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# Systems of Linear Equations in Three Variables

## Solve multiple equations by evaluating them in pairs

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Practice Systems of Linear Equations in Three Variables
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Solving Linear Systems in Three Variables

#### Example 2

Is the point, (-3, 2, 1), a solution to the system:

\begin{align*}x+y+z &= 0\\ 4x+5y+z &= -1?\\ 3x+2y-4z &=-8\end{align*}

Check to see if the point satisfies all three equations.

Equation 1: \begin{align*}(-3)+(2)+(1)=-3+2+1=0 \end{align*}

Equation 2: \begin{align*}4(-3)+5(2)+(1)=-12+10+1=-1 \end{align*}

Equation 3: \begin{align*}3(-3)+2(2)-4(1)=-9+4-4=-9 \neq -8 \end{align*}

Since the third equation is not satisfied by the point, the point is not a solution to the system.

#### Example 3

Solve the following system using linear combinations:

\begin{align*}5x-3y+z &= -1\\ x+6y-4z &=-17\\ 8x-y+5z &= 12\end{align*}

Combine the first and second equations to eliminate \begin{align*}z\end{align*}. Then combine the first and third equations to eliminate \begin{align*}z\end{align*}.

\begin{align*}& 4(5x-3y+z=-1) \quad \Rightarrow \quad 20x-12y+\cancel{4z}=-4\\ & \quad x+6y-4z=-17 \qquad \quad \ \ \underline{\;\;\;\; x+6y-\cancel{4z}=-17}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ 21x-6y=-21\end{align*}

Result from equations 1 and 2: \begin{align*}21x-6y=-21\end{align*}

\begin{align*}& -5(5x-3y+z=-1) \quad \Rightarrow \quad \ -25x+15y-\cancel{5z}=5\\ & \quad \quad 8x-y+5z=12 \qquad \qquad \ \ \underline{\qquad 8x-y+\cancel{5z}=12}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad -17x+14y=17\end{align*}

Result from equations 1 and 3: \begin{align*}-17x+14y=17\end{align*}

Now we have reduced our system to two equations in two variables. We can eliminate \begin{align*}y\end{align*} most easily next and solve for \begin{align*}x\end{align*}.

\begin{align*}& \quad \ 7(21x-6y=-21) \quad \Rightarrow \quad \ \ 147x-\cancel{42y}=-147\\ & 3(-17x+14y=17) \qquad \qquad \ \ \underline{ -51x+\cancel{42y}=51 \;\;\;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 96x=-96\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x=-1\end{align*}

Now find \begin{align*}y\end{align*}:

\begin{align*}21(-1)-6y &= -21\\ -21-6y &= -21\\ -6y &= 0\\ y &= 0\end{align*}

Finally, we can go back to one of the original three equations and use our \begin{align*}y\end{align*} and \begin{align*}z\end{align*} values to find \begin{align*}x\end{align*}.

\begin{align*}5(-1)-3(0) + z & = -1\\ -5 + z & = -1\\ z & = 4\end{align*}

Therefore the solution is (-1, 0, 4).

Don’t forget to check your answer by substituting the point into each equation.

Equation 1: \begin{align*}5(-1)-3(0)+(4)=-5+4=-1 \end{align*}

Equation 2: \begin{align*}(-1)+6(0)-4(4)=-1-16=-17 \end{align*}

Equation 3: \begin{align*}8(-1)-(0)+5(4)=-8+20=12 \end{align*}

#### Example 4

Solve the following system using linear combinations:

\begin{align*}2x+y-z &= 3\\ x-2y+z &= 5\\ 6x+3y-3z &= 6\end{align*}

Combine equations 1 and two to eliminate \begin{align*}z\end{align*}. Then combine equations 2 and 3 to eliminate \begin{align*}z\end{align*}.

\begin{align*}& 2x+y-\cancel{z}=3\\ & \underline{x-2y+\cancel{z}=5}\\ & \quad \ \ 3x-y=8\end{align*}

