Respuesta :
We need to solve the next system of linear equations:
[tex]\begin{gathered} x+2y+z=12 \\ 2x-2y-3z=-18 \\ 3x-5y+4z=20 \end{gathered}[/tex]Let's pair the equations to eliminate 1 variable.
The first pair, add both equations:
[tex]\begin{gathered} x+2y+z=12 \\ 2x-2y-3z=-18 \\ --------------- \\ (x+2x)+(2y-2y)+(z-3z)=12-18 \\ -------------------------- \\ 3x+0-2z=-6 \end{gathered}[/tex]The second pair:
[tex]\begin{gathered} 2x-2x-3z=-18 \\ 3x-5y+4z=20 \end{gathered}[/tex]Multiply the first equation by 5 and the second equation by -2:
[tex]\begin{gathered} 5(2x-2x-3z=-18) \\ -2(3x-5y+4z=20) \end{gathered}[/tex]Then, add both equations:
[tex]\begin{gathered} 10x-10y-15z=-90 \\ -6x+10y-8z=-40 \\ -------------- \\ (10x-6x)+(-10y+10y)+(-15z-8z)=(-90-40) \\ ---------------------------- \\ 4x+0-23z=-130 \end{gathered}[/tex]Now, solve the new system :
[tex]\begin{gathered} 3x+0-2z=-6 \\ 4x+0-23z=-130 \end{gathered}[/tex]Multiply the first equation by 4 and multiply the second equation by -3:
[tex]\begin{gathered} 4(3x+0-2z=-6) \\ -3(4x+0-23z=-130) \end{gathered}[/tex][tex]\begin{gathered} 12x-8z=-24 \\ -12x+69z=390 \end{gathered}[/tex]Add both equations:
[tex]\begin{gathered} (12x-12x)+(-8z+69z)=(-24+390) \\ -------------------------- \\ 0+61z=366 \end{gathered}[/tex]Solve for z:
[tex]\begin{gathered} z=\frac{366}{61} \\ \text{Then} \\ z=6 \end{gathered}[/tex]To solve for x, replace the z value on one equation:
[tex]\begin{gathered} 4x+0-23(6)=-130 \\ \text{Then} \\ 4x-138=-130 \\ 4=-130+138 \\ 4x=8 \\ x=\frac{8}{4} \\ x=2 \end{gathered}[/tex]Finally, replace the z and x values:
[tex]\begin{gathered} x+2y+z=12 \\ 2+2y+6=12 \\ \end{gathered}[/tex]Solve for y:
[tex]\begin{gathered} 8+2y=12 \\ 2y=12-8 \\ 2y=4 \\ y=\frac{4}{2} \\ y=2 \end{gathered}[/tex]Hence, the result for each variable are:
x= 2
y= 2
z=6
