Systems of Equations
System of Equations: Elimination Method
Let's solve the system: 2p + 4q = 2 5p + 3q = -9
Since there isn't a coefficient of 1 or -1 we will use the Elimination Method.
We will multiply the first equation by 5 to get: 10p + 20 q = 10
Multiply the second equation by -2 to get: -10p + -6q = 18
Solving Systems of Equations: Elimination
Now the system is: 10p + 20q = 10 -10p - 6q = 18 (Notice: coefficients of p are 10, -10)
You can now add these two equations together to get: 14q = 28
Now solve for q!
q = 2 Now substitute 2 for q in one of the original equations to solve for p
2p + 4(2) = 2
2p + 8 = 2 2p = -6 p = -3
The solution is an ordered pair: (-3, 2)
Remember that the goal of elimination is to create a system of equations with one variable that has coefficients that differ only by their sign.
You can multiply one or both equations by a constant to achieve this.
You will find a practice sheet with solutions in the module. Complete this sheet and when you are ready complete the assesment. You will also find this assessment in the module.