I solved this earlier but Im not sure if Im doing it right. Could someone help guide me through this?
Solve for y:
\(x = 7y\)
\(x+y+z=80\)
\(2x-y+z=105\)
x=7y
x+y+z=80
2x-y+z=105
Isolate and substitute
(7y)+y+z=80
8y+z=80
z=80-8y
2(7y)+y+(80-8y)=105
14y+y-8y=25
15y-8y=25
7y=25
y=25/7
x=7y
x+y+z=80
2x-y+z=105
Isolate and substitute
(7y)+y+z=80
8y+z=80
z=80-8y
2(7y)+y+(80-8y)=105
14y+y-8y=25
15y-8y=25
7y=25
y=25/7
x = 7y
x + y + z = 80
2x - y + z = 105 → z = 105 - 2x + y (3)
Subbing the first equation into the second, we have
7y + y + z = 80 → 8y + z = 80 (4)
And subbing the first into (3) we have
z = 105 - 2(7y) + y
z = 105 - 13y (5)
Put (5) into (4)
8y + 105 - 13y = 80
-5y = -25 → y = 5
And z = 105 - 13(5) = 105 - 65 = 40
And x = 7y = 7(5) = 35
So (x, y, z) = (35, 5, 40)