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Essential Question: How do you know when a system is ready to be solved using elimination?

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Substitution 1) Determine what’s the easiest variable to solve for. a) It could be x or y and it could be in either equation 2) Solve for that variable. 3) Substitute that variable in the other equation with the equation you just solved for. 4) Solve for the variable you have left 5) Use your answer to substitute back in and solve for the other variable 6) Put your answer in (x, y) format

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4x + 3y = 4 2x – y = 7 What is the easiest variable to solve for? ▪ ▪ “y” in the 2 nd equation y = 2x – 7 (hint: we’ll use this again in a second) In the 1 st equation, substitute “y” with “2x – 7”. Solve for x. 4x + 3= 4 (2x – 7) y 4x + 6x – 21 = 4 10x – 21 = 4 10x = 25 x = 2.5

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4x + 3y = 4 2x – y = 7 ▪ So far we know: x = 2.5 Substitute “2.5” for x in either equation, and solve for y. ▪ Hint: Use the equation where we got y by itself (y = 2x – 7), you’ll find it’s faster ▪ y = 2(2.5) – 7 ▪ y = -2 Solution is (2.5, -2)

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3x – y = 0 4x + 3y = 26 What is the easiest equation to solve for? ▪ ▪ “y” in the 1 st equation y = 3x In the 2 nd equation, substitute “y” with “3x”. Solve for x. 4x + 3 = 26 (3x) y 4x + 9x = 26 13x = 26 x = 2

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3x – y = 0 4x + 3y = 26 ▪ So far we know: x = 2 Substitute “2” for x in either equation, and solve for y. ▪ Hint: Use the equation where we got y by itself (y = 3x), you’ll find it’s faster ▪ y = 3(2) ▪ y = 6 Solution is (2, 6)

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To join a health club costs $100 for 2 months or $200 for 6 months. The cost to join includes both a monthly charge and an one-time initiation fee. Find both costs. Solution: Make two equations ▪ Let “m” be the number of months, and “f” be the initiation fee ▪ $100 = 2m + f ▪ $200 = 6m + f Solve them by substitution (next slide)

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100 = 2m + f 200 = 6m + f Solve for “f” in the 1 st equation (least amount of work) ▪ Substitute “100 – 2m” for “f” in the 2 nd equation ▪ Solve for m ▪ Substitute “25” for m in either equation, and solve for f ▪ f = 100 – 2m (use the converted equation) ▪ f = 100 – 2(25) = Solution is (m, f) = (25, 50) f = 100 – 2m 200 = 6m + (100 – 2m) m = 25 25 25 25 50 50 100 – 2m

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Your turn: Solve by substitution At a pizza place, a soda and two slices of pizza costs $10.25. A soda and four slices of pizza costs $18.75. Find the cost of both a slice of pizza as well as the cost of a soda. Soda: $1.75 Pizza: $4.25

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Assignment Page 128 1 – 15 (odd problems) Show your work

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Essential Question: How do you know when a system is ready to be solved using elimination?

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Elimination 1) Determine one variable to eliminate 2) Multiply (if necessary) to make the coefficients in front of those variables into opposites 3) Add the two equations together 4) Solve for the remaining variable 5) Substitute back into either equation to solve for the 2 nd variable 6) Put your answer into (x, y) format

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4x – 2y = 7 x + 2y = 3 Determine the easiest variable to eliminate ▪ Add the two equations together ▪ Solve for the remaining variable ▪ Substitute back into either equation to find the 2 nd variable ▪ Solution is “y”, because it’s coefficients are already 2 & -2 5x = 10 x = 2 (2) + 2y = 3 2y = 1 y = ½ (2, ½)

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Your turn: Solve by elimination -2x + 3y = 14 2x + 2y = 6 Determine the easiest variable to eliminate ▪ Add the two equations together ▪ Solve for the remaining variable ▪ Substitute back into either equation to find the 2 nd variable ▪ Solution is “x”, because it’s coefficients are already 2 & -2 5y = 20 y = 4 (-1, 4) x = -1

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Sometimes, you’ll have to multiply one or both of the equations by a number in order to eliminate 4x + 9y = 1 4x + 6y = -2 Both x terms have the coefficient “4” in front If you multiply all terms in the 2 nd equation by -1, you’ll be able to eliminate the “x” terms 4x + 9y = 1 → 4x + 9y = 1 4x + 6y = -2 ( -1) -4x – 6y= 2

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Your turn: Solve by elimination 2x – 3y = 4 2x – 5y = -6 Determine the easiest variable to eliminate ▪ Add the two equations together ▪ Solve for the remaining variable ▪ Substitute back into either equation to find the 2 nd variable ▪ Solution is “x”, because it’s coefficients are the same, but we need to flip one 2y = 10 y = 5 (9.5, 5) x = 9.5 2x – 3y = 4 -2x + 5y = 6

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Assignment Page 128 18 – 29 (all problems) Show your work

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Essential Question: How do you know when a system is ready to be solved using elimination?

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F ROM Y ESTERDAY : Sometimes, you’ll have to multiply one or both of the equations by a number in order to eliminate 4x + 9y = 1 4x + 6y = -2 Both x terms have the coefficient “4” in front If you multiply all terms in the 2 nd equation by -1, you’ll be able to eliminate the “x” terms 4x + 9y = 1 → 4x + 9y = 1 4x + 6y = -2 ( -1) -4x – 6y= 2

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An example of “or both” 3x + 7y = 15 5x + 2y = -4 1) Decide on a variable to eliminate 2) Multiply each equation by the coefficient in the opposite equation

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3x + 7y = 15 5x + 2y = -4 Eliminating “x” 3x + 7y = 15 5x + 2y = -4 Coefficients need to be opposite in order to eliminate, so multiply the 2 nd equation by -1 15x + 35y = 75 -15x – 6y = 12 x5 x3 → 15x + 35y = 75 → 15x + 6y = -12

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3x + 7y = 15 5x + 2y = -4 Same equation: eliminating “y” 3x + 7y = 15 5x + 2y = -4 Coefficients need to be opposite in order to eliminate, so multiply the 2 nd equation by -1 6x + 14y = 30 -35x – 14y = 28 x2 x7 → 6x + 14y = 30 → 35x + 14y = -28

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Your Turn: Solve by Elimination 2x + 4y = -4 3x + 5y = -3 Answer: (4, -3)

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Sometimes, you’ll eliminate both “x” and “y” terms. Just as with inequalities, the rules are the same about “no solution” (false statement) and “infinite solutions” (true statement). 2x – y = 3 -2x + y = -3 0 = 0True = Infinite Solutions

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An example of no solution -2x + 4y = 6x3-6x + 12y = 18 -3x + 6y = 8x2-6x + 12y = 16 Coefficients need to be opposite in order to eliminate, so multiply the 2 nd equation by -1 -6x + 12y = 18 6x – 12y = -16 0 = 2False = No Solution

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Your turn: Solve by elimination 2x – 3y = 18 -2x + 3y = -6 0 = 12 → no solution

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Assignment Page 129 30 – 41 (all problems) Show your work

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Elimination Using Addition and Subtraction. Solving Systems of Equations So far, we have solved systems using graphing and substitution. Solve the system.

Elimination Using Addition and Subtraction. Solving Systems of Equations So far, we have solved systems using graphing and substitution. Solve the system.

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