Solution of a system with 3 variables $$\begin{array}{rcrcrcr}
x & - & 2y & + & 3z & = & 7 \\\ 2x & + & y & + & z & = & 4 \\\ -3 x & + & 2y &- &2z & = & -10
\end{array}$$
I have no idea how to do this and my math book is just telling me to do it, and explains nothing. I have a problem containing 3 equations with 3 variables. What do I do? I tried to solve for x and y and then find z but that didn't work and I got the wrong answer. What do I do? My book tells me to "Multiply each side of equation by -1 and add the result to equation 2, also add equations 2 and 3" I have no idea what this means or why they do it, it is never explained in this book anywhere and they seem like completely arbirtrary number not dependent upon anything, like the author is doing it for fun.
 A: An organizational trick I wish I had learned sooner is to represent your system as a matrix, and then solve accordingly. Your system is given by
$$
\begin{bmatrix} 
1 & -2 & 3 & 7\\
2 & 1 & 1 & 4\\
-3 & 2 & -2 & -10\\
\end{bmatrix}
$$
Notice this the entries of the matrix are just the coefficients of $x,y,z$ and the sum of the equation in each row. Let's say you want to solve for $z$ by eliminating $x$ and $y$ from the third equation. Start by substracting twice the first equation from the second, and adding three times the first equation to the third to clear the $x$ from the first and second equations. That is, $$(2x+y+z)-2(x-2y+3z)=5y-5z=4-2(7)=-10$$ and $$3(x-2y+3z)+(-3x+2y-2z)=3(7)+(-10)=11.$$ So the trick is to find some multiple of the leading coefficient to add or subtract from another to get a $0$ coefficient for $x$. This gives in the matrix
$$
\begin{bmatrix} 
1 & -2 & 3 & 7\\
0 & 5 & -5 & -10\\
0 & -4 & 7 & 11\\
\end{bmatrix}
$$
Immediately you have your last two equations as a system in terms of only $y$ and $z$, namely $5y-5z=-10$ and $-4y+7z=11$, which you should be able to solve normally. Continuing in the manner above...
Since $2x+y+z-2(x-2y+3z)=5y-5z=4-2(7)=-10$, and similarly for the third row. Notice also that the second row now gives the equation $5y-5z=-10$, which is equivalent to $y-z=-2$. Simplifying the matrix gives
$$
\begin{bmatrix} 
1 & -2 & 3 & 7\\
0 & 1 & -1 & -2\\
0 & -4 & 7 & 11\\
\end{bmatrix}
$$
This nice thing about having $x$ cleared from the second equations means you can now clear $y$ from the third equations by manipulating the second equation, without having to worry about introducing a new $x$. So adding 4 times the second equation to the third, that is,
$4(y-z)+(-4y+7z)=4(-2)+11=3$, gives a new matrix
$$
\begin{bmatrix} 
1 & -2 & 3 & 7\\
0 & 1 & -1 & -2\\
0 & 0 & 3 & 3\\
\end{bmatrix}
$$
But this last row gives the equation $3z=3$, or $z=1$. Immediately from $y-z=-2$, you see $y=-2+z=-1$. You can then solve for $x$. This is a nice algorithm for solving a system of linear equations of reasonable size.
A: I assume you know how to solve a system of two linear equations such as
$$\left\{ 
\begin{array}{c}
-5x-5y=-5 \\ 
x+4y=-2%
\end{array}%
\right.\qquad(\ast)$$ 
but not how to represent it in terms of a matrix. I start with your system of 3 equations
$$\left\{ 
\begin{array}{c}
x-2y+3z=7 \\ 
2x+y+z=4 \\ 
-3x+2y-2z=-10.%
\end{array}%
\begin{array}{c}
\text{(eq. 1)} \\ 
\text{(eq. 2)} \\ 
\text{(eq. 3)}%
\end{array}%
\right.\qquad(\ast\ast)$$
In order to eliminate one of the variables $x,y,z$ you can replace one of the 3 equations as follows. Multiply another equation by an adequate multiplier and add the result termwise to it. Suppose you want to eliminate $z$. If you multiply eq. 2 by $2$ you obtain the equivalent equation
$$4x+2y+2z=8.$$
Now add it to eq. 3 to get the equation
$$\left( 4-3\right) x+\left( 2+2\right) y+\left( 2-2\right)
z=8-10\Leftrightarrow x+4y=-2,$$
which means you did eliminate the variable $z$. The multiplier $m=2$ was chosen so that $1$ (coeff. of $z$ in eq. 2) $\times m-2$ (coeff.of $z$ in eq.3)$=0$, i.e. $m=2$. Thus you can replace your system by the equivalent one
$$\left\{ 
\begin{array}{c}
x-2y+3z=7 \\ 
2x+y+z=4 \\ 
x+4y=-2%
\end{array}%
\begin{array}{c}
\text{(eq. 1)} \\ 
\text{(eq. 2)} \\ 
\text{(new eq. 3)}%
\end{array}%
\right.$$
Similarly you get
$$\left\{ 
\begin{array}{c}
\left( 1-6\right) x-\left( 2+3\right) y+\left( 3-3\right) z=7-12 \\ 
2x+y+z=4 \\ 
x+4y=-2%
\end{array}%
\begin{array}{c}
\text{(new eq. 1)} \\ 
\text{(eq. 2)} \\ 
\text{(new eq. 3)}%
\end{array}%
\right.$$ 
by multiplying eq. 2 by the multiplier $m=-3$ (so that $1\times c+3=0$, as above) and adding it to eq.1, which simplifies to
$$\left\{ 
\begin{array}{c}
-5x-5y=-5 \\ 
2x+y+z=4 \\ 
x+4y=-2%
\end{array}%
\begin{array}{c}
\text{(new eq. 1)} \\ 
\text{(eq. 2)} \\ 
\text{(new eq. 3)}%
\end{array}%
\right.$$ 
From these new eqs. 1 and 3, you can compute $x$ and $y$ (the initial system $(\ast)$ of 2 eqs. can be solved by this method too. Do you see how?). Inserting them in eq. 2 we obtain $z=4-2x-y$.
