Top 3 Methods for Solving Systems of Equations

Hey, guys! Welcome to this video over comparing different methods for solving a system of equations.

If you recall, a system of equations is when you have more than one equation with unknown variables in a given problem. So, in order to solve that problem, you need to be able to find the value of all the variables in each equation.

There are three different ways that you could do this: the substitution method, elimination method, and using an augmented matrix.

In this video, I’m assuming that you already know how to perform each method, so I want to spend a lot of time explaining not how to do them but rather when to use each method.

First, I will verbally tell you when to use each method, then I will write out three different examples, and we will decide together which method is most efficient for each system.

When to Use the Substitution Method

You should use the substitution method when one of the variables in one of your equations has already been isolated (it has a coefficient of 1).

When to Use the Elimination Method

You should use the elimination method when the same variables in all of the equations share the same coefficient, or when they share the same but negative coefficient.

When to Use an Augmented Matrix

You would use an augmented matrix when the substitution and elimination method are either impractical or impossible altogether.

Now, let’s look at three different systems, and use what we’ve just learned to think through which method is most useful for each system.

EXAMPLE SYSTEM #1
\(5x – 58y = -883\)
\(-5x + 2y = -13\)

EXAMPLE SYSTEM #2
\(9x + 4y = 65\)
\(x – 18y = -2\)

EXAMPLE SYSTEM #3
\(2x + 7y – 3z = 47\)
\(x – 4y + 8z = -33\)
\(7x + 2y +10z = 11\)

So, what we will do is go through each system, decide which method would be most efficient, and then solve with that method.

Elimination Method Example

Alright, let’s look at this first equation.

\(5x – 58y = -883\)
\(-5x + 2y = -13\)

Now, thinking back to the explanation I gave on when to use each method, notice what I said about elimination: “You should use the elimination method when the same variables in all of the equations share the same coefficient, or when they share the same but negative coefficient.”

Well this exact thing is true in the case of this particular system. So, let’s solve this system using elimination.

\(-56y = -896\)
\(y = 16\)

Now, we plug our y variable back into one of the original equations. I’ll plug it into the first.

\(5x – 58(16) = -883\)
\(5x – 928 = -883\)
\(5x = 45\)
\(x = 9\)

Great, so we’ve solved this system using elimination, because our same two variables had the same coefficient or when they share the same but negative coefficient (like in our case).

Substitution Method Example

Let’s move on to system #2.

\(9x + 4y = 65\)
\(x – 18y = -2\)

Alright, so again, let’s think back on what was said in our explanation on when to use each method. Recall what was said about substitution: “You should use the substitution method when one of the variables in one of your equations has already been isolated.”

Well, such is the case with this system. Our \(x\)-variable in our second equation has a coefficient of 1. So, let’s solve this system using substitution.

\(9(18y – 2) + 4y = 65\)
\(162y – 18 + 4y = 65\)
\(166y = 83\)
\(y = \frac{1}{2}\)
\(x = 18(\frac{1}{2}) – 2\)
\(x = 7\)

That was very simple to solve using substitution. Remember, the signifier to help you know when to use it is if one of the equations has a variable that is already isolated.

Augmented Matrix Example

Let’s look at our last system, system #3.

\(2x + 7y – 3z = 47\)
\(x – 4y + 8z = -33\)
\(7x + 2y +10z = 11\)

Remember, what we said about when to use an augmented matrix. Well, right now is a good time. Using elimination or substitution for that matter would take a lot more work than would using an augmented matrix.

So, let’s set up our matrix and solve.

\(\begin{bmatrix} \left.\begin{matrix} 2& 7& -3\\ 1& -4& 8\\ 7& 2& 10 \end{matrix}\right| & \begin{matrix} 47\\ -33\\ 11 \end{matrix} \end{bmatrix}\)
\(R_{1}\leftrightarrow R_{2}\)
\(\begin{bmatrix} \left.\begin{matrix} 1& -4& 8\\ 2& 7& -3\\ 7& 2& 10 \end{matrix}\right| & \begin{matrix} -33\\ 47\\ 11 \end{matrix} \end{bmatrix}\)
\(\begin{matrix} \\ -2R_{1}+R_{2}=R_{2}\\ -7R_{1}+R_{3}=R_{3} \end{matrix}\begin{bmatrix} \left.\begin{matrix} 1& -4& 8\\ 0& 15& -19\\ 0& 30& -46 \end{matrix}\right| & \begin{matrix} -33\\ 113\\ 242 \end{matrix} \end{bmatrix}\)
\(\begin{matrix} \\ \frac{R_{2}}{15}=R_{2}\\ \end{matrix}\begin{bmatrix} \left.\begin{matrix} 1& -4& 8\\ 0& 1& \frac{-19}{15}\\ 0& 30& -46 \end{matrix}\right| & \begin{matrix} -33\\ \frac{113}{15}\\ 242 \end{matrix} \end{bmatrix}\)
\(\begin{matrix} 4R_{2}+R_{1}=R_{1}\\ \\ -30R_{2}+R_{3}=R_{3} \end{matrix}\begin{bmatrix} \left.\begin{matrix} 1& 0& \frac{44}{15}\\ 0& 1& \frac{-19}{15}\\ 0& 0& -8 \end{matrix}\right| & \begin{matrix} \frac{-43}{15}\\ \frac{113}{15}\\ 16 \end{matrix} \end{bmatrix}\)
\(\begin{matrix} \\ \\ \frac{R_{3}}{-8}=R_{3} \end{matrix}\begin{bmatrix} \left.\begin{matrix} 1& 0& \frac{44}{15}\\ 0& 1& \frac{-19}{15}\\ 0& 0& 1 \end{matrix}\right| & \begin{matrix} \frac{-43}{15}\\ \frac{113}{15}\\ -2 \end{matrix} \end{bmatrix}\)
\(\begin{matrix} -\frac{44}{15}R_{3}+R_{1}=R_{1}\\ \frac{19}{15}R_{3}+R_{2}=R_{2}\\ \end{matrix}\begin{bmatrix} \left.\begin{matrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{matrix}\right| & \begin{matrix} 3\\ 5\\ -2 \end{matrix} \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}\)
\(x=3\),\(y=5\),\(z=-2\)

I hope that this video over the comparison of methods for solving systems was helpful for you.

See you guys next time!

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