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.

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

Short 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-by-step solution

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 $2$

role="math" localid="1646975408325" $\begin{array}{rcl}2(x+2y-4z)& =& 2·27\\ 2x+4y-8z& =& 54...\left(4\right)\end{array}$

Now add the fourth equation with the first equation

role="math" localid="1646975401792" $\begin{array}{rcl}2x-4y+z+2x+4y-8z& =& -15+54\\ 4x-7z& =& 39...\left(5\right)\end{array}$

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 role="math" $z$

role="math" localid="1646975712602" $\begin{array}{rcl}4x-7z& =& 39\\ 4x& =& 7z+39\\ x& =& \frac{7}{4}z+\frac{39}{4}...\left(8\right)\end{array}$

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 $z$.

role="math" localid="1646975925653" $\begin{array}{rcl}x+2y-4z& =& 27\\ \frac{7}{4}z+\frac{39}{4}+2y-4z& =& 27\\ 2y& =& 4z-\frac{7}{4}z+27-\frac{39}{4}\\ 2y& =& \frac{16z-7z}{4}+\frac{108-39}{4}\\ 2y& =& \frac{9}{4}z+\frac{69}{4}\\ y& =& \frac{9}{8}z+\frac{69}{8}\end{array}$

So the ordered triple is given as

role="math" localid="1646975974006" $\left(\frac{7}{4}z+\frac{39}{4},\frac{9}{8}z+\frac{69}{8},z\right)$where $z$ is any real number.