Solving systems of three linear equations

You can use substitution, elimination, or graphing to solve a system of three linear equations

In this lesson we’ll look at how to solve systems of three equations with three different variables.

Remember that a solution to a system of equations is the set of numbers that makes all of the equations true. If a three variable system has a solution, it’ll have a solution for each variable.

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Step-by-step example of solving a system of three linear equations

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Solving the system of three equations using elimination

Example

Solve the system of equations.

$-x-5y+z=17$

$-5x-5y+5z=5$

$2x+5y-3z=-10$

So that we can stay organized, let’s number the equations.

[1]  $-x-5y+z=17$

[2]  $-5x-5y+5z=5$

[3]  $2x+5y-3z=-10$

Notice that equations [1] and [3] have $-5y$ and $5y$. We can add these two equations to cancel out a $y$-term.

$(-x-5y+z)+(2x+5y-3z)=17+-10$

Remove parentheses and combine like terms.

$-x-5y+z+2x+5y-3z=17+-10$

$-x+2x-5y+5y+z-3z=17+-10$

$x-2z=7$

You might have also noticed that equations [2] and [3] have $-5y$ and $5y$. So let’s add those together to get another $x$ and $z$ equation.

$(-5x-5y+5z)+(2x+5y-3z)=5+-10$

Remove parentheses and combine like terms.

$-5x-5y+5z+2x+5y-3z=5+-10$

$-5x+2x-5y+5y+5z-3z=5+-10$

$-3x+2z=-5$

Let’s add these new equations together because one has $2z$ and one has $-2z$.

$x-2z=7$

$-3x+2z=-5$

$(x-2z)+(-3x+2z)=7+-5$

Remove parentheses and combine like terms.

$x-2z+-3x+2z=7-5$

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

$-2x=2$

$x=-1$

Choose one of the new equations to plug in $x=-1$ to solve for $z$. We’ll choose $x-2z=7$.

$-1-2z=7$

$-2z=8$

$z=-4$

Choose an original equation to plug in $x=-1$ and $z=-4$ to solve for $y$. We’ll choose $-x-5y+z=17$.

$-(-1)-5y+(-4)=17$

Simplify and solve for $y$.

$1-5y-4=17$

$-5y+1-4=17$

$-5y-3=17$

$-5y=20$

$y=-4$

The solution is $(-1,-4,-4)$ or $x=-1$, $y=-4$ and $z=-4$.

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Let’s do one more.

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If a three variable system has a solution, it’ll have a solution for each variable.

Deciding which method to use to solve the system, and what to do when the system has no solution

Example

Use any method to find the unique solution to the system of equations.

$3a-3b+4c=-23$

$a+2b-3c=25$

$4a-b+c=25$

So that we can stay organized, let’s number the equations.

[1]  $3a-3b+4c=-23$

[2]  $a+2b-3c=25$

[3]  $4a-b+c=25$

None of the terms have the same coefficients so we’ll need to multiply an equation to eliminate a variable. Let’s multiply equation [2] by $3$ so we can eliminate the $a$ term by subtracting it from equation [1]. 

Let’s call the new form of equation [2], equation [2a].

$3(a+2b-3c=25)$

[2a]  $3a+6b-9c=75$

Now let’s subtract [2a] from equation [1] to eliminate the $a$ term.

$(3a-3b+4c)-(3a+6b-9c)=-23-75$

Distribute the minus sign through the parentheses.

$3a-3b+4c-3a-6b+9c=-98$

$3a-3a-3b-6b+4c+9c=-98$

[1.2a]  $-9b+13c=-98$

We’ll call this new equation [1.2a].

Now we need to get another equation with only $b$ and $c$ terms. Let’s use equations [2] and [3].

This time we need to multiply equation [2] by $4$ so we can subtract it from equation [3] and get rid of the $a$ term. We’ll call the new form of equation [2], [2b].

$4(a+2b-3c=25)$

[2b]  $4a+8b-12c=100$

Now we subtract equation [2b] from equation [3] and call the result [3.2b].

[3]  $4a-b+c=25$

$(4a-b+c)-(4a+8b-12c)=25-100$

Distribute the minus sigh through the parentheses.

$4a-b+c-4a-8b+12c=25-100$

Simplify.

$4a-4a-b-8b+c+12c=25-100$

[3.2b]  $-9b+13c=-75$

Now use the two equations with $b$ and $c$ to solve.

[1.2a]  $-9b+13c=-98$

[3.2b]  $-9b+13c=-75$

Now we can subtract the two equations, and we get

$(-9b+13c)-(-9b+13c)=-98-(-75)$

$0=-23$

Isn’t that impossible?

If something like this happens it means that there are no real solutions that can make the three equations true.

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