Hmm
what you do is try to eliminate 1 variable in 2 equaions first
we can elminate x's in first and 2nd equation
add first and 2nd equation
x+3y-2z=10
<u>-x-2y+z=-7 +</u>
0x+y-z=3
y-z=3
multiply 2nd equation by 3 and add to last one
-3x-6y+3z=-21
<u>3x+9y-5z=28 +</u>
0x+3y-2z=7
3y-2z=7
we now have
y-z=3
3y-2z=7
multiply first equation by -2 and add to 2nd
-2y+2z=-6
<u>3y-2z=7 +</u>
y+0z=1
y=1
now we can sub back
y-z=3
1-z=3
minus 1
-z=2
times -1
z=-2
sub baack into any equation
x+3(1)-2(-2)=10
x+3+4=10
x+7=10
minus 7
x=3
x=3
y=1
z=-2
(3,1,-2)
$560.673 is how much that nice chain would cost
-33=-11*3
-8=-11+3
so
(x-11)(x+3)
Answer:
4x(x+2)
Step-by-step explanation:
