## Tuesday, February 28, 2012

Suppose we wish to solve a system of two equations in two variables by the elimination method or substitution method.  Which method is easiest to use depends on the way the equations are written and personal preference.

In the system     x = 2y + 5       it's easiest to subsitute 2y + 5 for x in the second equation.
2x + 3y = 10

But in the system    x + 6y = 12   it's easiest to eliminate one of the two variables first.
5x + 6y = 20

A rule of thumb that I use to determine which method to use is if either equation in the system is already solved for one variable in terms of the other (see first system above), use the substitution method.

Solving the first sytem using substitution, we get

2(2y+ 5) + 3y = 10

4y + 10 + 3y = 10

7y + 10 = 10

7y = 0

y = 0

Substitute 0 for y in the first equation to get x = 2(0) + 5 = 5

Therefore, the solution is x = 5, y = 0.

Solving the second system, it's easiest to eliminate the y variable by subtracting the second equation from the first. This gives us

-4x = -8

x = 2

Substituting 2 for x in the first equation of the system gives us

2 + 6y = 12

6y = 10

y = 5/3

Therefore, the solution is  x = 2, y = 5/3

1. I have found a primitive solution to the general
form. using abstraction and substitution that can
provide a shortcut it just requires you to arrange
the equations into the general form
Ax + By = C
Dx + Ey = F

Given the two equations
Ax + By = C
Dx + Ey = F
The values for x and y that satisfy both equations is
x = (CE -BF) / (AE -BD) y = (CD-AF)/(-AE+BD)

you can find more information and examples at math journal

2. John, yes thanks for the information. I know that method as well, using determinants and matrices. In fact, I explain this method in my book, Algebra Simplified Intermediate & Advanced.