February 9th February 9th ## Solving Systems of Equations - Elimination Using MultiplicationAfter studying this lesson, you will be able to: - Solve systems of equations by elimination using multiplication.
Sometimes when we try to add equations together, no variables will cancel out. What we can do is to multiply one or both of the equations by some number so that a variable will cancel out when we add the equations together. Then we can follow the regular steps of addition (elimination): 1. Add the equations together to eliminate one variable. (Write the equations one over the other and add them together...add straight down.) 2. Solve for the remaining variable. 3. Substitute the solution into one of the original equations and solve for the other variable.
Solve 3x + 6y = -6, 5x - 2y = 14 Line up the equations and add straight down to eliminate a variable: 3x + 6y = -6
No variable cancels, so we need to use multiplication to force a variable to cancel. If we multiply the second equation by 3, then the y-variables will cancel:
Now we substitute x = 2 into the first equation and solve. 3 (2) + 6y = -6 6 + 6y = -6 6y = -12 y = -2 The solution is (2, -2)
Solve 3x + 4y = -25, 2x - 3y = 6 Line up the equations and add straight down to eliminate a variable: 3x + 4y = -25
No variable cancels, so we need to use multiplication to force a variable to cancel. If we multiply the first equation by 3 and second equation by 4, then the y-variables will cancel:
Now we substitute x = -3 into the first equation and solve. 3 (-3) + 4y = -25 -9 + 4y = -25 4y = -16 y = -4 The solution is (-3, -4)
Solve 7x -5y = 76, 4x + y = 55 Line up the equations and add straight down to eliminate a variable: 7x - 5y = 76 4x + y = 55 No variable cancels, so we need to use multiplication to force a variable to cancel. If we multiply the second equation by 5, then the y-variables will cancel:
Now we substitute x = 13 into the first equation and solve. 7(13) - 5y = 76 91 - 5y = 76 -5y = -15 y = 3 The solution is (13, 3) |

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