Substitution:
1. Solve one of the equations for one of the variables in terms of an expression involving the other variable.
2. Substitute the expression derived in step 1 for the "solved for" variable in the second equation.
3. Solve the resulting single variable equation.
4. Substitute the value derived in step 3 for the appropriate variable in either of the original equations and solve for the remaining variable.
Elimination.
1. Multiply one of the equations by a factor that will result in the coefficient on one of the variables being the additive inverse of the coefficient on that same variable in the second equation. You may have to find two multipliers, one for each equation, in order to result in additive inverse coefficients.
2. Add the two equations, term by term. Since you have created the additive inverse situation for one of the variables, the resulting coefficient on that variable will be zero. That variable is eliminated (hence the name of the method).
3. Solve the resulting single variable equation.
4. Take the result of step 3 and substitute for the appropriate variable in either of the original equations, then solve for the remaining variable.
In either case, if at any point you obtain a triviality such as 0 = 0, then you have a consistent/dependent system (the graph is two identical lines and the solution set has infinite elements). If you end up with an absurdity, such as 7 = 0, then you have an inconsistent system (the graph is two parallel lines and the solution set is empty). If you get discrete values for each of your variables, then you have a consistent/independent system (the graphs intersect at a point, and that ordered pair is your solution set)
John
My calculator said it, I believe it, that settles it
