that answer is B because first u have to solve for Z in 2x-3y+Z=19
Z will be Z=-19-2x+3y
Answer:
B. (-2,6,3)
Step-by-step explanation:
First we will cancel the z-variable in the first two equations. We will do this by adding the second equation to the first:
[tex]\left \{ {{2x-3y+z=-19} \atop {+(5x+y-z=-7)}} \right. \\\\7x-2y=-26[/tex]
Next we cancel the z-variable in the bottom two equations. We will do this by subtracting the bottom equation from the middle one:
[tex]\left \{ {{5x+y-z=-7} \atop {-(-x+6y-z=35)}} \right. \\\\6x-5y=-42[/tex]
We can now take these equations without z as a system:
[tex]\left \{ {{7x-2y=-26} \atop {6x-5y=-42}} \right.[/tex]
We will make the coefficients of y the same by multiplying the top equation by 5 and the bottom by 2:
[tex]\left \{ {{5(7x-2y=-26)} \atop {2(6x-5y=-42)}} \right. \\\\\left \{ {{35x-10y=-130} \atop {12x-10y=-84}} \right.[/tex]
Next we subtract the bottom equation from the top:
[tex]\left \{ {{35x-10y=-130} \atop {-(12x-10y=-84)}} \right. \\\\23x=-46[/tex]
Divide both sides by 23:
23x/23 = -46/23
x = -2
Substitute this into the first equation without z:
7(-2)-2y = -26
-14-2y = -26
Add 14 to each side:
-14-2y+14 = -26+14
-2y = -12
Divide both sides by -2:
-2y/-2 = -12/-2
y = 6
Substitute both x and y into our first original equation:
2(-2)-3(6)+z = -19
-4-18+z = -19
-22+z = -19
Add 22 to each side:
-22+z+22 = -19+22
z = 3
