91Ó°ÊÓ

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.

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

role="math" localid="1646976573461" $$\begin{gathered} 3(x-4y+3z)=3·15 \\ 3x-12y+9z=45...\left(4\right) \end{gathered}$$

Now add the fourth equation with the second equation.

role="math" localid="1646976655356" $\begin{array}{rcl}3x-12y+9z+(-3x+y-5z)& =& 45+(-5)\\ 3x-12y+9z-3x+y-5z& =& 45-5\\ -11y+4z& =& 40...\left(5\right)\end{array}$

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.

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

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.