Result from equations 1 and 2: \begin{align*}3x-y=8\end{align*}

\begin{align*}& 3(x-2y+z=5) \quad \Rightarrow \quad 3x-6y+\cancel{3z}=15\\ & 6x+3y-3z=6 \qquad \qquad \underline{6x+3y-\cancel{3z}=6\;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad 9x-3y=21\end{align*}

Result from equations 2 and 3: \begin{align*}9x-3y=21\end{align*}

Now we have reduced our system to two equations in two variables. Now we can combine these two equations and attempt to eliminate another variable.

\begin{align*}& -3(3x-y=8) \quad \Rightarrow \ \quad -9x+3y=-24\\ & \quad \ 9x-3y=21 \qquad \qquad \underline{\;\;\; 9x-2y=21 \;\;}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 0= -3\end{align*}

Since the result is a false equation, there is no solution to the system.

### Review

1. Is the point, (2, -3, 5), the solution to the system:

\begin{align*}2x+5y-z &= \text{-}16\\ 5x-y-3z &=\text{-}2\\ 3x+2y+4z &= 20\end{align*}

1. Is the point, (-1, 3, 8), the solution to the system:

\begin{align*}8x+10y-z &= 14\\ 11x+4y-3z &=\text{-}23\\ 2x+3y+z &= 10\end{align*}

1. Is the point, (0, 3, 5), the solution to the system:

\begin{align*}5x-3y+2z &= 1\\ 7x+2y-z &= 1\\ x+4y-3z &= \text{-}3\end{align*}

1. Is the point, (1, -1, 1), the solution to the system:

\begin{align*}x-2y+2z &= 5\\ 6x+y-4z &= 1\\ 4x-3y+z &= 8\end{align*}

Solve the following systems in three variables using linear combinations.

1. .
\begin{align*}3x-2y+z &= 0\\ 4x+y-3z &= \text{-}9\\ 9x-2y+2z &= 20\end{align*}
1. .
\begin{align*}11x+15y+5z &= 1\\ 3x+4y+z &= \text{-}2\\ 7x+13y+3z &= 3\end{align*}
1. .
\begin{align*}2x+y+7z &= 5\\ 3x-2y-z &= \text{-}1\\ 4x-y+3z &= 5\end{align*}
1. .
\begin{align*}x+3y-4z &= \text{-}3\\ 2x+5y-3z &= 3\\ \text{-}x-3y+z &= \text{-}3\end{align*}
1. .
\begin{align*}3x-2y-5z &= \text{-}8\\ 3x+2y+5z &= \text{-}8\\ 6x+4y-10z &= \text{-}16\end{align*}
1. .
\begin{align*}x+2y-z &= \text{-}1\\ 2x+4y+z &= 10\\ 3x-y+8z &= 6\end{align*}
1. .
\begin{align*}x+y-z &= \text{-}3\\ 2x-y-z &= 6\\ 4x+y+z &= 0\end{align*}
1. .
\begin{align*}4x+y+3z &= 8\\ 8x+2y+6z &= 15\\ 3x-3y-z &= 5\end{align*}
1. .
\begin{align*}2x+3y-z &= \text{-}1\\ x+2y+3z &= \text{-}4\\ \text{-}x+y-2z &= 3\end{align*}
1. .
\begin{align*}x-3y+4z & = 14\\ \text{-}x+2y-5z & = \text{-}13\\ 2x+5y-3z & =\text{-}5\end{align*}
1. .
\begin{align*}x+y+z &= 3\\ x+y-z &= 3\\ 2x+2y+z &= 6\end{align*}

To see the Review answers, open this PDF file and look for section 3.12.

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### Vocabulary Language: English

Consistent

A system of equations is consistent if it has at least one solution.

eliminating a variable

Eliminating the variable means making the coefficient of the variable equal to zero thereby removing that variable from that particular system and reducing the number of unknown quantities.

Inconsistent

A system of equations is inconsistent if it has no solutions.

linearly independent

Three equations are linearly independent if each equation cannot be produced by a linear combination of the other two. Systems that are linearly independent will have just one solution.

scaling a row

Scaling a row means multiplying every coefficient in the row by any number you choose (besides zero). This can be helpful for getting coefficients to match so that they can be eliminated.

system of equations

A system of equations is a set of two or more equations.