Final presentation
- 1. Learning Algebra Solving Linear Systems of Equations By: Mr. Patrick Paez
- 2. The first thing we need to do is learn about what systems of linear equations are. Systems of linear equations are two or more linear equations of the same variables. A point that lies on both lines would be the solution to the system.
- 3. Let’s look at an example of a system of linear equations. x + 2y = 7 3x – 2y = 5 Since we are just dealing with an x – y coordinate system, or a Cartesian coordinate system, linear equations will either intersect or not intersect, making them parallel. If they do intersect, the point of intersection is the solution to the system.
- 4. There are three different ways to find solutions for these systems of linear equations. They are: • Graphing • Substitution • Elimination
- 5. Let’s start with solving by graphing. As stated earlier, the solution of two linear equations is the point of intersection. This point can be easily found by following three simple steps. 1) Put both equations in y-intercept form. 2) Graph each equation and estimate the point of intersection. 3) Check the coordinates algebraically by substituting into the original equations. GRAPHING
- 6. Here is an example of solving a system of linear equations by graphing. You have the following two equations. Use a graph to solve the system of equations. Eq 1: x + 2y = 7 x + 2y = 7 -x -x 2y = -x + 7 /2 /2 /2 y = -(1/2)x + 7/2 Eq 2: 3x – 2y = 5 3x - 2y = 5 -3x -3x -2y = -3x + 5 /-2 /-2 /-2 y = (3/2)x - 5/2
- 7. Now let’s see how this looks on a graph As we can see on the graph, the intersection is (3,2) y = -(1/2)x + 7/2 y = (3/2)x - 5/2
- 8. The point, or coordinate, (3,2) would be the solution to that particular system of equations. Remember to substitute the point (3,2) into the original equations to check your answer. (3) + 2(2) = 7 3 + 4 = 7 7 = 7 3(3) – 2(2) = 5 9 – 4 = 5 5 = 5 Now, using the graphing method, try to find the solution for the following systems of equations. Problem 1 -x + y = -7 x + 4y = -8 Problem 2 y = -x + 4 y = 2x - 8
- 9. Problem 1 The two lines intersect at point (4,-3) Don’t forget to check your answer: y = x - 7 y = -(1/4)x - 2 -x + y = -7 -(4) + (-3) = -7 -7 = -7 x + 4y = -8 (4) + 4(-3) = -8 -8 = -8
- 10. Problem 2 The two lines intersect at point (4,0) Don’t forget to check your answer: y = -x + 4 0 = -(4) + 4 0 = 0 y = 2x - 8 0 = 2(4) - 8 0 = 0
- 11. Substitution Graphing is a good way to solve these systems. However, there will be some problems where graphing will get a little difficult. Good thing there is another strategy you can use. Here are the steps for solving by substitution. 1) Solve one of the equations for one variable. 2) Substitute the expression from step 1 into the other equation and solve for the other variable. 3) Substitute the value from step 2 into the revised equation from step 1 and solve.
- 12. Example 1: y = 3x + 2 x + 2y = 11 Step 1: Solve one equation for a variable. Equation 1 is already solved for y. Step 2: Substitute 3x + 2 for y in equation 2 and solve for x. x + 2y = 11 x + 2(3x + 2) = 11 x + 6x + 4 = 11 7x = 7 x = 1
- 13. Example 1 (continued): Step 3: Substitute 1 for x in the original equation 1 to find the value of y. y = 3x + 2 y = 3(1) + 2 y = 5 Finish your work by checking your answers in equation 2. x + 2y = 11 1 + 2(5) = 11 1 + 10 = 11 11 = 11
- 14. Elimination Again, some equations may be a little harder to solve when using graphing or substitution. Solving by elimination can be one of the easier methods to use given the right situation. These are the three steps. 1) Add or subtract the equations to eliminate one variable. 2) Solve the resulting equation for the other variable. 3) Substitute in either original equation to find the value of the eliminated variable.
- 15. 2x + 3y = 11 -2x + 5y = 13 Step 1: Add the equations to eliminate one variable. 2x + 3y = 11 -2x + 5y = 13 8y = 24 Step 2: Solve for y 8y = 24 y = 3 Example 1:
- 16. Example 1 (continued): Step 3: Substitute 3 for y in either equation and solve for x. 2x + 3y = 11 2x + 3(3) = 11 2x + 9 = 11 2x = 2 x = 1 Finish your work by checking your answers in equation 2. -2x + 5y = 13 -2(1) + 5(3) = 13 -2 + 15 = 13
- 17. Elimination: Part 2 Not all equations may be set up as nicely as the previous example. In some equations, you may have to multiply an equation by a constant so you can add or subtract the equations to eliminate one variable.
- 18. Example 2: 6x + 5y = 19 2x + 3y = 5 Step 1: Multiply the bottom equation by -3 to make eliminating x possible. 6x + 5y = 19 -3(2x + 3y = 5) 6x + 5y = 19 -6x – 9y = -15 Step 2: Add the two equations together to eliminate x. -4y = 4
- 19. Example 2 (continued): Step 3: Solve for y -4y = 4 y = -1 Step 4: Substitute -1 for y in either equation and solve for x. 6x + 5y = 19 6x + 5(-1) = 19 6x = 24 x = 4
- 20. Finish your work by checking your answers in equation 2. 2x + 3y = 5 2(4) + 3(-1) = 5 8 - 3 = 5 5 = 5 Thank you for your time…and don’t forget to study!!!