Question

How do you solve the system $-4x-5y-z=18$, $-2x-5y-2z=12$, and $-2x+5y+2z=4$?

Answer 1

Finding the solution to the given system of equations:

Step 1: Finding the value of $x$.

The given system of equations is:

$-4x-5y-z=18$ $...\left(1\right)$

$-2x-5y-2z=12$ $...\left(2\right)$

$-2x+5y+2z=4$ $...\left(3\right)$

Adding the equations $\left(2\right)$ and $\left(3\right)$, we get:

$$\begin{gathered} -2x-5y-2z-2x+5y+2z=12+4 \\ \Rightarrow -4x=16 \\ \Rightarrow x=\frac{16}{-4} \\ \therefore x=-4 \end{gathered}$$

Step 2: Substituting the value of $x$.

Substituting the value of $x$ in the equations $\left(1\right)$ and $\left(3\right)$, we get:

$\begin{array}{rcl}-4x-5y-z& =& 18\\ & \Rightarrow & -4\times \left(-4\right)-5y-z=18\\ & \Rightarrow & -5y-z=18-16\\ & \Rightarrow & -5y-z=2...\left(4\right)\end{array}$

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

Step 3: Finding the value of $z$.

Adding the equations $\left(4\right)$ and $\left(5\right)$, we get:

$\begin{array}{rcl}-5y-z+5y+2z& =& 2-4\\ \therefore z& =& -2\end{array}$

Step 4: Finding the value of $y$

Substituting $x=-4$ and $z=-2$ in equation $\left(1\right)$, we get:

$\begin{array}{rcl}-4x-5y-z& =& 18\\ & \Rightarrow & -4\times \left(-4\right)-5y-\left(-2\right)=18\\ & \Rightarrow & 16-5y+2=18\\ & \Rightarrow & 5y=18-18\\ \therefore y& =& 0\end{array}$

Therefore, the solution of the given system of equations is $x=-4,y=0,z=-2$.