Solution

Substitute the coordinates of $(ten, y) = (−3, 3)$ into both inequalities.

Table $\PageIndex{4}$
$\color{Cerulean}{Check :}\:\:\color{YellowOrange}{(-3,3)}$
Inequality $1$: $\brainstorm{aligned} 2 ten + 6 y \leq 6 \\ two ( \colour{Cerulean}{- 3}\color{blackness}{ )} + half-dozen ( \colour{Cerulean}{three}\color{black}{ )} \leq 6 \\ - 6 + 18 \leq half dozen \\ 12 \leq six \:\:\color{cherry}{✗}\terminate{aligned}$	Inequality $ii$: $\begin{aligned} - \frac { one } { 3 } ten - y & \leq three \\ - \frac { i } { 3 } ( \color{Cerulean}{- 3}\color{blackness}{ )} - (\color{Cerulean}{ iii}\color{black}{ )} & \leq 3 \\ one - 3 & \leq 3 \\ - 2 & \leq 3 \:\:\color{Cerulean}{✓}\finish{aligned}$

Respond

$(-3,3)$ is not a solution; it does not satisfy both inequalities.

We can graph the solutions of systems that contain nonlinear inequalities in a similar manner. For example, both solution sets of the following inequalities can exist graphed on the same prepare of axes:

$\left\{ \brainstorm{array} { l } { y < \frac { one } { two } 10 + 4 } \\ { y \geq x ^ { ii } } \end{array} \correct.$

And the intersection of both regions contains the region of simultaneous ordered pair solutions.

From the graph, nosotros wait the ordered pair $(1, three)$ to solve both inequalities.

Table $\PageIndex{5}$
$\colour{Cerulean}{Check :}\:\:\color{black}{(one,3)}$
Inequality $1$: $\brainstorm{array} { l } { y < \frac { 1 } { 2 } x + 4 } \\ { \color{Cerulean}{3}\color{black}{ <} \frac { 1 } { 2 } ( \colour{Cerulean}{one}\color{blackness}{ )} + iv } \\ { iii < 4 \frac { i } { 2 } \quad } \:\:\color{Cerulean}{✓}\finish{array}$	Inequality $2$: $\begin{assortment} { fifty } { y \geq x ^ { two } } \\ { \colour{Cerulean}{3}\color{black}{ \geq} ( \color{Cerulean}{i}\color{black}{ )} ^ { 2 } } \\ { 3 \geq 1 } \:\:\color{Cerulean}{✓}\terminate{array}$

Graphing Solutions to Systems of Inequalities

Solutions to a system of inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems we graph the solution sets of the inequalities on the aforementioned set of axes and determine where they intersect. This intersection, or overlap, will define the region of mutual ordered pair solutions.

Example $\PageIndex{two}$:

Graph the solution set: $\left\{ \brainstorm{array} { l } { - 2 10 + y > - four } \\ { 3 x - vi y \geq half-dozen } \end{array} \right.$.

Solution

To facilitate the graphing process, we commencement solve for $y$.

$\left\{ \brainstorm{assortment} { l 50 } { - 2 10 + y > - iv } \\ { three ten - 6 y \geq 6 } \end{array} \right. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > 2 x - four } \\ { y \leq \frac { 1 } { 2 } x - 1 } \stop{assortment} \right.$

For the first inequality, we utilise a dashed purlieus defined by $y = 2x − 4$ and shade all points higher up the line. For the second inequality, we use a solid purlieus defined by $y = \frac{ane}{ 2} x − ane$ and shade all points below. The intersection is darkened.

Now we nowadays our solution with only the intersection shaded.

Respond:

Example $\PageIndex{3}$:

Graph the solution prepare: $\left\{ \begin{array} { l } { - 3 x + ii y > 6 } \\ { half dozen x - 4 y > 8 } \finish{assortment} \right.$.

Solution

Nosotros brainstorm by solving both inequalities for $y$.

$\left\{ \begin{array} { l } { - 3 x + ii y > vi } \\ { vi x - 4 y > eight } \cease{array} \right. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > \frac { 3 } { 2 } ten + 3 } \\ { y < \frac { 3 } { 2 } x - 2 } \end{array} \right.$

Because of the strict inequalities, we will use a dashed line for each boundary. For the starting time inequality shade all points above the boundary and for the second inequality shade all points below the boundary.

