 # Part 2.  Review…  Solve the following system by elimination:  x + 2y = 1 5x – 4y = -23  (2)x + (2)2y = 2(1)  2x + 4y = 2 5x – 4y = -23  7x = -21.

## Presentation on theme: "Part 2.  Review…  Solve the following system by elimination:  x + 2y = 1 5x – 4y = -23  (2)x + (2)2y = 2(1)  2x + 4y = 2 5x – 4y = -23  7x = -21."— Presentation transcript:

Part 2

 Review…  Solve the following system by elimination:  x + 2y = 1 5x – 4y = -23  (2)x + (2)2y = 2(1)  2x + 4y = 2 5x – 4y = -23  7x = -21  x = -3  x + 2y = 1  -3 + 2y = 1  2y = 4  y = 2  (-3, 2)

 What do you think you should do to solve this system?  4x + 5y = 7 6x – 2y = -18  (2)4x + (2)5y = (2)7  8x + 10y = 14  (5)6x – (5)2y= (5)-18  30x – 10y = -90 Can you multiply one of the equations to eliminate a variable? We will need to multiply BOTH equations to eliminate one of the variables. Since the “y” has opposite signs, what can we multiply the 1 st equation by and what can we multiply the 2 nd equation by to eliminate the “y”s? We can multiply the 1 st equation by 2 so it will become “10y.” We can multiply the 2 nd equation by 5 so it will become “-10y.”

  8x + 10y = 14 30x – 10y = -90  38x = - 76  x = -2  4x + 5y = 7  4(-2) + 5y = 7  -8 + 5y = 7  5y = 15  y = 3 Now we have the system set up so we can solve it. Solve the system by adding. Now substitute “-2” into one of the ORIGINAL equations and solve for “y.” The solution to the system is (-2, 3).

  3x + 2y = 10 2x + 5y = 3  (2)3x + (2)2y = (2)10 (-3)2x + (-3)5y = (-3)3  6x + 4y = 20 -6x – 15y = -9  -11y = 11  y = -1 Solve the following system by elimination. What do we need to multiply each equation by in order to eliminate one of the variables? Let’s eliminate the “x” variable. Multiply the 1 st equation by 2 to get “6x.” Multiply the 2 nd equation by -3 to get “-6x.” Remember, the signs will need to have opposite values. 3x + 2y = 10 3x + 2(-1) = 10 3x – 2 = 10 3x = 12 x = 4 The solution to the system is (4, -1).

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