Question

please help!
Use the substitution method to solve the following system of equations.

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

(–1, 2, –1)

(1, –2, 1)

(–1, –2, 1)

(1, –2, –1)

Answer 1

\[\begin{cases}4x-y-3z=-3\\x-2y-2z=-3\\
x+y+3z=-2\end{cases}\]
Solve for one variable in terms of the other two:
\[\begin{cases}x=\frac{y+3z-3}{4}\\\\x=2y+2z-3\\\\
x=-y-3z-2\end{cases}\]
First two equations give you
\[\frac{y+3z-3}{4}=2y+2z-3\\
y+3z-3=8y+8z-12\\
-7y-5z=-9\\
\color{red}{7y+5z=9}\]
and the last two equations give you
\[2y+2z-3=-y-3z-2\\
\color{red}{3y+5z=1}\]
Solve the red system, then from there solve for x.

Answer 2

how do i solve the red system?

Answer 3

Subtract 3y+5z=1 from 7y+5z=9 . The result is 4y = 8.

Answer 4

Or you can keep up with the substitution method:
\[\begin{cases}7y+5z=9\\
3y+5z=1\end{cases}\]
Solving the second equation for \(5z\), you have
\[5z=1-3y\]
Substituting into the first equation, you get
\[7y+(1-3y)=9\\
4y=8\]
Either way works

Answer 5

im so confused

Answer 6

what is z i have only heard of x,y