Wednesday, August 28, 2013

Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 40

Solve the system of equations $\begin{equation}
\begin{aligned}

2x + 3y - z =& 0 \\
x - 4y + 2z =& 0 \\
3x - 5y - z =& 0

\end{aligned}
\end{equation}
$. If the system is inconsistent or has dependent equations, say so.


$
\begin{equation}
\begin{aligned}

4x + 6y - 2z =& 0
&& 2 \times \text{ Equation 1}
\\
x - 4y + 2z =& 0
&& \text{Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

5x + 2y \phantom{+2z} =& 0
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-2x - 3y + z =& 0
&& -1 \times \text{ Equation 1}
\\
3x - 5y - z =& 0
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

x - 8y \phantom{+z} =& 0
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5x + 2y =& 0
&& \text{New Equation 2}
\\
x -8y =& 0
&& \text{New Equation 3}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

20x + 8y =& 0
&&4 \times \text{ New Equation 2}
\\
x - 8y =& 0
&&
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

21x \phantom{+8y} =& 0
&& \text{Add}
\\
x =& 0
&& \text{Divide each side by $21$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5(0) + 2y =& 0
&& \text{Substitute } x = 0 \text{ in New Equation 2}
\\
2y =& 0
&& \text{Multiply}
\\
y =& 0
&& \text{Divide each side by $2$}

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

2(0) + 3(0) - z =& 0
&& \text{Substitute } x = 0 \text{ and } y = 0 \text{ in Equation 1}
\\
0 + 0 - z =& 0
&& \text{Multiply}
\\
z =& 0
&& \text{Divide each side by $-1$}

\end{aligned}
\end{equation}
$


The ordered triple is $\displaystyle \left( 0,0,0 \right)$.

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