As we can come across, in that location is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.

Respond:

$\varnothing$

Example $\PageIndex{iv}$:

Graph the solution set: $\left\{ \brainstorm{assortment} { 50 } { y \geq - 4 } \\ { y < x + 3 } \\ { y \leq - three 10 + three } \cease{array} \right.$

Solution

Begin by graphing the solution sets to all three inequalities.

Afterward graphing all three inequalities on the aforementioned set of axes, we determine that the intersection lies in the triangular region pictured below.

Respond:

The graph suggests that $(−1, 1)$ is a simultaneous solution. As a check, we could substitute that indicate into the inequalities and verify that it solves all three conditions.

Table $\PageIndex{6}$
$\color{Cerulean}{Bank check:}\:\:\color{black}{(-ane,1)}$
Inequality $i$: $\brainstorm{array} { l } { y \geq - 4 } \\ { \color{Cerulean}{ane}\color{black}{ \geq} - four }\:\:\color{Cerulean}{✓} \end{assortment}$	Inequality $2$: $\brainstorm{array} { fifty } { y < ten + 3 } \\ { \color{Cerulean}{1}\color{blackness}{ <}\color{Cerulean}{ - 1}\colour{black}{ +} 3 } \\ { 1 < 2 } \:\:\color{Cerulean}{✓} \end{assortment}$	Inequality $3$: $\begin{array} { fifty } { y \leq -3x + three } \\ { \color{Cerulean}{one}\color{black}{\leq} -3(\color{Cerulean}{ - 1}\color{black}{ )+} 3 } \\ { 1 \leq 3+3 } \\{1 \leq 6}\:\:\color{Cerulean}{✓} \cease{assortment}$

Employ the same technique to graph the solution sets to systems of nonlinear inequalities.

Example $\PageIndex{5}$:

Graph the solution prepare: $\left\{ \brainstorm{array} { 50 } { y < ( x + 1 ) ^ { 2 } } \\ { y \leq - \frac { i } { ii } x + three } \end{array} \right.$.

Solution

The showtime inequality has a parabolic boundary. This boundary is a horizontal translation of the basic office $y = ten^{2}$ to the left $1$ unit. Because of the strict inequality, the boundary is dashed, indicating that it is not included in the solution set up. The 2d inequality is linear and will be graphed with a solid boundary. Solution sets to both are graphed beneath.

After graphing the inequalities on the aforementioned set of axes, we decide that the intersection lies in the region pictured below.

Respond:

Exercise $\PageIndex{1}$

Graph the solution set up: $\left\{ \begin{assortment} { fifty } { y \geq - | x + 1 | + iii } \\ { y \leq 2 } \end{array} \right.$.

Reply: $5906eac1da9f9dd2ddcfa5b88d2fde17.png$
Effigy $\PageIndex{13}$

www.youtube.com/v/9z87e7Iw9JE

Fundamental Takeaways

To graph solutions to systems of inequalities, graph the solution sets of each inequality on the same gear up of axes and determine where they intersect.
You tin check your reply by choosing a few values inside and out of the shaded region to see if they satisfy the inequalities or not. While this is not a proof, doing so will requite a good indication that you have graphed the correct region.

Practice $\PageIndex{2}$

Make up one's mind whether or not the given point is a solution to the given organization of inequalities.

1. $(-2,i)$;

$\left\{ \begin{assortment} { fifty } { y > 3 x + 5 } \\ { y \leq - x + one } \end{array} \correct.$

2. $(-one,-iii)$;

$\left\{ \brainstorm{assortment} { l } { y \geq 3 x - 1 } \\ { y < - 2 x } \end{array} \right.$

three. $(-2,-ane)$;

$\left\{ \begin{assortment} { l } { ten - 2 y > - 1 } \\ { iii x - y < - 3 } \finish{array} \right.$

four. $(0,-5)$;

