3 3 systems of inequalities answer key

Graphically, a system with no solution is represented by three planes with no point in common. Now we present our solution with only the intersection shaded. 3. The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. A. IMAG0006.JPG In this example, the ordered pair (4, 7) is the solution to the system of linear equations. Have questions or comments? Adopted a LibreTexts for your class? Key Takeaways. IMAG0010.JPG The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. H. 7. The ninth graders are hosting the next school dance. J. We can graph the solutions of systems that contain nonlinear inequalities in a similar manner. IMAG0008.JPG Check solutions to systems of inequalities with two variables. Graph the solution set: $\left\{ \begin{array} { l } { y < ( x + 1 ) ^ { 2 } } \\ { y \leq - \frac { 1 } { 2 } x + 3 } \end{array} \right.$. As we can see, there is no intersection of these two shaded regions. IMAG0012.JPG 11. Graph solution sets of systems of inequalities. For the first inequality, we use a dashed boundary defined by $y = 2x − 4$ and shade all points above the line. IMAG0013.JPG You can check your answer by choosing a few values inside and out of the shaded region to see if they satisfy the inequalities or not. IMAG0009.JPG Key Takeaways Key Points. Determine whether or not the given point is a solution to the given system of inequalities. Solutions to a system of inequalities are the ordered pairs that solve all the inequalities in the system. IMAG0010.JPG You can check your answer by choosing a few values inside and out of the shaded region to see if they satisfy the inequalities or not. We begin by solving both inequalities for $y$. $\left\{ \begin{array} { l } { x < 0 } \\ { y < 0 } \end{array} \right.$. Begin by graphing the solution sets to all three inequalities. 10. J. Answer Key. IMAG0022.JPG IMAG0011.JPG $\left\{ \begin{array} { l } { y > 3 x + 5 } \\ { y \leq - x + 1 } \end{array} \right.$, $\left\{ \begin{array} { l } { y \geq 3 x - 1 } \\ { y < - 2 x } \end{array} \right.$, $\left\{ \begin{array} { l } { x - 2 y > - 1 } \\ { 3 x - y < - 3 } \end{array} \right.$, $\left\{ \begin{array} { c } { 5 x - y \geq 5 } \\ { 3 x + 2 y < - 1 } \end{array} \right.$, $\left\{ \begin{array} { l } { - 8 x + 5 y \geq 3 } \\ { 2 x - 3 y < 0 } \end{array} \right.$, $\left\{ \begin{array} { l } { 2 x - 9 y < - 1 } \\ { 3 x - 6 y > - 2 } \end{array} \right.$, $\left\{ \begin{array} { c } { 2 x - y \geq - 1 } \\ { x - 3 y < 6 } \\ { 2 x - 3 y > - 1 } \end{array} \right.$, $\left\{ \begin{array} { c } { - x + 5 y > 10 } \\ { 2 x + y < 1 } \\ { x + 3 y < - 2 } \end{array} \right.$, $\left\{ \begin{array} { l } { y + 4 \geq 0 } \\ { \frac { 1 } { 2 } x + \frac { 1 } { 3 } y \leq 1 } \\ { - 3 x + 2 y \leq 6 } \end{array} \right.$, $\left\{ \begin{array} { l } { y \leq - \frac { 3 } { 4 } x + 2 } \\ { y \geq - 5 x + 2 } \\ { y \geq \frac { 1 } { 3 } x - 1 } \end{array} \right.$, $ \left\{ \begin{array} { l } { y \geq x ^ { 2 } + 1 } \\ { y < - 2 x + 3 } \end{array} \right.$, $\left\{ \begin{array} { l } { y < ( x - 1 ) ^ { 2 } - 1 } \\ { y > \frac { 1 } { 2 } x - 1 } \end{array} \right.$, $\left\{ \begin{array} { l } { y < 0 } \\ { y \geq - | x | + 4 } \end{array} \right.$, $\left\{ \begin{array} { l } { y < | x - 3 | + 2 } \\ { y \geq 2 } \end{array} \right.$. IMAG0004.JPG 8. B. IMAG0020.JPG IMAG0007.JPG A. Consider the point $(−1, 0)$ on the solid boundary defined by $y = 2x + 2$ and verify that it solves the original system: $\begin{array} { l } { y > x - 2 } \\ { \color{Cerulean}{0}\color{black}{ >}\color{Cerulean}{ -1}\color{black}{ -} 2 } \\ { 0 > -3 }\:\: \color{Cerulean}{✓} \end{array}$, $\begin{array} { l } { y \leq 2 x + 2 } \\ { \color{Cerulean}{0}\color{black}{ \leq} 2 (\color{Cerulean}{ -1}\color{black}{ )} + 2 } \\ { 0 \leq 0 } \:\:\color{Cerulean}{✓} \end{array}$. For the second inequality, we use a solid boundary defined by $y = \frac{1}{ 2} x − 1$ and shade all points below. Try the free Mathway calculator and problem solver below to practice various math topics. $\left( - \frac { 1 } { 2 } , - 5 \right)$; $\left\{ \begin{array} { l } { y \leq - 3 x - 5 } \\ { y > ( x - 1 ) ^ { 2 } - 10 } \end{array} \right.