A system of equations 3×3 is a system consisting of three equations with three variables. There are several methods to solve a system of equations 3×3, such as graphing, substitution, and Gaussian elimination.

In this article, we will learn how to solve systems of equations 3×3 using the substitution and elimination methods. We will look at several solved exercises to learn about this topic.

## How to solve systems of equations 3×3?

To solve a system of three equations with three variables, we can use one of several methods. Here are two common approaches:

**Gaussian elimination:**This method consists of adding or subtracting equations to eliminate variables, one at a time until the system is in what is known as row-echelon reduced form. Once the system is in this form, it is easy to solve for the variables.**Cramer’s rule:**This method consists of expressing the solution in terms of determinants of certain matrices. To use Cramer’s rule, it is necessary to be able to calculate determinants, which can be a little more complicated than the methods used in Gaussian elimination.

Here is an example of how to solve a system of three equations with three variables using Gaussian elimination:

Suppose we have the following system:

$$\begin{cases} 3x + 4y – 2z = 0\\2x – 3y + 4z = 11\\x – 2y + 3z = 7 \end{cases}$$

We can start by eliminating the variable *x* from the second and third equations. To do this, we multiply the second equation by 1 and the third equation by -2, and then add the resulting equations:

$$\begin{cases} 2x – 3y + 4z = 11\\-2x + 4y -6z = -14 \end{cases}$$

____________________

$latex y-2z=-3$

Then, we can eliminate the variable *x* from the first and third equations by multiplying the first equation by 1 and the third equation by -3, and then adding the resulting equations:

$$\begin{cases} 3x+4y-2z=0\\-3x+6y-9z=-21 \end{cases}$$

____________________

$latex 10y-11z=-21$

Finally, we can multiply the equation $latex y-2z=-3$ by -10 and add these two equations to eliminate the *y* variable:

$$\begin{cases} -10y+20z=30\\10y-11z=-21 \end{cases}$$

____________________

$latex 9z=9$

Solving this equation for *z*, we find that $latex z = 1$.

Substituting this value into the equation $latex y-2z=-3$, we can solve for *y*: $latex y = -1$.

Substituting these values back into the original equations, we can solve for *x*: $latex x = 2$.

Therefore, the solution for the system is $latex x = 2$, $latex y = -1$ and $latex z = 1$.

## Systems of equations 3×3 – Examples with answers

**EXAMPLE 1**

Find the solution to the following system of three equations with three unknowns by the substitution method:

$$\begin{cases} x+y-2z=1\\2x-4y+z=0\\2y-3z=-1 \end{cases}$$

##### Solution

We number the equations as (i), (ii), (iii):

$$\begin{cases} \text{(i)}~~ x+y-2z=1\\\text{(ii)}~~ 2x-4y+z=0\\ \text{(iii)}~~2y-3z=-1 \end{cases}$$

From the third (iii), we isolate z:

$latex z=\frac{2}{3}y+\frac{1}{3} $

Substitute z in (ii) to obtain (ii’):

$$2x-4y+\frac{2}{3}y+\frac{1}{3}=\ 2x-\frac{10}{3}y+\frac{1}{3}=0 \; \; \text{(ii’)}$$

We also substitute z in (i) to obtain (i’):

$$ x+y-2(\frac{2}{3}y+\frac{1}{3})=\, x-\frac{1}{3}y-\frac{2}{3}=1 \; \; \text{(i’)} $$

From the above equation (i’) we isolate *y*: $latex y=3x-5 $.

Substituting *y* in (ii’), we get: $latex 17-8x=0$. Subtracting, we are left with:

$$ x=\frac{17}{8}$$

Substitute the found value of *x* in (i’) and isolate *y*: $latex \frac{17}{8}-\frac{1}{3}y-\frac{2}{3}=1$

to obtain:

$$y=\frac{11}{8}$$

Finally, we substitute *x* and *y* in the equation (i) and solve for z:

$$\frac{17}{8}+\frac{11}{8}-2z=1$$

$$z=\frac{5}{4}$$

Summarizing, the solution to the system is:

$$\left[ x=\frac{17}{8},y=\frac{11}{8},z=\frac{5}{4}\right] $$

**EXAMPLE **2

**EXAMPLE**

We have the following system of equations of three linear equations with three unknowns:

$$\begin{cases} \text{(i)}~~x+y+z=6 \\ \text{(ii)}~~x-y+2z=5 \\ \text{(iii)}~~x-y-3z=-10 \end{cases}$$

Show, by the elimination method that the solution is: $latex \left[ x=1,y=2,z=3 \right] $.

##### Solution

Subtracting (i) minus (ii) we obtain: $latex 2y-z=1 $

Subtracting (ii) minus (iii) gives: $latex 5z=15$.

It follows that: **$latex z=3 $**

Substituting in the penultimate expression, we are left with: $latex 2y-3=1 $, whose solution is: $latex y=2 $.

Finally, we substitute the values obtained in equation (i): $latex x+2+3=6$.