$\left\{ \begin{assortment} { c } { 5 x - y \geq 5 } \\ { 3 x + 2 y < - 1 } \terminate{array} \right.$

5. $(-\frac{1}{2} ,0)$;

$\left\{ \begin{assortment} { 50 } { - viii x + 5 y \geq 3 } \\ { 2 x - three y < 0 } \terminate{array} \right.$

6. $(-i, \frac{1}{2})$;

$\left\{ \begin{assortment} { l } { two ten - 9 y < - i } \\ { three ten - 6 y > - 2 } \end{array} \correct.$

7. $(-one,-2)$;

$\left\{ \begin{array} { c } { 2 x - y \geq - 1 } \\ { x - three y < 6 } \\ { two x - 3 y > - i } \end{array} \right.$

8. $(-v,two)$;

$\left\{ \begin{array} { c } { - x + five y > 10 } \\ { 2 x + y < ane } \\ { 10 + iii y < - two } \stop{array} \right.$

9. $(0,3)$;

$\left\{ \begin{array} { l } { y + 4 \geq 0 } \\ { \frac { i } { two } x + \frac { 1 } { 3 } y \leq 1 } \\ { - 3 x + two y \leq 6 } \finish{array} \right.$

10. $(1,i)$;

$\left\{ \begin{array} { l } { y \leq - \frac { 3 } { four } x + ii } \\ { y \geq - 5 x + two } \\ { y \geq \frac { 1 } { 3 } x - 1 } \cease{array} \right.$

xi. $(-1,2)$;

$ \left\{ \begin{array} { l } { y \geq 10 ^ { 2 } + 1 } \\ { y < - 2 x + 3 } \end{array} \right.$

12. $(4,5)$;

$\left\{ \begin{array} { fifty } { y < ( x - ane ) ^ { 2 } - 1 } \\ { y > \frac { ane } { two } x - 1 } \end{array} \correct.$

13. $(-2,-3)$;

$\left\{ \begin{array} { l } { y < 0 } \\ { y \geq - | x | + 4 } \stop{assortment} \right.$

14. $(1,2)$;

$\left\{ \begin{array} { 50 } { y < | x - iii | + 2 } \\ { y \geq 2 } \end{array} \right.$

15. $\left( - \frac { ane } { two } , - 5 \right)$;

$\left\{ \begin{array} { l } { y \leq - three x - five } \\ { y > ( 10 - 1 ) ^ { two } - ten } \terminate{assortment} \right.$

sixteen. $(-4,one)$

$\left\{ \begin{array} { l } { x \geq - 5 } \\ { y < ( ten + 3 ) ^ { ii } - ii } \cease{array} \right.$

17. $\left( - \frac { 3 } { ii } , \frac { one } { 3 } \right)$;

$\left\{ \brainstorm{array} { l } { 10 - 2 y \leq iv } \\ { y \leq | 3 ten - 1 | + 2 } \end{array} \right.$

18, $\left( - iii , - \frac { iii } { 4 } \right)$;

$\left\{ \brainstorm{array} { l } { three x - 4 y < 24 } \\ { y < ( ten + 2 ) ^ { 2 } - 1 } \finish{array} \right.$

19. $(4,2)$;

$\left\{ \begin{array} { l } { y < ( x - 3 ) ^ { 2 } + 1 } \\ { y < - \frac { 3 } { 4 } 10 + v } \finish{array} \right.$

twenty. $(\frac{5}{2}, i)$

$\left\{ \brainstorm{assortment} { l } { y \geq - ane } \\ { y < - ( ten - ii ) ^ { ii } + three } \end{array} \correct.$

Answer

1. Yep

3. Yep

5. Aye

7. Aye

9. Yes

11. Yes

13. No

15. Yep

17. Yes

19. No

Exercise $\PageIndex{3}$

Graph the solution set.

