VIEWS: 0 PAGES: 3 CATEGORY: Preschool POSTED ON: 9/2/2010 Public Domain
Algebra Concepts and Applications Chapter 13 Lesson 13-3 Examples Use substitution to solve each system of equations. 1 y = -3x x + 2y = 10 The first equation tells you that y is equal to -3x. So, substitute -3x for y in the second equation. Then solve for x. x + 2y = 10 x + 2(-3x) = 10 Replace y with -3x. x - 6x = 10 -5x = 10 -5x 10 -5 = -5 Divide each side by -5. x = -2 Now substitute -2 for x in either equation and solve for y. Choose the equation that is easier for you to solve. y = -3x y = -3(-2) Replace x with -2. y=6 The solution of this system of equations is (-2, 6). You can see from the graph that the solution is correct. You can also check by substituting (-2, 6) into each of the original equations. 2 x - y = -4 x = 3y + 5 Substitute 3y + 5 for x in the first equation. Then solve for y. x - y = -4 3y + 5 - y = -4 Replace x with 3y + 5. 2y + 5 = -4 2y + 5 - 5 = -4 - 5 Subtract 5 from each side. 2y = -9 2y -9 2 =2 Divide each side by 2. 9 y = -2 Algebra Concepts and Applications Chapter 13 9 Now substitute -2 for y into either equation and solve for x. x = 3y + 5 9 9 x = 3-2 + 5 Replace y with -2. 27 x=-2 +5 17 x=-2 17 9 The solution of this system of equations is - 2 , -2. Check by replacing (x, y) 17 9 with - 2 , -2 in each equation. Example 3 -x + y = 7 3x - y = -5 Solve the first equation for y since the coefficient of y is 1. -x + y = 7 y = x + 7 Next, find the value of x by Now substitute 1 for x in either substituting x + 7 for y in the second equation and solve for y. equation. -x + y = 7 3x - y = -5 -1 + y = 7 Replace x with 1. 3x - (x + 7) = -5 -1 + y + 1 = 7 + 1 3x - x - 7 = -5 y =8 2x - 7 = -5 2x - 7 + 7 = -5 + 7 2x = 2 2x 2 2 =2 x =1 The solution is (1, 8). Examples Use substitution to solve each system of equations. 4 y = -5x + 1 5x + y = -2 Find the value of x by substituting -5x + 1 for y in the second equation. 5x + y = -2 5x + (-5x + 1) = -2 Replace y with -5x + 1. 1 = -2 The statement 1 = -2 is false. This means that there are no ordered pairs that are solutions to both equations. Compare the slope-intercept forms of the equations, y = -5x + 1 and y = -5x - 2. Notice that the graphs of these equations have the same slope but different y-intercepts. Thus, the lines are parallel, and the system has no solution. Algebra Concepts and Applications Chapter 13 5 y = -4x + 1 8x + 2y = 2 8x + 2y =2 8x + 2(-4x + 1) =2 Replace y with -4x + 1. 8x - 8x + 2 =2 Distributive Property 2 =2 The statement 2 = 2 is true. This means that an ordered pair for a point on either line is a solution to both equations. The system has infinitely many solutions. Example 6 The Garden Center sells bags of wild flower seed that contain part seed and Gardening Link part sand to aid in planting. One seed mix contains 35% seed and a second type contains 55% seed. How much of each type of mix should be added to make 500 pounds of a 45% seed mix? Explore Let a = the number of pounds of the 35% seed. Let b = the number of pounds of the 55% seed. 35% Seed 55% Seed 45% Seed Total Pounds a b 500 Pounds of Seed 0.35a 0.55b 0.45(500) Plan Write two equations to represent the information. a + b = 500 total pounds 0.35a + 0.55b = 0.45(500) total pounds of seed Solve Use substitution to solve this system. Since a + b = 500, a = 500 - b. 0.35a + 0.55b = 0.45(500) 0.35(500 - b) + 0.55b = 0.45(500) Replace a with 500 - b. 175 - 0.35b + 0.55b = 225 Distributive Property 175 + 0.2b - 175 = 225 - 175 Subtract 175 from each side. 0.2b = 50 0.2b 50 0.2 = 0.2 Divide each side by 0.2. b = 250 Now substitute 250 for b in either equation and solve for a. a + b = 500 a + 250 = 500 Replace b with 250. a + 250 - 250 = 500 - 250 Subtract 250 from each side. a = 250 So, 250 pounds of 35% seed mix and 250 pounds of 55% seed mix should be used. Examine Check by substituting (250, 250) into the original equations. The solution is correct.