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Solve Systems of Linear Inequalities

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Presentation on theme: "Solve Systems of Linear Inequalities"β€” Presentation transcript:

1 Solve Systems of Linear Inequalities

2 Linear Inequalities Recall that a linear equation is one of the form 𝑦=π‘šπ‘₯+𝑏 The graph of a linear equation is a line; π‘š represents the slope of the line, while 𝑏 is the 𝑦-intercept For the first part of this lesson you will learn how to interpret a linear inequality Then, you will learn how to solve a system of linear inequalities (in two unknowns)

3 Linear Inequalities There are 4 different forms of linear inequalities: 𝑦>π‘šπ‘₯+𝑏 𝑦<π‘šπ‘₯+𝑏 𝑦β‰₯π‘šπ‘₯+𝑏 π‘¦β‰€π‘šπ‘₯+𝑏 We can determine what the graph of such inequalities will look like if we check a few values of π‘₯, first Let’s do this for the linear inequality 𝑦>2π‘₯βˆ’1; we will find some values for 𝑦=2π‘₯βˆ’1 in a table

4 Linear Inequalities *Find the corresponding values of 𝑦; plot the points as open circles; do NOT draw a line! This animation will show you what to do next 𝒙 π’š=πŸπ’™βˆ’πŸ (𝒙,π’š) βˆ’2 𝑦=2 βˆ’2 βˆ’1=βˆ’5 (βˆ’2,βˆ’5) βˆ’1 1 2

5 Linear Inequalities Note that all along the graph of 𝑦=2π‘₯βˆ’1 will be open circles since our inequality is 𝑦>2π‘₯βˆ’1 rather than 𝑦β‰₯2π‘₯βˆ’1 *To indicate that a point is not included in a set of numbers on the number line, we marked it with an open circle *To indicate that the points on a line are not included in a set of ordered pairs, we will draw a dashed line *Draw a dashed line through the points

6 Linear Inequalities *You found that all values greater than the 𝑦-value at a particular π‘₯ are above the line; the points that are solutions to the inequality 𝑦>2π‘₯βˆ’ 1 are all above the line *Shade the area above the dashed line; this area includes all π‘₯,𝑦 pairs such that 𝑦>2π‘₯βˆ’1 *As you may correctly guess, when the inequality is less-than, the shaded area lies __________ *Also, for β‰₯ or ≀, we will use a solid line This is summarized in the table on the next slide

7 Linear Inequalities Line Shade 𝑦> Dashed Above the line 𝑦<
Below the line 𝑦β‰₯ Solid 𝑦≀

8 Guided Practice Graph the following linear inequalities. 𝑦< 1 3 π‘₯+1
𝑦β‰₯βˆ’2π‘₯ π‘¦β‰€βˆ’ 1 2 π‘₯βˆ’2 𝑦>π‘₯+3

9 Guided Practice

10 Guided Practice

11 Guided Practice

12 Guided Practice

13 Special Cases If the slope of a line is zero, its linear equation is 𝑦=0β‹…π‘₯+𝑏, but this is just 𝑦=𝑏 The graph of such a linear equation is a horizontal line As an inequality, follow the same rules from the table given previously An example is given on the next slide

14 Special Cases

15 Special Cases Since division by zero is not defined, then a line for which the π‘₯- coordinate is the same for every point has an undefined slope *The equation for such a line has the form π‘₯=π‘Ž, where π‘Ž is the π‘₯- intercept *These are all vertical lines *To graph π‘₯>π‘Ž or π‘₯β‰₯π‘Ž, shade to the right of the line *To graph π‘₯<π‘Ž or π‘₯β‰€π‘Ž, shade to the left of the line An example is given on the next slide

16 Special Cases

17 Systems of Linear Inequalities
A system of linear equations in two variables may look like the following 4π‘₯βˆ’π‘¦=βˆ’7 π‘₯+3𝑦=6 You should recognize that such a system can be solved by the elimination method; it can also be solved by solving for one variable, then substituting for that variable in the other equation The solution, if it exists, is the point (π‘₯,𝑦) where the lines intersect

18 Systems of Linear Inequalities
A system of linear inequalities in two variables may look like the following 4π‘₯βˆ’π‘¦<βˆ’7 π‘₯+3𝑦β‰₯6 *We cannot solve this system algebraically; we must graph each inequality *The solution, if it exists, is that area of the coordinate plane that contains π‘₯,𝑦 pairs that are solutions to both linear inequalities Graph the two inequalities and highlight the area representing the solution

19 Systems of Linear Inequalities

20 Guided Practice Graph the system of inequalities. 4π‘₯+𝑦β‰₯βˆ’2 π‘₯+𝑦β‰₯1
π‘₯βˆ’2𝑦>βˆ’4 3π‘₯βˆ’π‘¦>3 π‘₯+2π‘¦β‰€βˆ’2 π‘₯>2

21 Guided Practice Graph the system of inequalities.
𝑦> 3 2 π‘₯βˆ’1 𝑦<βˆ’ 1 2 π‘₯+3 𝑦β‰₯3π‘₯+1 𝑦>3π‘₯βˆ’1 𝑦<βˆ’π‘₯βˆ’3 𝑦≀2π‘₯+3

22 Concentrate!


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