Q. 10 - 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.    

$\left\{\begin{array}{l}x-4y+3z=15\\ -3x+y-5z=-5\\ -7x-5y-9z=10\end{array}\right.$

Step-by-Step Solution

Answer

The given system is inconsistent.

Step 1. Given Information

We are given a system of equations $\left\{\begin{array}{l}x-4y+3z=15\\ -3x+y-5z=-5\\ -7x-5y-9z=10\end{array}\right.$.

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

Step 2. Eliminate x using the first and second equation

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

$$\begin{gathered} 3(x-4y+3z)=3·15 \\ 3x-12y+9z=45 ...\left(4\right) \end{gathered}$$

Now add the fourth equation with the second equation.

$\begin{array}{rcl}3x-12y+9z+(-3x+y-5z)& =& 45+(-5)\\ 3x-12y+9z-3x+y-5z& =& 45-5\\ -11y+4z& =& 40 ...\left(5\right)\end{array}$

Step 3. Eliminate x using the first and third equation

Multiply both sides of the first equation by $7$.

$$\begin{gathered} 7(x-4y+3z)=7·15 \\ 7x-28y+21z=105 ...\left(6\right) \end{gathered}$$

Now add the sixth equation with the third equation.

$\begin{array}{rcl}7x-28y+21z+(-7x-5y-9z)& =& 105+10\\ 7x-28y+21z-7x-5y-9z& =& 115\\ -33y+12y& =& 115\\ 3(-11x+4y)& =& 115\\ -11x+4y& =& \frac{115}{3} ...\left(7\right)\end{array}$

Step 4. Find the solution

Subtract the fifth equation from the seventh equation

$\begin{array}{rcl}-11x+4y-(-11x+4y)& =& \frac{115}{3}-40\\ -11x+4y+11x-4y& =& \frac{115-120}{3}\\ 0& =& -\frac{5}{3}\end{array}$

This is a false statement.

So the given system of equation has no solution and thus is inconsistent.