1. Solve for the variable 1. 5x – 4 = 2x + 8 2. -4(x + 2) + 3x = 2

2. Solution: (-2, 5). x = -2 2x + y = 1.... y = -2x + 1...... Wait! It’s not in slope- intercept form!

3.  x = -2 2x + y = 1 What makes a system easy to solve by graphing? …when both equations are in slope-intercept form.

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5.  Do you notice anything about one of the equations? x = -2 2x + y = 1 x = -2 2(-2) + y = 1 -4 + y = 1 +4 =+4 y = 5 Do you know the value of x? (-2 Do you know the value of y?, 5)

6. Solution: (-2, 5). x = -2 2x + y = 1.... y = -2x + 1......

7. Coordinate Algebra with Support UNIT QUESTION: How do I justify and solve the solution to a system of equations or inequalities? Standard: MCC9-12.A.REI.1, 3, 5, 6, and 12 Today’s Question: When is it better to use substitution than elimination for solving systems? Standard: MCC9-12.A.REI.6

8.  Substitution

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11. 1)One equation will be ISOLATED (it will have either x or y by itself) or can be solved for x or y easily. 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 equation from Step 1 and solve. 4)Your SOLUTION is the ordered pair formed by x & y. 5)CHECK the solution in each of the original equations.

12. ( 16) x = -4 3x + 2y = 20 Why would substitution be a good method to use? x = -4 3(-4) + 2y = 20 -12 + 2y = 20 +12 =+12 2y = 32 __ __ 2 2 y = 16 xy 3x + 2y = 20 3x + 2(16) = 20 3x + 32 = 20 - 32 = -32 3x = -12 __ __ 3 3 x = -4 -4,

13. x = -4 (-4, 16) 3x + 2y = 20 3(-4) + 2(16) = 20 -12 + 32 = 20 20 = 20 We are correct!

14. (2 ) y = x – 1 x + y = 3 Why would substitution be a good method to use? y = x – 1 x +x – 1 = 3 2x – 1 = 3 + 1 = +1 2x = 4 __ __ 2 2 x = 2 xy y = x – 1 y = 2 – 1 y = 1 x + y = 3 2 + y = 3 y = 1, 1

15. y = x - 1 1 = 2 - 1 1 = 1 x + y = 3 2 + 1 = 3 The values work in both equations, so we are correct! y = x – 1 (2, 1) x + y = 3 x y 3 = 3

16. (-2 3x + 2y = -12 y = x - 1 3x + 2(x – 1) = -12 3x + 2x – 2 = -12 5x – 2 = -12 + 2 = + 2 ___ ____ 5x = -10 5 5 x = -2 x y y = x - 1 y = -2 - 1 y = -3, -3) Does it matter which equation we use to substitute x? Why would we want to use y = x – 1 instead of 3x + 2y = -12?

17. 3x + 2y = -12 3 (-2) + 2 (-3) = -12 -6 + -6 = -12 y = x - 1 -3 = -2 - 1 The values work in both equations, so we are correct! -3 = -3 3x + 2y = -12 (-2, -3) y = x – 1 x y -12 = -12

18. 22) x = 1/2y – 3 4x – y = 10 x = 1/2y – 3 4(1/2y – 3) – y = 10 2y – 12 – y = 10 y – 12 = 10 +12 = +12 y = 22 y x Does it matter which equation we use to substitute y? x = ½(22) – 3 x = 11– 3 x = 8 (8,

19. x = ½ y - 3 8 = ½ (22) - 3 8 = 11 - 3 4x – y = 10 4(8) – 22 = 10 The values work in both equations, so we are correct! 32 – 22 = 10 8 = 8 x = 1/2y – 3 (8, 22) 4x – y = 10 x y 10 = 10

20. No Solution x = -5y + 4 3x + 15y = -1 x = -5y + 4 3(-5y + 4) + 15y = -1 -15y + 12 + 15y = -1 12 = -1 Does 12 = -1? What is our answer?

21. x = -5y + 4 No solution 3x + 15y = -1 There aren’t any values to substitute and check. So, what do we do?

22. Anytime the answer is No Solution…. x = -5y + 4 3x + 15y = -1 x = -5y + 4 3(-5y + 4) + 15y = -1 -15y + 12 + 15y = -1 12 = -1 Does 12 = -1? Just go back and check your work and make sure there is no solution.

23. -1) 2x – 5y = 29 x = -4y + 8 2(-4y + 8) – 5y = 29 - 8y + 16 – 5y = 29 - 13y + 16 = 29 -16 = -16 -13y = 13 ___ ___ -13 -13 y = -1 x y Does it matter which equation we use to substitute y? x = -4y + 8 x = -4(-1) + 8 x = 4 + 8 x = 12 (12,

24. 2x – 5y = 29 2 (12) – 5 (-1) = 29 24 – (-5) = 29 x = -4y + 8 12 = -4 (-1) + 8 The values work in both equations, so we are correct! 24 + 5 = 29 2x – 5y = 29 (12, -1) x = -4y + 8 x y 29 = 29 12 = 4 + 8 12 = 12

26. 1)One equation will be ISOLATED (it will have either x or y by itself) or can be solved for x or y easily. 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 equation from Step 1 and solve. 4)Your SOLUTION is the ordered pair formed by x & y. 5)CHECK the solution in each of the original equations.