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Bellringer

Bellringer. On piece of paper… Put your name Graph the system 2x – 2y = 4 y – x = 6 State whether the solution is Independent, Dependent, or Inconsistent. 3-2 Solving Systems Algebraically.

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Bellringer

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  1. Bellringer On piece of paper… • Put your name • Graph the system 2x – 2y = 4 y – x = 6 • State whether the solution is Independent, Dependent, or Inconsistent

  2. 3-2 Solving Systems Algebraically M11.D.2.1.4: Write and/or solve systems of equations using graphing, substitution, and/or elimination

  3. Objectives Solving Systems by Substitution Solving Systems by Elimination

  4. x + 3y = 12 –2x + 4y = 9 Solving Systems by Substitution Solve the system by substitution. Step 1:  Solve for one of the variables. Solving the first equation for x is the easiest. x + 3y = 12 x = –3y + 12 Step 2:  Substitute the expression for x into the other equation. Solve for y. –2x + 4y = 9 –2(–3y + 12) + 4y = 9 Substitute for x. 6y – 24 + 4y = 9 Distributive Property 6y + 4y = 33 y = 3.3 Step 3:  Substitute the value of y into either equation. Solve for x. x = –3(3.3) + 12 x = 2.1 The solution is (2.1, 3.3).

  5. 2 p + s = 10.25 4 p + s = 18.75 Relate:  2 • price of a slice of pizza + price of a soda = $10.25 4 • price of a slice of pizza + price of a soda = $18.75 Define:  Let p = the price of a slice of pizza.     Let s = the price of a soda. Write: Real World Example At Renaldi’s Pizza, a soda and two slices of the pizza–of–the–day costs $10.25. A soda and four slices of the pizza–of–the–day costs $18.75. Find the cost of each item. 2p + s = 10.25 Solve for one of the variables. s = 10.25 – 2p

  6. Continued (continued) 4p + (10.25 – 2p) = 18.75 Substitute the expression for s into the other equation. Solve for p. p = 4.25 2(4.25) + s = 10.25 Substitute the value of p into one of the equations. Solve for s. s = 1.75 The price of a slice of pizza is $4.25, and the price of a soda is $1.75.

  7. 3x + y = –9 –3x – 2y = 12 3x + y = –9 –3x – 2y = 12 Two terms are additive inverses, so add. –y = 3 Solve for y. Solving by Elimination Use the elimination method to solve the system. y = –3 3x + y = –9 Choose one of the original equations. 3x + (–3) = –9 Substitute y.Solve for x. x = –2 The solution is (–2, –3).

  8. 2m + 4n = –4 3m + 5n = –3 2m + 4n = –4 10m + 20n = –20 1 3m + 5n = –3–12m–20n = 12 2 –2m = –8 Add. Multiply by 5. 1 2 Multiply by –4. Solving an Equivalent System Solve the system by elimination. To eliminate the n terms, make them additive inverses by multiplying. m = 4 Solve for m. 2m + 4n = –4 Choose one of the original equations. 2(4) + 4n = –4 Substitute for m. 8 + 4n = –4 4n = –12 Solve for n. n = –3 The solution is (4, –3).

  9. –3x + 5y = 6 6x – 10y = 0 –6x + 10y = 12 6x – 10y = 0 Multiply the first line by 2 to make the x terms additive inverses. 0 = 12 Solving a System Without a Unique Solution Solve each system by elimination. a. –3x + 5y = 6 6x – 10y = 0 Elimination gives an equation that is always false. The two equations in the system represent parallel lines. The system has no solution.

  10. –3x + 5y = 6 6x – 10y = –12 –6x + 10y = 12 6x + 10y = –12 Multiply the first line by 2 to make the x terms additive inverses. 0 = 0 {(x, y)| y = x + } 3 5 6 5 Continued (continued) b. –3x + 5y = 6 6x – 10y = –12 Elimination gives an equation that is always true. The two equations in the system represent the same line. The system has an infinite number of solutions:

  11. Homework Pg 128 & 129 # 1,2,18,19,30,31

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