Solve Systems of Equations by Elimination

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Solve Systems of Equations by Elimination

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Presentation transcript:

Solve Systems of Equations by Elimination Part 2

3rd: Then plug -3 in for x and solve for y 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 3rd: Then plug -3 in for x and solve for y x + 2y = 1 -3 + 2y = 1 2y = 4 y = 2 (-3, 2) 1st: If we multiply the top equation by “2” then the “y”s will cancel out. 2nd: Now we can add the equations vertically, and solve for x

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

Now we have the system set up so we can solve it. 8x + 10y = 14 30x – 10y = -90 38x = - 76 x = -2 4x + 5y = 7 4(-2) + 5y = 7 -8 + 5y = 7 5y = 15 y = 3 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).

Solve the following system by elimination. 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 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 1st equation by 2 to get “6x.” Multiply the 2nd 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).