Respuesta :

ramalmlki

Answer:

[tex]\boxed{x = 4, ~y = -1,~ and~z = 3}[/tex]

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Step-by-step explanation:

(eq-1) → x + y + z = 6

(eq-2) → x - y + z = 8

(eq-3) → x + y - z = 0

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Eliminate eq-1 and eq-3

[tex]x + y + z = 6[/tex]

x + y - z = 0    

[tex]2z = 6[/tex]

[tex]z = \frac{6}{2}[/tex]

[tex]z = 3[/tex]

.

Eliminate eq-1 and qe-2

[tex]x + y + z = 6[/tex]

x - y + z = 8    

[tex]2y = -2[/tex]

[tex]y = \frac{-2}{2}[/tex]

[tex]y = -1[/tex]

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Substitute the value of z and y to on of equations

[tex]x + y + z = 6[/tex]

[tex]x + (-1) + 3 = 6[/tex]

[tex]x + 2 = 6[/tex]

[tex]x = 6 - 2[/tex]

[tex]x = 4[/tex]

absor201

Answer:

The solution for the following system is: [tex]\mathbf{x=4, y=-1, z=3}[/tex]

Step-by-step explanation:

We need to find the solution to the following system.

[tex]x+y+z=6 \ and \ x-y+z=8 \ and \ x+y-z=0[/tex]

Let:

[tex]x+y+z=6 --eq(1)\\ x-y+z=8 --eq(2)\\ x+y-z=0--eq(3)[/tex]

Adding eq(1) and eq(2)

[tex]x+y+z=6 \\x-y+z=8 \\-------\\2x+2z=14 --eq(4)[/tex]

Adding eq(2) and eq(3)

[tex]x-y+z=8 \\ x+y-z=0\\-------\\2x=8\\x=8/2\\x=4[/tex]

We get value of x=4

Now, putting value of x in eq(4) to find value of z

[tex]2x+2z=14\\Put x=4\\2(4)+2z=14\\8+2z=14\\2z=14-8\\2z=6\\z=6/2\\z=3[/tex]

We get value of z=3

Now, putting value of x and z in eq(1) to find value of y

[tex]x+y+z=6\\Put \ x=4, z=3\\4+y+3=6\\y+7=6\\y=6-7\\y=-1[/tex]

So, value of y=-1

The solution for the following system is: [tex]\mathbf{x=4, y=-1, z=3}[/tex]

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