$, $\left\{ \begin{array} { l } { x \geq - 5 } \\ { y < ( x + 3 ) ^ { 2 } - 2 } \end{array} \right.$. Now consider the point $(2, 0)$ on the dashed boundary defined by $y = x − 2$ and verify that it does not solve the original system: $\begin{array} { l } { y > x - 2 } \\ { \color{Cerulean}{0}\color{black}{ >}\color{Cerulean}{ 2}\color{black}{ -} 2 } \\ { 0 > 0 } \:\:\color{red}{✗}\end{array}$, $\begin{array} { l } { y \leq 2 x + 2 } \\ {\color{Cerulean}{ 0}\color{black}{ \leq} 2 (\color{Cerulean}{ 2}\color{black}{ )} + 2 } \\ { 0 \leq 6 } \:\:\color{Cerulean}{✓}\end{array}$. The intersection is darkened. 18, $\left( - 3 , - \frac { 3 } { 4 } \right)$; $\left\{ \begin{array} { l } { 3 x - 4 y < 24 } \\ { y < ( x + 2 ) ^ { 2 } - 1 } \end{array} \right.$, $\left\{ \begin{array} { l } { y < ( x - 3 ) ^ { 2 } + 1 } \\ { y < - \frac { 3 } { 4 } x + 5 } \end{array} \right.$, $\left\{ \begin{array} { l } { y \geq - 1 } \\ { y < - ( x - 2 ) ^ { 2 } + 3 } \end{array} \right.$, 17. It is known as the origin, or point (0, 0). Determine the region of the plane that is the solution of the system. IMAG0015.JPG IMAG0016.JPG 5x â 3 â¦ Try the given examples, or type in your own problem and check your answer with the â¦ IMAG0015.JPG Substitution of a variable into another equation is usually the best method for solving nonlinear systems of equations. While this is not a proof, doing so will give a good indication that you have graphed the correct region. Determine whether or not $(-3,3)$ is a solution to the following system: $\left\{ \begin{aligned} 2 x + 6 y \leq 6 \\ - \frac { 1 } { 3 } x - y \leq 3 \end{aligned} \right.$. Because of the strict inequalities, we will use a dashed line for each boundary. A system of inequalities33 consists of a set of two or more inequalities with the same variables. F. 9. $\begin{aligned} 2 x + 6 y \leq 6 \\ 2 ( \color{Cerulean}{- 3}\color{black}{ )} + 6 ( \color{Cerulean}{3}\color{black}{ )} \leq 6 \\ - 6 + 18 \leq 6 \\ 12 \leq 6 \:\:\color{red}{✗}\end{aligned}$, $\begin{aligned} - \frac { 1 } { 3 } x - y & \leq 3 \\ - \frac { 1 } { 3 } ( \color{Cerulean}{- 3}\color{black}{ )} - (\color{Cerulean}{ 3}\color{black}{ )} & \leq 3 \\ 1 - 3 & \leq 3 \\ - 2 & \leq 3 \:\:\color{Cerulean}{✓}\end{aligned}$. IMAG0013.JPG To graph solutions to systems of inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect. IMAG0002.JPG Therefore, there are no simultaneous solutions. Therefore, to solve these systems we graph the solution sets of the inequalities on the same set of axes and determine where they intersect. IMAG0007.JPG Solution sets to both are graphed below. $\left( - \frac { 3 } { 2 } , \frac { 1 } { 3 } \right)$; $\left\{ \begin{array} { l } { x - 2 y \leq 4 } \\ { y \leq | 3 x - 1 | + 2 } \end{array} \right.$. After graphing the inequalities on the same set of axes, we determine that the intersection lies in the region pictured below. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. To facilitate the graphing process, we first solve for $y$. Click here to let us know! 33A set of two or more inequalities with the same variables. Solution: Given, 3x â 2 < 2x + 1. Algebra 1 : Inequalities Lesson 12: Systems of Inequalities Word Problems (Answer Key) 1. This PDF book include mcgraw hill answer key algebra 2 tests conduct. $\begin{array} { l } { y \geq - 4 } \\ { \color{Cerulean}{1}\color{black}{ \geq} - 4 }\:\:\color{Cerulean}{✓} \end{array}$, $\begin{array} { l } { y < x + 3 } \\ { \color{Cerulean}{1}\color{black}{ <}\color{Cerulean}{ - 1}\color{black}{ +} 3 } \\ { 1 < 2 } \:\:\color{Cerulean}{✓} \end{array}$, $\begin{array} { l } { y \leq -3x + 3 } \\ { \color{Cerulean}{1}\color{black}{\leq} -3(\color{Cerulean}{ - 1}\color{black}{ )+} 3 } \\ { 1 \leq 3+3 } \\{1 \leq 6}\:\:\color{Cerulean}{✓} \end{array}$. \quad\Rightarrow\quad \left\{ \begin{array} { l } { y > 2 x - 4 } \\ { y \leq \frac { 1 } { 2 } x - 1 } \end{array} \right.\). Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. They would like to make at least a $500 profit from selling tickets. $\left\{ \begin{array} { l } { y \geq \frac { 2 } { 3 } x - 3 } \\ { y < - \frac { 1 } { 3 } x + 3 } \end{array} \right.$, $\left\{ \begin{array} { l } { y \geq - \frac { 1 } { 4 } x + 1 } \\ { y < \frac { 1 } { 2 } x - 2 } \end{array} \right.$, $\left\{ \begin{array} { l } { y > \frac { 2 } { 3 } x + 1 } \\ { y > \frac { 4 } { 3 } x - 5 } \end{array} \right.$, $\left\{ \begin{array} { l } { y \leq - 5 x + 4 } \\ { y < \frac { 4 } { 3 } x - 2 } \end{array} \right.$, $\left\{ \begin{array} { l } { x - y \geq - 3 } \\ { x + y \geq 3 } \end{array} \right.$, $\left\{ \begin{array} { l } { 3 x + y < 4 } \\ { 2 x - y \leq 1 } \end{array} \right.$, $\left\{ \begin{array} { l } { - x + 2 y \leq 0 } \\ { 3 x + 5 y < 15 } \end{array} \right.$, $\left\{ \begin{array} { c } { 2 x + 3 y < 6 } \\ { - 4 x + 3 y \geq - 12 } \end{array} \right.$, $\left\{ \begin{array} { l } { 3 x + 2 y > 1 } \\ { 4 x - 2 y > 3 } \end{array} \right.$, $\left\{ \begin{array} { l } { x - 4 y \geq 2 } \\ { 8 x + 4 y \leq 3 } \end{array} \right.$, $\left\{ \begin{array} { l } { 5 x - 2 y \leq 6 } \\ { - 5 x + 2 y < 2 } \end{array} \right.$, $\left\{ \begin{array} { l } { 12 x + 10 y > 20 } \\ { 18 x + 15 y < - 15 } \end{array} \right.$, $\left\{ \begin{array} { l } { x + y < 0 } \\ { y + 4 > 0 } \end{array} \right.$, $\left\{ \begin{array} { l } { x > - 3 } \\ { y < 1 } \end{array} \right.$, $\left\{ \begin{array} { l } { 2 x - 2 y < 0 } \\ { 3 x - 3 y > 3 } \end{array} \right.$, $\left\{ \begin{array} { l } { y + 1 \leq 0 } \\ { y + 3 \geq 0 } \end{array} \right.$. Legal. This point does not satisfy both inequalities and thus is not included in the solution set. For the first inequality shade all points above the boundary and for the second inequality shade all points below the boundary. Construct a system of linear inequalities that describes all points in the fourth quadrant. This boundary is a horizontal translation of the basic function $y = x^{2}$ to the left $1$ unit. Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line. Use the same technique to graph the solution sets to systems of nonlinear inequalities. Graph the solution set: $\left\{ \begin{array} { l } { y \geq - | x + 1 | + 3 } \\ { y \leq 2 } \end{array} \right.$. To verify this, we can show that it solves both of the original inequalities as follows: $\begin{array} { l } { y > x - 2 } \\ { \color{Cerulean}{2}\color{black}{ >}\color{Cerulean}{ 3}\color{black}{ -} 2 } \\ { 2 > 1 }\:\: \color{Cerulean}{✓} \end{array}$, $\begin{array} { l } { y \leq 2 x + 2 } \\ { \color{Cerulean}{2}\color{black}{ \leq} 2 (\color{Cerulean}{ 3}\color{black}{ )} + 2 } \\ { 2 \leq 8 } \:\:\color{Cerulean}{✓} \end{array}$. And the intersection of both regions contains the region of simultaneous ordered pair solutions. Graph the solution set: $\left\{ \begin{array} { l } { - 2 x + y > - 4 } \\ { 3 x - 6 y \geq 6 } \end{array} \right.$. | Attachments: IMAG0009.JPG H. 5. This Long Division Worksheet the number of digits for the divisors and quotients may be varied from 1 to 3. Upon completing this section you should be able to: Graph two or more linear inequalities on the same set of coordinate axes. We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. IMAG0014.JPG B. Since (3,2) checks in both equations, it is the solution to the system. $\left\{ \begin{array} { l } { x > 0 } \\ { y > 0 } \end{array} \right.$, 19. B. This intersection, or overlap, will define the region of common ordered pair solutions. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. For example, both solution sets of the following inequalities can be graphed on the same set of axes: $\left\{ \begin{array} { l } { y < \frac { 1 } { 2 } x + 4 } \\ { y \geq x ^ { 2 } } \end{array} \right.$. Construct a system of linear inequalities that describes all points in the second quadrant. 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