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Lesson 10 Solving system of equations by substitution August 6, 2014 Warm up Lesson 10 1. Find the x-intercept, y-intercept and slope of the following equation. -3y + 6x + 9 = 0 2. Translate the following equation 2 units down and 3 units to the right. 4. To install a security system in my house there was an initial fee of $115 plus the $35 monthly fee. Write an equation that relates time (x) to the amount of money spent (y). 2y – 3x = 4 3. Graph the following equation 5. Graph the following equation y = -2x 2y + 3x = 4 y y x 6. Change title of homework from Lesson 1 to Lesson 10. x Answers to Lesson 9 1. -12 2. 2 3. -2 4. -8 5. Y = - 4x + 29 6. y = 4x + 2 7. y = 2x 8. Y = - ½ x – 1 9. y = - 3/2x + 7/2 11. Y = 2x – 14 12. slope: - 2/3, y-int: 10 14. Y = - 2 x + 2 15. y = -3 17. Y = 8x + 7 20. y = 2x + 3 18. 19. 10. y = -2/3 x - 2 13. – 5/2 16. y = 3x + 16 21. (100 p)n 100 22. B Consistent systems. Consistent systems Exactly one solution y Lines intersect and may or may not be perpendicular Different slopes x Inconsistent system • No solution • Lines are parallel • Same slopes, different y intercepts y x Dependent system • Infinite number of solutions. • Lines are coincident • Same slopes, same y intercepts y x Solving system of equations by substitution x–y=2 6x – y = -4 4x – 3y = 11 2x + 2y = 15 1. Pick one equation and solve for one variable. x–y=2 6x – y = -4 x=y+2 6x = y – 4 y = 6x + 4 2. Using the other equation, substitute 3. Solve for the remaining variable. 4x – 3y = 11 2x + 2y = 15 4(y + 2) – 3y = 11 2x + 2(6x+4) = 15 2x + 12x + 8 = 15 4y + 8 – 3y = 11 14x + 8 = 15 y + 8 = 11 y=3 4. Substitute the value found in step 3, into equation used in step 1 6x – y = - 4 x–y=2 6(1/2) – y = -4 x–3=2 3 –y = -4 x =5 (5, 3) y=7 14x = 7 x=½ ( ½ , 7) 5. Check you answer using the same equation in step 2. 4x – 3y = 11 2x + 2y = 15 4(5) – 3(3) = 11 2(1/2) + 2(7) = 15 20 – 9 = 11 1 + 14 = 15 11 = 11 15 = 15 Therefore the solution that will answer both of these equations is (5,3) Therefore the solution that will answer both of these equations is ( ½ ,7) Solving system of equations by substitution 2x – y = 16 -x + 2y = -8 1. Pick one equation and solve for one variable. 5. Check your answer with equation used in step 2. 2x – y = 16 -x + 2y = -8 2y = x -8 2(8) – (0) = 16 2y + 8 = x 16 – 0 = 16 16 = 16 2. Using the other equation, substitute 2x – y = 16 2(2y + 8) - y = 16 3. Solve for the remaining variable. 4y + 16 – y = 16 3y = 0 y=0 -x + 2y = -8 4. Substitute the value found in step 3, into equation used in step 1 -x + 2(0) = -8 - x = -8 x=8 (8, 0) Solving system of equations by substitution 5x – 3y = -11 1. Pick one equation and solve for one variable. x – 2y = 2 5. Check your answer with equation used in step 2. x – 2y = 2 5x – 3y = -11 x = 2y + 2 5(-4) – 3(-3) = -11 -20 + 9 = -11 2. Using the other equation, substitute 3. Solve for the remaining variable. 5x – 3y = -11 - 11 = - 11 5(2y + 2) – 3y = -11 10y + 10 – 3y = -11 7y + 10 = -11 7y = -21 y = -3 x – 2y = 2 4. Substitute the value found in step 3, into equation used in step 1 x – 2(-3) = 2 x+6=2 x = -4 (-4, -3) On your own. Solve each system of equation below by substitution method. 4. 1. y = 2x (4, 8) y = 2x – 5 4x – y = 7 x + y = 12 (1,-3) 5. 2. -2x + 3y = 14 2x – 3y = 12 x = 4y + 1 (9,2) X + 2y = 7 (-1,4) 6. 3. x–y=3 y = -x + 5 x – 4y = 10 (6,-1) 6x + 4y = 13 (2 ½ , - ½ ) Homework • Lesson 10 show work to get credit