$\left\{ \begin{array} { l } { y \geq \frac { 2 } { iii } 10 - 3 } \\ { y < - \frac { 1 } { 3 } x + 3 } \stop{array} \right.$
$\left\{ \brainstorm{array} { l } { y \geq - \frac { one } { iv } ten + 1 } \\ { y < \frac { one } { 2 } x - 2 } \end{assortment} \correct.$
$\left\{ \begin{array} { fifty } { y > \frac { ii } { 3 } x + one } \\ { y > \frac { four } { iii } x - 5 } \end{array} \right.$
$\left\{ \begin{array} { 50 } { y \leq - five x + 4 } \\ { y < \frac { 4 } { 3 } 10 - 2 } \end{assortment} \right.$
$\left\{ \begin{array} { l } { x - y \geq - 3 } \\ { x + y \geq 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { 3 x + y < 4 } \\ { two 10 - y \leq ane } \end{array} \right.$
$\left\{ \begin{assortment} { l } { - x + 2 y \leq 0 } \\ { 3 x + 5 y < 15 } \end{array} \right.$
$\left\{ \begin{array} { c } { 2 x + 3 y < vi } \\ { - iv x + iii y \geq - 12 } \end{array} \right.$
$\left\{ \begin{assortment} { l } { 3 x + 2 y > one } \\ { 4 x - 2 y > three } \stop{array} \right.$
$\left\{ \begin{array} { l } { x - 4 y \geq ii } \\ { eight 10 + 4 y \leq 3 } \end{assortment} \right.$
$\left\{ \begin{array} { fifty } { five x - ii y \leq half dozen } \\ { - 5 x + two y < 2 } \end{array} \correct.$
$\left\{ \begin{array} { l } { 12 10 + 10 y > 20 } \\ { 18 10 + 15 y < - 15 } \terminate{array} \correct.$
$\left\{ \begin{array} { l } { x + y < 0 } \\ { y + 4 > 0 } \cease{assortment} \right.$
$\left\{ \begin{array} { l } { ten > - iii } \\ { y < 1 } \end{array} \right.$
$\left\{ \begin{array} { fifty } { ii x - ii y < 0 } \\ { 3 10 - three y > three } \end{assortment} \right.$
$\left\{ \brainstorm{array} { l } { y + ane \leq 0 } \\ { y + 3 \geq 0 } \end{array} \correct.$
Construct a arrangement of linear inequalities that describes all points in the kickoff quadrant.
Construct a system of linear inequalities that describes all points in the second quadrant.
Construct a organization of linear inequalities that describes all points in the third quadrant.
Construct a system of linear inequalities that describes all points in the fourth quadrant.

Answer

seven.

11.

13.

15. $\varnothing$

17. $\left\{ \brainstorm{array} { l } { x > 0 } \\ { y > 0 } \terminate{array} \right.$

xix. $\left\{ \brainstorm{array} { 50 } { 10 < 0 } \\ { y < 0 } \end{assortment} \right.$

Practice $\PageIndex{4}$

Graph the solution fix.

