Question

Can someone solve these 2 problems ? 9 and 8

Answer 1

I just answered that on someone elses. the last one is 4x+y+5z=3

answer the other one on the online algebra calcultor 

Answer 2

$\begin{bmatrix}x+y+z=5\\ x-y+2z=-4\\ 4x+y+z=2\end{bmatrix} \\\\ *x+y+z=5 \\y=5-x-z \\\\ *4x+y+z=2 \\4x+(5-x-z)+z=2 \\4x+5-x-z+z=2 \\4x-x+5=2 \\3x+5=2 \\3x=2-5 \\3x=-3 \\\\x= \frac{-3}{3} \\\\x=-1$

$*x-y+2z=-4 \\x-(5-x-z)+2z=-4 \\x-5+x+z+2z=-4 \\x+x+z+2z-5=-4 \\2x+3z-5=-4 \\2x+3z=-4+5 \\2x+3z=1 \\2(-1)+3z=1 \\-2+3z=1 \\3z=1+2 \\3z=3 \\\\z= \frac{3}{3} \\\\z=1$

$*x+y+z=5 \\(-1)+y+(1)=5 \\-1+1+y=5 \\y=5$

x = -1
y = 5
z = 1