Solve Systems of Equations by Elimination

Methods to Solve Systems of Equations: Graphing (y = mx + b) Substitution Graphing with x- and y- intercepts. Remember, to Solve a System of Equations, you are finding where the lines INTERSECT.

How to solve systems of equations by Elimination: This method is used when the equations are in standard form: Ax + By = C Set up the equations so the variables line up. 2x + y = 4 x – y = 2 Notice that the x’s & y’s line up on top of each other. Notice that the “y” terms in both equations are opposites of each other. If we add this system together, the “y” terms will cancel out and we can solve for “x”.

How to solve systems of equations by Elimination: 2x + y = 4 + x – y = 2 3x = 6 x = 2 Substitute the “x” value into either equation and solve for “y.” 2x + y = 4 2(2) + y = 4 4 + y = 4 y = 0 Add the equations vertically. Get “x” alone. Divide. This is the first part of the ordered pair for your solution. (2, ___) The solution to the system is (2, 0).

Solve the system of equations using the elimination method. x + 3y = 2 -x + 2y = 3 5y = 5 y = 1 Substitute the “y” value into either equation and solve for “x.” x + 3y = 2 x + 3(1) = 2 x + 3 = 2 x = -1 Add the equations vertically. Get “y” alone. Divide. This is part of the ordered pair for your solution. (___, 1) -1 The solution to the system is (-1, 1).

Solve the system of equations using the elimination method. 2x – y = 2 4x + 3y = 24 In these equations, neither the “x” or the “y” will cancel out, we will have to do an extra step first. THINK of what you can multiply the first equation by so that one of the variables could cancel out. If we multiply the top equation by “3” then the “y”s will cancel out. (3)2x – (3)y = (3)2 6x – 3y = 6 4x + 3y = 24 10x = 30 x = 3 2x – y = 2 2(3) – y = 2 6 – y = 2 -y = -4 y = 4 Now we can add the equations vertically. Divide by “-1” in order to get “y” alone. Get “x” alone. Divide. (3, __) 4 The solution to the system is (3, 4).

The solution to the system is (5, -4). Solve the system of equations using the elimination method. 4x + 3y = 8 x - 2y = 13 In these equations, neither the “x” or the “y” will cancel out, we will have to do an extra step first. THINK of what you can multiply one of the equations by so that one of the variables could cancel out. If we multiply the bottom equation by “-4” then the “x”s will cancel out. (-4)x – (-4)2y = (-4)13 4x + 3y = 8 -4x + 8y = -52 11y = -44 y = -4 4x + 3y = 8 4x + 3(-4)= 8 4x – 12 = 8 4x = 20 x = 5 Now we can add the equations vertically. Get “y” alone. Divide. (__, -4) 5 The solution to the system is (5, -4).

Solve Systems of Equations by Elimination Part 2

Review… (-3, 2) 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? 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).