Q. 9 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 1–10, solve each system of equations using the method of substitution or the method of elimination. If the system has no solution, say that it is inconsistent.   

$\left\{\begin{array}{l}2x-4y+z=-15\\ x+2y-4z=27\\ 5x-6y-2z=-3\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system is $\left(\frac{7}{4}z+\frac{39}{4},\frac{9}{8}z+\frac{69}{8},z\right)$ where $z$ is any real number.

Step 1. Given Information

We are given a system of equations $\left\{\begin{array}{l}2x-4y+z=-15\\ x+2y-4z=27\\ 5x-6y-2z=-3\end{array}\right.$.

We need to solve the system of using the method of substitution or the method of elimination. 

Step 2. Eliminate y using first and second equation

Multiply both sides of the second equation by $2$

$\begin{array}{rcl}2(x+2y-4z)& =& 2·27\\ 2x+4y-8z& =& 54 ...\left(4\right)\end{array}$

Now add the fourth equation with the first equation

$\begin{array}{rcl}2x-4y+z+2x+4y-8z& =& -15+54\\ 4x-7z& =& 39 ...\left(5\right)\end{array}$

Step 3. Eliminate y using the second and the third equation

Multiply both sides of the second equation by $3$.

$\begin{array}{rcl}3(x+2y-4z)& =& 3·27\\ 3x+6y-12z& =& 81 ...\left(6\right)\end{array}$

Now add the sixth equation and the third equation.

$\begin{array}{rcl}5x-6y-2x+3x+6y-12z& =& -3+81\\ 8x-14z& =& 78\\ 4x-7z& =& 39 ...\left(7\right)\end{array}$

Step 4. Write x in terms of z

As the two equations formed: equations $5$ and $7$ are the same. So we cannot find the exact values of the variables $x$ and $z$.

So we express $x$ in terms of $z$

$\begin{array}{rcl}4x-7z& =& 39\\ 4x& =& 7z+39\\ x& =& \frac{7}{4}z+\frac{39}{4} ...\left(8\right)\end{array}$

Step 5. Express y in terms of z

Substitute $\begin{array}{rcl}& & \frac{7}{4}z+\frac{39}{4} \end{array}$ for $x$ in the second equation and write $y$ in terms of $z$.

$\begin{array}{rcl}x+2y-4z& =& 27\\ \frac{7}{4}z+\frac{39}{4} +2y-4z& =& 27\\ 2y& =& 4z-\frac{7}{4}z+27-\frac{39}{4} \\ 2y& =& \frac{16z-7z}{4}+\frac{108-39}{4}\\ 2y& =& \frac{9}{4}z+\frac{69}{4} \\ y& =& \frac{9}{8}z+\frac{69}{8}\end{array}$

So the ordered triple is given as

$\left(\frac{7}{4}z+\frac{39}{4},\frac{9}{8}z+\frac{69}{8},z\right)$ where $z$ is any real number.