(z + 15)(z + 3)
x=3 y=2 z=6
y=84-72=12 z=84-52=32 x=84-12-32=40
-3y + z = 12
(2+x+y+6+z) += 9x,y and z are variables, there are 3.
10
(z + 15)(z + 3)
x=3 y=2 z=6
If th equestion meant: (x+y+z)^2The expansion is:(x+y+z)^2= x^2+2xy+y^2+2yz+z^2+2zx
y=84-72=12 z=84-52=32 x=84-12-32=40
-3y + z = 12
z = -4 x-5 y+2
(2+x+y+6+z) += 9x,y and z are variables, there are 3.
I used the matrix method to find the answer: x=4, y=-7, z=-5.
7 (seven)
12+1z
-2