$\left\{ \begin{array} { l } { y \geq - \frac { 1 } { 2 } x + 3 } \\ { y \geq \frac { three } { ii } x - 3 } \\ { y \leq \frac { 3 } { 2 } x + 1 } \cease{array} \right.$
$\left\{ \begin{array} { l } { y \leq - \frac { 3 } { 4 } x + 2 } \\ { y \geq - five ten + 2 } \\ { y \geq \frac { one } { 3 } x - 1 } \stop{assortment} \correct.$
$\left\{ \begin{assortment} { l } { 3 ten - 2 y > 6 } \\ { 5 x + 2 y > 8 } \\ { - 3 10 + iv y \leq 4 } \terminate{assortment} \right.$
$\left\{ \begin{assortment} { fifty } { 3 x - 5 y > - 15 } \\ { 5 x - 2 y \leq 8 } \\ { x + y < - 1 } \cease{array} \right.$
$\left\{ \brainstorm{array} { l } { 3 x - 2 y < - 1 } \\ { five x + ii y > 7 } \\ { y + ane > 0 } \terminate{array} \correct.$
$\left\{ \brainstorm{array} { l } { 3 ten - 2 y < - 1 } \\ { 5 x + 2 y < 7 } \\ { y + ane > 0 } \end{array} \right.$
$\left\{ \begin{array} { 50 } { 4 10 + 5 y - 8 < 0 } \\ { y > 0 } \\ { x + 3 > 0 } \cease{assortment} \right.$
$\left\{ \begin{array} { l } { y - 2 < 0 } \\ { y + two > 0 } \\ { 2 x - y \geq 0 } \cease{array} \correct.$
$\left\{ \begin{assortment} { l } { \frac { 1 } { 2 } x + \frac { 1 } { 2 } y < 1 } \\ { x < 3 } \\ { - \frac { i } { ii } ten + \frac { one } { 2 } y \leq 1 } \end{array} \right.$
$\left\{ \begin{assortment} { l } { \frac { ane } { 2 } 10 + \frac { 1 } { three } y \leq 1 } \\ { y + 4 \geq 0 } \\ { - \frac { one } { 2 } x + \frac { 1 } { three } y \leq one } \end{array} \correct.$
$\left\{ \begin{assortment} { 50 } { y < x + ii } \\ { y \geq 10 ^ { 2 } - three } \terminate{array} \right.$
$\left\{ \begin{assortment} { l } { y \geq x ^ { 2 } + i } \\ { y > - \frac { 3 } { 4 } x + 3 } \end{array} \right.$
$\left\{ \begin{array} { l } { y \leq ( x + 2 ) ^ { 2 } } \\ { y \leq \frac { 1 } { 3 } x + 4 } \end{array} \correct.$
$\left\{ \begin{array} { l } { y < ( x - 3 ) ^ { ii } + i } \\ { y < - \frac { 3 } { 4 } 10 + five } \end{array} \right.$
$\left\{ \begin{array} { l } { y \geq - 1 } \\ { y < - ( x - 2 ) ^ { two } + 3 } \stop{array} \correct.$
$\left\{ \brainstorm{assortment} { 50 } { y < - ( x + 1 ) ^ { 2 } - 1 } \\ { y < \frac { three } { 2 } x - 2 } \finish{array} \right.$
$\left\{ \begin{array} { 50 } { y \leq \frac { 1 } { three } x + 3 } \\ { y \geq | 10 + iii | - 2 } \end{array} \right.$
$\left\{ \begin{assortment} { l } { y \leq - x + 5 } \\ { y > | x - 1 | + ii } \end{array} \right.$
$\left\{ \begin{assortment} { l } { y > - | 10 - 2 | + five } \\ { y > ii } \terminate{array} \right.$
$\left\{ \begin{array} { l } { y \leq - | x | + 3 } \\ { y < \frac { 1 } { 4 } x } \end{array} \right.$
$\left\{ \begin{array} { l } { y > | x | + ane } \\ { y \leq 10 - 1 } \end{assortment} \right.$
$\left\{ \brainstorm{array} { 50 } { y \leq | 10 | + one } \\ { y > 10 - one } \stop{array} \right.$
$\left\{ \begin{array} { 50 } { y \leq | x - iii | + 1 } \\ { x \leq 2 } \end{array} \right.$
$\left\{ \begin{array} { 50 } { y > | ten + one | } \\ { y < x - two } \stop{array} \right.$
$\left\{ \begin{assortment} { l } { y < 10 ^ { three } + two } \\ { y \leq ten + 3 } \terminate{array} \right.$
$\left\{ \begin{array} { fifty } { y \leq 4 } \\ { y \geq ( x + iii ) ^ { 3 } + ane } \end{assortment} \right.$
$\left\{ \brainstorm{array} { fifty } { y \geq - 2 x + vi } \\ { y > \sqrt { x } + three } \finish{array} \right.$
$\left\{ \brainstorm{array} { l } { y \leq \sqrt { ten + 4 } } \\ { x \leq - one } \end{assortment} \right.$
$\left\{ \brainstorm{array} { l } { y \leq - 10 ^ { 2 } + 4 } \\ { y \geq x ^ { 2 } - 4 } \stop{array} \right.$
$\left\{ \brainstorm{array} { l } { y \geq | x - i | - 3 } \\ { y \leq - | x - 1 | + 3 } \end{array} \right.$

Answer

one.

seven.

nine.

11.

13.

15.

17.

19.

21. $\varnothing$

23.

25.

27.

29.

Footnotes

³³A prepare of two or more inequalities with the same variables.