And we get: **$latex x=1 $.**

**EXAMPLE **3

**EXAMPLE**

Having the following system of equations

$$\begin{cases} \text{(i)}~~2x+3y+z=1 \\ \text{(ii)}~~6x-2y-z=-14 \\ \text{(iii)}~~3x+y-z=1 \end{cases}$$

Find the values of *x*, *y*, *z* that satisfy the system.

##### Solution

We will apply the elimination method.

Multiply equation (i) by 3 and equation (ii) by -1 and add them together:

$$3(2x+3y+z)-(6x-2y-z)=3(1)-(-14) $$

*x* is canceled to obtain: $latex 11y+4z=17 $ (I)

Multiply equation (ii) by 1 and equation (iii) by -2 and add them together:

$$ (6x-2y-z)-2(3x+y-z)=(-14)-2(1) $$

*x* is eliminated to obtain: $latex z-4y=-16 $ (II)

Subtract (I) minus 4(II):

$$(11y+4z)-4(z-4y)=(17)-4(-16)$$

z is eliminated to have: $latex 27y=81 $ (III)

From eq.(III) we solve for *y*: $latex 27y=81 $, and we obtain: $latex y=3 $.

Substitute the obtained value of *y* in (II): $latex z-4(3)=-16 $, whose solution is: $latex z=-4 $.

Finally, we substitute the values obtained for *z* and *y* in (i): $latex 2x+3(3)+(-4)=1$.

By subtracting *x*: $latex 2x+3(3)+(-4)=1$, we obtain: $latex x=-2$.

The solution to the system is: $latex \left[ x=-2,y=3,z=-4\right] $

**EXAMPLE **4

**EXAMPLE**

We have the following system of three equations with three unknowns:

$$\begin{cases} \text{(i)}~~\frac{x}{2}+\frac{y}{2}-\frac{z}{3}=3 \\ \text{(ii)}~~ \frac{x}{3}+\frac{y}{6}-\frac{z}{2}=-5 \\ \text{(iii)}~~\frac{x}{6}-\frac{y}{3}+\frac{z}{6}=0 \end{cases}$$

Obtain the solution by the substitution method.

##### Solution

From (iii) we solve for *z*: $latex \frac{x}{6}-\frac{y}{3}+\frac{z}{6}=0 $,

And we get: $latex z=2y-x $ (iii’)

Substituting in (ii): $latex \frac{x}{3}+\frac{y}{6}-\frac{2y-x}{2}=-5 $

Now, we isolate *y*: $latex y=x+6 $ (ii’)

Substituting (ii’) in (iii’): $latex z=2(x+6)-x=\; x+12 $, that is: $latex z=x+12 $ (iv)

The values obtained in (ii’) and (iv) are used in equation (i):

$$\frac{x}{2}+\frac{x+6}{2}-\frac{x+12}{3}=3 $$

And the variable *x* is isolated: $latex x=6$.

Now substitute the value obtained in (ii’) and solve for *y*: $latex y=6+6=12$.

Finally, we substitute the value of *x* in (iv): $latex z=6+12=18$.

In summary, the solution to the system of equations is: $latex \left[ x=6,y=12,z=18 \right] $

**EXAMPLE **5

**EXAMPLE**

Find the solutions to the following system of equations:

$$\begin{cases} \text{(i)}~~x+y=5 \\ \text{(ii)}~~ x+z=6 \\ \text{(iii)}~~ y+z=7 \end{cases}$$

##### Solution

Subtracting (i) minus (ii) we obtain:

$latex y-z=-1$ (ii’)

Subtracting the above equation (ii’) minus (iii) gives:

$$-2z=-8$$

It follows that: $latex z=4$.

Substituting the value obtained for z in eq.(iii) we are left with:

$latex y+4=7$, from which we obtain: $latex y=3$.

Finally, the obtained value of *y* is substituted in (i):

$latex x+3=5$, from which we obtain: $latex x=2$.

In summary, the system of equations has the following solution:

$$\left[ x=2,y=3,z=4\right] $$

**EXAMPLE **6

**EXAMPLE**

Solve the following system of equations:

$$\begin{cases} \text{(i)}~~3x+2y=2 \\ \text{(ii)}~~ 2y+2z=\frac{3}{2} \\ \text{(iii)}~~x+4z=\frac{4}{3} \end{cases}$$

##### Solution

We will apply the elimination method.

Subtracting equation (ii) from (i):

$$(3x+2y=2)-\left(2y+2z= \frac{3}{2} \right)$$

==> $latex 3x-2z = \dfrac{1}{2}$ (I)

2(I) + (iii):

$$2(3x-2z=\frac{1}{2})+\left(x+4z=\frac{4}{3}\right)$$

==> $latex 7x=\frac{7}{3}$

==> $latex x=\frac{1}{3}$

Substituting $latex x=\frac{1}{3}$ in (i):