A: If you know how to solve two equations in two unknowns, here is how I'd approach it without have to figure out any weird or tricky substitutions.
Multiply the first equation by $-2$ and add it to the second. The point is to eliminate the $x$ term in the equation. Now you have 
$x-2y+3z=7$
$5y-5z=-10$
$-3x+2y-2z=-10$
Now do the same thing with the third equation. Multiply the first by 3 and add to the third to get 
$x-2y+3z=7$
$5y-5z=-10$
$-4y+7z=11$
Now the last two equations are a system of two equations and two unknowns. Solve those like you normally do, and you get $y=$ something and $z=$ something. You can now plug those into the first equation to solve for $x$, and you get the solution.
A: What we would like to do is to transform
$$\begin{array}{rcrcrcr}
x & - & 2y & + & 3z & = & 7 \\\ 2x & + & y & + & z & = & 4 \\\ -3 x & + & 2y &- &2z & = & -10
\end{array}$$
into something of this form.
$$\begin{array}{rcrcrcr}
x &  &  &  &  & = & 3 \\\  &  & y &  &  & = & 12 \\\  &  &  & &z & = & 21
\end{array}$$
I find it easier by labeling equations as E1, E2, E3, etc..
Our starting point is this$\dots$
$$\begin{array}{rrcrcrcr}
(E1)& x & - & 2y & + & 3z & = & 7 \\\ (E2) & 2x & + & y & + & z & = & 4 \\\ (E3) & -3 x & + & 2y &- &2z & = & -10
\end{array}$$
as I solve the systems of equations, move towards a solution, the system of equations will go through several generations. My first goal is to eliminate the $z$ from (E1) and (E2). We do this by using (E3) as a pivot row, as follows
$$\begin{array}{r}
2(E1) + 3(E3) \\\  2(E2)+(E3) \\\  (E3)
\end{array}$$
Notice that (E3) is used in calculating the 2nd generation of all 3 rows. Performing these operations gives us
$$\begin{array}{rrcrcrcr}
(E1)& -7x & + & 2y &  &  & = & -16 \\\ (E2) & x & + & 4y &  &  & = & -2 \\\ (E3) & -3 x & + & 2y &- &2z & = & -10
\end{array}$$
Next, we want to eliminate the $y$s from (E1) and (E3) by using the new (E2) as the pivot row. The operations we wish to perform to produce the 3rd generation are 
$$\begin{array}{r}
(E2) - 2(E1) \\\  (E2) \\\  (E2)-2(E3)
\end{array}$$
giving us
$$\begin{array}{rrcrcrcr}
(E1)& 15x &  &  &  &  & = & 30 \\\ (E2) & x & + & 4y &  &  & = & -2 \\\ (E3) & 7 x &  &  &+ &4z & = & 18
\end{array}$$
Eliminating the $x$s from (E2) and (E3), where use (E1) as the pivot row.
$$\begin{array}{r}
\frac{1}{15}(E1) \\\  \frac{1}{4}\bigg((E2)-\frac{1}{15}(E1)\bigg) \\\  \frac{1}{4}\bigg((E3)-\frac{7}{15}(E1)\bigg)
\end{array}$$
giving us
$$\begin{array}{rrcrcrcr}
(E1)& x &  &  &  &  & = & 2 \\\ (E2) &  &  & y &  &  & = & -1 \\\ (E3) &  &  &  & &z & = & 1
\end{array}$$
This method is called Elimination. You can Google Gaussian Elimination with Back-Substitution for a computationally more efficient method.
A: I have a more simple solution using matrices :
You'll have 3 matrices : $A = 
\begin{bmatrix} 
1 & -2 & 3\\
2 & 1 & 1\\
-3 & 2 & -2\\
\end{bmatrix}; B= 
\begin{bmatrix}
7\\
4\\
10\\
\end{bmatrix}$ and $X =
\begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}
$
You can notice you have one with your coefficients, another one with every lines' result, and the last one with your unknowns.
What we know is that $A*X = B$ ; so we can write $X = A^{-1} * B$
You simply calculate your uknowns.
You should obtain $X = 
\begin{bmatrix}
\frac {-2}{7}\\
\frac {9}{7}\\
\frac {23}{7}\\
\end{bmatrix}
$. So you end up with 
$\left\{ 
\begin{array}{c}
x =\frac {-2}{7} \\ 
y = \frac {9}{7} \\
z = \frac {23}{7}
\end{array}%
\right.$
