Question

Use the substitution method to solve the following system of equations.

4x – y – 3z = –3
x – 2y – 2z = –3
x + y + 3z = –2

Answer 1

i hate these they take forever

Answer 2

method is long and tedious, eliminate one variable and then substitute back
it is much easier to cheat
http://www.wolframalpha.com/input/?i=4x+%E2%80%93+y+%E2%80%93+3z+%3D+%E2%80%933%2C+x+%E2%80%93+2y+%E2%80%93+2z+%3D+%E2%80%933+%2Cx+%2B+y+%2B+3z+%3D+%E2%80%932

Answer 3

Yeah i just don't get it all...

Answer 4

oh no that is wrong

Answer 5

subtract third equation from second get \[-3y-5z=-1\] or
\[3y+5z=1\]

Answer 6

then multiply the second equation by -4 and add to the first
\[4x – y – 3z = –3 \]
\[-4x-4 y -12z = 8\]
\[-5y-15z=5\] or
\[y+3z=-1\]

Answer 7

now we have two equation
\[y+3z=-1\]
\[3y+5z=1\]
now it is easier to solve
you get \(y=2,z=-1\) and substitute back to solve for \(x\)

Answer 8

You will get x=-1