$latex 3(\frac{1}{3})+2y=2$ ==> $latex 2y+1=2$,

getting: $latex y=\frac{1}{2}$

The obtained value of *y* is substituted in (ii): $latex 2(\frac{1}{2})+2z=3/2$

and we get: $latex z=\frac{1}{4}$

Summarizing, the solution to the system of equations is:

$$\left[ x=\frac{1}{3},y=\frac{1}{2},z=\frac{1}{4}\right] $$

**EXAMPLE **7

**EXAMPLE**

Find the values of the variables *u*, *v,* and *w* in the following system of equations:

$$\begin{cases} \text{(i)}~~\frac{1}{u}+\frac{1}{v}=5 \\ \text{(ii)}~~ \frac{1}{u}+\frac{1}{w}=6 \\ \text{(iii)}~~\frac{1}{v}+\frac{1}{w}=7 \end{cases}$$

##### Solution

We start by making the following variable change:

$$x=\frac{1}{u}$$

$$y=\frac{1}{v}$$

$$z=\frac{1}{w}$$

To obtain:

$$\begin{cases} \text{(i)}~~x+y=5 \\ \text{(ii)}~~ x+z=6 \\ \text{(iii)}~~y+z=7 \end{cases}$$

The solution of this system is: $latex \left[ x=2,y=3,z=4\right] $ (see example 5).

The change of variable previously made is now reversed:

$latex u=\frac{1}{x}$ sustituyendo el *x=2 *se obtiene: $latex u=\frac{1}{2}$

$latex v=\frac{1}{y}$ sustituyendo el *y=3* se obtiene: $latex v=\frac{1}{3}$

$latex w=\frac{1}{z}$ sustituyendo el *z=4* se obtiene: $latex w=\frac{1}{4}$

**EXAMPLE **8

**EXAMPLE**

Solve the following system of equations:

$$\begin{cases} \text{(i)}~~ \frac{3}{u}+\frac{2}{v}=2 \\ \text{(ii)}~~\frac{2}{v}+\frac{2}{w}=\frac{3}{2} \\ \text{(iii)}~~ \frac{1}{u}+\frac{4}{w}=\frac{4}{3} \end{cases}$$

##### Solution

The following variable change is made:

$$ x=\frac{1}{u}$$

$$y=\frac{1}{v}$$

$$z=\frac{1}{w}$$

To obtain:

$$\begin{cases} \text{(i)}~~ 3x+2y=2 \\ \text{(ii)}~~2y+2z=\frac{3}{2} \\ \text{(iii)}~~ x+4z=\frac{4}{3} \end{cases}$$

and its solution is: $latex \left[ x=\frac{1}{3},y=\frac{1}{2},z=\frac{1}{4}\right] $ (see example 6)

The change of variable previously made is now reversed:

$latex u=\frac{1}{x}$ substituting *x=1/3* we get: $latex u=3$

$latex v=\frac{1}{y}$ substituting *y=1/2* we get: $latex v=2$

$latex w=\frac{1}{z}$ substituting *z=1/4* we get: $latex w=4$

**EXAMPLE **9

**EXAMPLE**

Find the values of x, y, and z that simultaneously satisfy the three given equations:

$$\begin{cases} \text{(i)}~~x+y=1 \\ \text{(ii)}~~y+z=-1 \\ \text{(iii)}~~x+z=-6 \end{cases}$$

##### Solution

Subtracting (i) minus (ii) :

$latex (x+y=1)-(y+z=-1)$ we get: $latex x-z=2$ (ii’)

Adding the above equation (ii’) plus (iii) :

$latex(x-z=2)+(x+z=-6)$ we get: $latex 2x=-4$

It follows that: $latex x=-2$.

Substituting the value obtained for *x* in eq.(i), we are left with:

$latex -2+y=1$, we get: $latex y=3$

Finally, the obtained value of *y* is substituted in (ii):

$latex 3+z=-1$, we get: $latex z=-4$

In summary, the system of equations has the following solution:

$$\left[ x=-2,y=3,z=-4\right] $$

**EXAMPLE **10

**EXAMPLE**

Solve the following system of equations:

$$\begin{cases} \text{(i)}~~ 5x-3z=2 \\ \text{(ii)}~~ -y+2z=-5 \\ \text{(iii)}~~ x+2z=8 \end{cases}$$

##### Solution

We will apply the elimination method.

2(i) + 3(ii):

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

==> $latex 10x-3y=-11$ (I)

(ii) – (iii):

$$(-y+2z=-5)-(x+2z=8)$$

==> $latex x+y=13$ (II)

(I)+3(II):

$$(10x-3y=-11)+3(x+y=13)$$

which reduces to: $latex13x=28$, obtaining: $latex x=\frac{28}{13}$

Substituting $latex x=\frac{28}{13}$ in (iii): $latex \frac{28}{13}+2z=8$, we find the value of *z*: $latex z=\frac{38}{13}$

The obtained value of *z* is substituted in (ii): $latex -y+2(\frac{38}{13})=-5$, and we are left with: $latex y=\frac{141}{13}$

Summarizing, the solution to the system of equations is:

$$\left[ x=\frac{28}{13},y=\frac{141}{13},z=\frac{38}{13}\right]$$

## Systems of equations 3×3 – Practice problems

#### Find the value of x in the system of equations: $$\begin{cases} 2x-2y+z=-5 \\ 3x+y+3z=-1 \\ 4x-y-2z=-12 \end{cases}$$

Write the value of x in the input box.

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