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

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

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

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

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

 Substitution

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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.

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

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

(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 = = +1 2x = 4 __ __ 2 2 x = 2 xy y = x – 1 y = 2 – 1 y = 1 x + y = y = 3 y = 1, 1

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

(-2 3x + 2y = -12 y = x - 1 3x + 2(x – 1) = -12 3x + 2x – 2 = -12 5x – 2 = = + 2 ___ ____ 5x = x = -2 x y y = x - 1 y = 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?

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

22) x = 1/2y – 3 4x – y = 10 x = 1/2y – 3 4(1/2y – 3) – y = 10 2y – 12 – y = 10 y – 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,

x = ½ y = ½ (22) = x – 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

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

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

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

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

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

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.