Systems of Linear Equations with 3 Variables: Substitution

If you see a system of equations (such as the one below) that would be very confusing to solve with normal substitution, you can use the Elimination Method. These are our example equations:

$(8 = x + y + z)$, $(11 = 2x + y + 3z)$, $(17 = 3x + 2y + 2z)$

The Elimination Method uses the fact that if $2$ equations have a term in common, such as $4x$, we can subtract one equation from the other, cancelling out the common term's variable.

The best idea would be to use the first and second equations, $(8 = x + y + z)$ and $(11 = 2x + y + 3z)$, because they both have the same $y$ term. It doesn't matter which equation we subtract, but in this example we will subtract the first one from the second because $8$ is smaller than $11$.

Doing this, we get:

$(11 = 2x + y + 3z)$ - $(8 = x + y + z)$ = $(3 = x + 2z)$

Now we have a $2$-variable equation, because the $y$ terms cancelled out, and now we are left with only $x$ and $z$. The next step is to repeat this same method again to get another equation with these same variables, so we can use substitution the rest of the way.

OK then. To get another equation with only $x$ and $z$, we need to get another pair of equations (not the same ones we used earlier) from our original set of three, which have a common $y$ term also.

There is no pair of equations from our original set other than the pair we used earlier which share a common term. The way to scale an equation without making it false is to multiply the whole equation by a number. Here is an example with our first equation.

$3(8 = x + y + z)$ = $(24 = 3x + 3y + 3z)$

In this example we multiplied our first equation by $3$. Now all three terms have coefficients of $3$.

Since the coefficients of all the terms in our first equation are $1$, this equation would be easy to multiply things by, and maybe we should use it. We cannot use the first and second equations again, so we might as well use the first and third equations.

$2(8 = x + y + z)$ = $(16 = 2x + 2y + 2z)$

Since the third equation's $y$ - term had a coefficient of $2$, the first equation was multiplied by $2$ to make its $y$ term's coefficient $2$ also. Now, let's subtract the first equation from the third.

$(17 = 3x + 2y + 2z)$ - $(16 = 2x + 2y + 2z)$ = $(1 = x)$

We concluded that $x$ was equal to $1$!

From the other equation $(3 = x + 2z)$ that we got before, we know that $z$ is also equal to $1$, and therefore $y$ is equal to $6$.

You can check these answers with the original equations at the top of this explanation, because this method works! The Elimination method is good for many-variable equations.

Combine the terms (...a lot)

First off: to solve a system of equation with $n$ variables using algebraic methods, you need to have at least $n$ equations. For example, if you have two variables you'll need two equations, if you have three variables you'll need three equations...and so forth.

So let's say we have three variables, $x, y,$ and $z$. We're also going to need three equations:

A) $x - y = z$

B) $3z + 4y = 2x + 8$

C) $y + 2z = 2 - x$

There's a lot to work with right now, so we want to use the substitution method to figure out what the variables are in terms of each other until we can construct one equation with one variable, and then use that one to figure out all the others.

First Step

There's a clear place to start here, so let's begin: what is $z$ in terms of $x$ and $y$?

The answer to this question is actually right above in equation A! $x - y = z$, so we know that we can replace $z$ in the other equations with $x - y$.

Now, let's start to figure out $y$ in terms of $x$. What is expression C if we replace $z$ with $x - y$? Hit continue when you've figured it out.

$y + 2(x - y) = 2 - x$

$y + 2x - 2y = 2 - x$

$2x - y = 2 - x$

$2x - y = 2 - x$

Now, we have both sides fully simplified... but there's a term with $x$ in it on both sides! It should be pretty easy to figure out what $y$ is in terms of $x$ from here. Isolate $y$ on one side of the equation, and then hit continue.

$2x - y = 2 - x$

$-y = 2 - x - 2x$

$-y = 2 - 3x$

$y = -2 + 3x$

Awesome! Now that we know that $z = x - y$ and $y = -2 + 3x$, we can figure out what $z$ is in terms of $x$, too.

