Gradient and Intercept_ JHS Unity School.pptx

About intercept

1. GRADIENT AND INTERCEPT

2. Solving Systems of Linear Equations by Graphing Concept: Solving systems of equations in two variables. EQ: How can I manipulate equations to solve a system of equations? Vocabulary : System of equations, Inconsistent, Consistent (independent/dependent.)

3. Introduction The solution to a system of equations is the point or points that make both equations true. Systems of equations can have one solution, no solutions, or an infinite number of solutions. There are various methods to solving a system of equations. One is the graphing method . On a graph, the solution to a system of equations can be easily seen. The solution to the system is the point of intersection , the point at which two lines cross or meet .

4. Key Concepts, continued Intersecting Lines Parallel Lines Same Line One solution No solutions Infinitely many solutions Consistent Independent Inconsistent Consistent Dependent

5. Guided Practice Example 1 Graph the system of equations. Then determine whether the system has one solution, no solution, or infinitely many solutions. If the system has a solution, name it.

6. Guided Practice: Example 1, continued Solve each equation for y . The first equation needs to be solved for y . The second equation ( y = – x + 3) is already in slope-intercept form. 4 x – 6 y = 12 Original equation –6 y = 12 – 4 x Subtract 4 x from both sides. Divide both sides by –6. Write the equation in slope-intercept form ( y = mx + b ).

7. Guided Practice: Example 1, continued Graph both equations using the slope-intercept method. The y -intercept of is –2. The slope is . The y -intercept of y = – x + 3 is 3. The slope is –1.

8. Guided Practice: Example 1, continued 3. Observe the graph. The lines intersect at the point (3, 0). This appears to be the solution to this system of equations.

9. Guided Practice: Example 1, continued 4 x – 6 y = 12 First equation in the system 4(3) – 6(0) = 12 Substitute (3, 0) for x and y . 12 – 0 = 12 Simplify. 12 = 12 This is a true statement. y = – x + 3 Second equation in the system (0) = –(3) + 3 Substitute (3, 0) for x and y . 0 = –3 + 3 Simplify. 0 = 0 This is a true statement.

10. Guided Practice: Example 1, continued The system has one solution, (3, 0). This system is consistent because it has at least one solution and it is independent because it only has 1 solution.

11. Guided Practice Example 2 Graph the system of equations. Then determine whether the system has one solution, no solution, or infinitely many solutions. If the system has a solution, name it.

12. Guided Practice: Example 2, continued Solve each equation for y .

13. Guided Practice: Example 2, continued Graph both equations using the slope-intercept method. The y -intercept of both equations is 1. The slope of both equations is 2.

14. Guided Practice: Example 2, continued Observe the graph. The graphs of y = 2x +1 and -8x + 4y = 4 are the same line. There are infinitely many solutions to this system of equations.

15. Guided Practice: Example 2, continued

16. Guided Practice: Example 2, continued The system has infinitely many solutions. This system is consistent because it has at least one solution and it is dependent because it has more than one solution.

17. Guided Practice Example 3 Graph the system of equations. Then determine whether the system has one solution, no solution, or infinitely many solutions. If the system has a solution, name it.

18. Guided Practice: Example 3, continued Solve each equation for y . The first equation needs to be solved for y . The second equation ( y = 3 x – 5) is already in slope-intercept form. –6 x + 2 y = 8 Original equation 2 y = 8 + 6 x Add 6 x to both sides. y = 4 + 3 x Divide both sides by 2. y = 3 x + 4 Write the equation in slope-intercept form ( y = mx + b ).

19. Guided Practice: Example 3, continued Graph both equations using the slope-intercept method. The y -intercept of y = 3 x + 4 is 4. The slope is 3. The y -intercept of y = 3 x – 5 is –5. The slope is 3.

20. Guided Practice: Example 3, continued Observe the graph. The graphs of –6 x + 2 y = 8 and y = 3 x – 5 are parallel lines and never cross.

21. Guided Practice: Example 3, continued The system has no solutions because there are no values for x and y that will make both equations true. This system is inconsistent because it has no solutions.

22. Slope Problems

23. Slope Problem Examples Determine a value for x such that the line through the points has the given slope . Let's use the slope formula and plug in what we know. ( x 1 , y 1 ) ( x 2 , y 2 ) You can cross-multiply to find a fraction-free equation for x to solve.

24. Example when you have a point and the slope A point on a line and the slope of the line are given. Find two additional points on the line. -1 Remember that slope is the change in y over the change in x. The slope is 2 which can be made into the fraction (0,-3) So this point is on the line also. You can see that this point is changing (adding) 2 to the y value of the given point and changing (adding) 1 to the x value. +2 +1 To find another point on the line, repeat this process with your new point (0,-3) +1 +2 (1,-1) (-1,5)

25. y intercept slope Example of given an equation, find the slope and y intercept Find the slope and y intercept of the given equation and graph it. First let's get this in slope-intercept form by solving for y . -3 x + 4 -3 x + 4 -4 -4 Now plot the y intercept From the y intercept, count the slope Change in y Change in x Now that you have 2 points you can draw the line

26. Example of how to find x and y intercepts to graph a line The x-intercept is where a line crosses the x axis (6,0) (-1,0) (2,0) What is the common thing you notice about the x-intercepts of these lines? The y value of the point where they cross the axis will always be 0 To find the x-intercept when we have an equation then, we will want the y value to be zero.

27. Now let's see how to find the y-intercept The y-intercept is where a line crosses the y axis (0,4) (0,1) (0,5) What is the common thing you notice about the y-intercepts of these lines? The x value of the point where they cross the axis will always be 0 To find the y-intercept when we have an equation then, we will want the x value to be zero.

28. Let's look at the equation 2x – 3y = 12 Find the x-intercept. We'll do this by plugging 0 in for y 2x – 3(0) = 12 Now solve for x. 2x = 12 2 2 x = 6 So the place where this line crosses the x axis is (6, 0)

29. 2x – 3y = 12 Find the y-intercept. We'll do this by plugging 0 in for x 2(0) – 3y = 12 Now solve for y. -3y = 12 -3 -3 y = - 4 So the place where this line crosses the y axis is (0, -4) We now have enough information to graph the line by joining up these points (6,0) (0,- 4)