Solve the system by substitution.5x - 2y + 3z = 62x - 4y - 3z = 14x + 6y - 8z = 12x= , y=, z=

We can isolate the variable x from the last equation by subtracting -6y and adding 8z to both sides, that is,
[tex]x=12-6y+8z\ldots(A)[/tex]Now, we can substitute this result into the first and second equations. It leads,
[tex]\begin{gathered} 5(12-6y+8z)-2y+3z=6 \\ 2(12-6y+8z)-4y-3z=14 \end{gathered}[/tex]By distributing the numbers into the parentheses, we have
[tex]\begin{gathered} 60-30y+40z-2y+3z=6 \\ 24-12y+16z-4y-3z=14 \end{gathered}[/tex]By collecting similar terms, we have
[tex]\begin{gathered} 60-32y+43z=6 \\ 24-16y+13z=14 \end{gathered}[/tex]Now, by moving the number 60 to the right hand side of first equation and the number 24 to the right hand side of the second equaton, we have
[tex]\begin{gathered} -32y+43z=-54 \\ -16y+13z=-10 \end{gathered}[/tex]Now, let's isolate the variable y from the second equation. It yields,
[tex]\begin{gathered} -16y=-10-13z \\ \text{then} \\ y=\frac{-10}{-16}+\frac{-13z}{-16} \end{gathered}[/tex]which gives
[tex]y=0.625+0.8125z\ldots(B)[/tex]Now, we can substutite this result into the first equation (-32y+43z=-54), that is,
[tex]-32(0.625+0.8125z)+43z=-54[/tex]which gives
[tex]-20-26z+43z=-54[/tex]or equivalently,
[tex]\begin{gathered} -20+17z=-54 \\ 17z=-34 \end{gathered}[/tex]then
[tex]\begin{gathered} z=\frac{-34}{17} \\ z=-2 \end{gathered}[/tex]So, we have obtained the firs result z=-2. In order to find y, we can substitute our last result into equation (B) from above and get
[tex]\begin{gathered} y=0.625+0.8125(-2) \\ y=0.625-1.625 \\ y=-1 \end{gathered}[/tex]and by substituting z=-2 and y=-1 into equation (A), we have
[tex]x=12-6(-1)+8(-2)[/tex]which gives
[tex]\begin{gathered} x=12+6-16 \\ x=12-10 \\ x=2 \end{gathered}[/tex]Therefore, the solution of the system is:
[tex]\begin{gathered} x=2 \\ y=-1 \\ z=-2 \end{gathered}[/tex]