If $z = x - y$ and $y = -2 + 3x$, what is $z$ in terms of $x$? Fully simplify your answer.

$z = x - y$

$z = x - (-2 + 3x)$

$z = x + 2 - 3x$

$z = 2 - 2x$

Now, we know everything in terms of $x$! Since we haven't used equation B yet, we can plug everything into it so that everything is in terms of $x$.

Try it on your own first!

Using the equation $3z + 4y = 2x + 8$, and the fact that $z = 2 - 2x$ and $y = -2 + 3x$, find a numerical value for $x$. If necessary, give your answer as a fully-simplified fraction.

$3z + 4y = 2x + 8$

$3(2 - 2x) + 4(-2 + 3x) = 2x + 8$

$6 + 6x - 8 + 12x) = 2x + 8$

$-2 + 6x = 2x + 8$

$6x - 2x = 8 + 2$

$4x= 10$

$x= \frac{10}{4}$

$x= \frac{5}{2}$

So, the solution for $x$ is $\frac{5}{2}$.

Using our numerical value for $x$ and the fact that $y = -2 + 3x$, find a numerical value for $y$. If necessary, give your answer as a fully-simplified fraction.

$y = -2 + 3x$

$y = -2 + 3(\frac{5}{2})$

$y = -2 + \frac{15}{2}$

$y = -\frac{4}{2} + \frac{15}{2}$

$y = -\frac{4}{2} + \frac{15}{2}$

$y = \frac{11}{2}$

So, the solution for $y$ is $\frac{11}{2}$.

We've got $x$ and $y$--now just $z$ is left. Using any of the equations we know, find the numerical value for $z$. If necessary, give your answer as a fully-simplified fraction.

I chose to just use $z = 2 - 2x$, so:

$z = 2 - 2(\frac{5}{2})$

$z = 2 - \frac{10}{2}$

$z = 2 - 5$

$z = -3$

So, the solution for $z$ is $-3$.

And we're done!

Finally, we know that $x = \frac{5}{2}$, $y = \frac{11}{2}$, and $z = -3$. If you plug these numbers into the original equations, they'll all still be true (and, in fact, that's the best way to check your work)!

So finally, a short summary in steps:

1. Find all the variables in terms of one of the variables using all but one of your equations.

2. Use the one remaining equation to find the numerical value of your chosen variable.

3. Use that number to figure out the solution to all the rest of the variables!

In both of the following examples by patrickJMT, you will have the chance to walk through examples of how to solve systems of equations with $3$ variables, using the process of elimination by addition.

(Video) Solving a System of Equations with $3$ Variables Using Elimination by Addition

by patrickJMT

In this video by patrickJMT, you can see an example of using elimination to solve a system of equations with $3$ variables.

(Video) Solving a System of Equations with $3$ Variables Using Elimination by Addition

by patrickJMT

This video by patrickJMT walks you through another example of solving a system of equations with $3$ variables using elimination.

Systems of Linear Equations: Substitution with 3 Variables

Solving a system of linear equations with 3 variables can seem really overwhelming at first, but with a few steps, you'll be able to tackle any problem!

In this example, we solve a system of three linear equations using the substitution method.

Steps:

1. Start with the System: We have three equations with variables $x$, $y$, and $z$.
2. Isolate a Variable: Choose an equation to isolate $z$ in terms of $x$ and $y$.
3. Substitute: We substitute this expression for $z$ into the other two equations. This reduces the system to two equations with two variables.
4. Solve for One Variable: After substituting, we solve for $y$ first, and then use that value to find $x$. Find the Remaining Variable: Finally, we substitute our values of $x$ and $y$ back into the equation for $z$. Conclusion: The solution to the system is the ordered triple $(x, y, z)$, representing the values that satisfy all three equations simultaneously.

By following these steps, you can tackle any system of linear equations with 3 variables using the substitution method!