Given the system of equations:
[tex]\begin{gathered} \begin{cases}{8-y+3z=8} \\ {3x+y_-2z{2x+4y+z=0}\end{cases}}\end{cases} \\ \\ \end{gathered}[/tex]
x - y + 3z = 8
3x + y - 2z = -2
2x + 4y + z = 0
Let's solve the system of equations using elimination method.
Take the first two equations:
x - y + 3z = 8
3x + y - 2z = -2
Since the y variables are opposite of each other, let's eliminate y by adding both equations:
x - y + 3z = 8
+ 3x + y - 2z = -2
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4x + z = 6
Rewrite the equation for z.
Subtract 4x from both sides:
4x - 4x + z = 6 - 4x
z = 6 - 4x
Also, eliminate y in the second and third equations:
3x + y - 2z = -2
2x + 4y + z = 0
Multiply the equation 2 by -4:
-4(3x + y - 2z = -2)
2x + 4y + z = 0
-12x - 4y + 8z = 8
2x + 4y + z = 0
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-10x + 9z = 8
We now have the results:
4x + z = 6
-10x + 9z = 8
Multiply the top equation by -9:
-36x - 9z = -54
-10x + 9z = 8
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-46x = -46
x = 1
Substitute 1 for x into any of the equations:
4x + z = 6
4(1) + z = 6
z = 6 - 4
z = 2
Substitute 2 for z, and 1 for x in any equation and solve for y:
2x + 4y + z = 0
2(1) + 4y + 2 = 0
4 + 4y = 0
4y = -4
y = -1
Therefore, we have the solutions:
x = 1, y = -1, z = 2
