# System of Equations, Part 2

So last time I solved a system of two equations using the substitution method where the information from one equation is inserted into the other equation. This is the method of choice if it is easy to solve for one of the unknowns. However, the example that I used also lends itself well to the other method: Elimination.

The elimination method, like the substitution method, uses the two equations to generate one equation with one unknown which can be solved. The system I worked on last time was:

x + y = 108
xy = 38

The elimination method is simply adding the two equations together with the goal of eliminating one of the unknowns. So let’s add these two equations:

x + y = 108
xy = 38
2x = 146

Notice that I now have an equation with one unknown. Dividing both sides of this new equation by 2 gives:

2x/2 = 146/2 ⟹ x = 73

the same answer as before. Now use this value for x in one of the original equations. Let’s use the first one:

x + y = 108 ⟹ y = 108 – x = 108 – 73 = 35

So we get (thankfully) the same solution as before. But this example is rather contrived in that the y‘s were conveniently of opposite signs in the given equations. So let’s consider the following system:

2x + 3y = 51
3x + 2y = 49

Adding these two equations together will just give us another equation with two unknowns. But just like we do with single variable equations, we can modify one or both of these. Let me take the first equation, and multiply it by – 2. Why I am doing this will soon be revealed:

-2(2x + 3y) = 51(-2) ⟹ -4x -6y = -102

I will now multiply the second equation by 3:

3(3x + 2y) = 49(3) ⟹ 9x + 6y = 147

Now let’s repeat the system, replacing the equations with the new ones:

-4x -6y = -102
9x + 6y = 147

Notice that if I now add these equations, the y variable will disappear:

-4x -6y = -102
9x + 6y = 147
5x = 45 ⟹ x = 45/5 = 9

I will now substitute this partial solution into the original second equation:

3x + 2y = 49 ⟹ 3(9) + 2y = 49 ⟹ 27 + 2y = 49 ⟹ 2y = 49 – 27
⟹ 2y = 22 ⟹ y = 11

So x = 9 and y = 11 solves both of these equations.

Both methods, substitution and elimination, can either be used to solve a system of equations, but one method may be less work than the other.

Posted on Categories Algebra, Pre-VCE