Simultaneous equations

**derekv** · August 1st 2007, 01:51 AM

Solve for x,y,and z. Q.1 4x - y + 3z = 9 3x + y - z = 5 2x + 3y - 2z = 1 Q.2 2a + 4b + 5c = 2 5a + 3b - 2c = 13 3a - 2b - 3c = 0

**tukeywilliams** · August 1st 2007, 01:58 AM

I would write these as matrices:

$\begin{bmatrix} 4 & -1 & -3 \\ 3 & 1 & -1 \\ 2 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 1 \end{bmatrix}$ and then row reduce to find $x,y,z$ .

$\begin{bmatrix} 2 & 4 & 5 \\ 5 & 3 & -2 \\ 3 & -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 13\\ 0 \end{bmatrix}$

**derekv** · August 1st 2007, 02:03 AM

Originally Posted by tukeywilliams

I would write these as matrices:

$\begin{bmatrix} 4 & -1 & -3 \\ 3 & 1 & -1 \\ 2 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 1 \end{bmatrix}$ and then row reduce to find $x,y,x$ .

$\begin{bmatrix} 2 & 4 & 5 \\ 5 & 3 & -2 \\ 3 & -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 13\\ 0 \end{bmatrix}$

Could you explain what you mean please? I don't understand.

**topsquark** · August 1st 2007, 03:43 AM

Originally Posted by derekv

Solve for x,y,and z. Q.1 4x - y + 3z = 9 3x + y - z = 5 2x + 3y - 2z = 1

If you don't know how to do row reduction, how about substitution?
$4x - y + 3z = 9$
$3x + y - z = 5$
$2x + 3y - 2z = 1$

Take the second equation and solve it for z:
$z = 3x + y - 5$

Now insert this value of z into the other two equations:
$4x - y + 3(3x + y - 5) = 9 \implies 13x + 2y = 24$
$2x + 3y - 2(3x + y - 5) = 1 \implies -4x + y = -9$

So we need to solve
$13x + 2y = 24$
$-4x + y = -9$

Now solve the bottom equation for y:
$y = 4x - 9$

and insert this value into the top equation:
$13x + 2(4x - 9) = 24 \implies 21x = 42$

So we have
$21x = 42$

or
$x = 2$ .

Thus
$y = 4(2) - 9 = -1$

and
$z = 3(2) + (-1) - 5 = 0$

Thus the solution to the system is (x, y, z) = (2, -1, 0).

-Dan

**Soroban** · August 1st 2007, 04:48 AM

Hello, derekv!

I'll solve #1 by Elimination . . . it takes a while, though.

1) Solve for x, y and z.

. . $\begin{array}{cccc}4x - y + 3z & = & 9 & {\color{blue}[1]} \\ 3x + y - z & = & 5 & {\color{blue}[2]} \\ 2x + 3y - 2z & = & 1 & {\color{blue}[3]}\end{array}$

Add ${\color{blue}[1]}$ and ${\color{blue}[2]}$ : . $7x + 2z \:=\:14\;\;{\color{blue}[4]}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[2]} by 3:} & 9x+ 3y - 3z & =& 15 \\<br /> \text{Subtract {\color{blue}[3]}:} & 2x + 3y - 2z & = & 1\end{array}$
. . and we have: . $7x - z \:=\:14\;\;{\color{blue}[5]}$

$\begin{array}{cccc}\text{We have {\color{blue}[4]}:} & 7x + 2z& = & 14 \\<br /> \text{Subtract {\color{blue}[5]}:} & 7x - z & = & 14 \end{array}$
. . and we have: . $3z \,=\,0\quad\Rightarrow\quad\boxed{z\,=\,0}$

Substitute into ${\color{blue}[5]}$ : . $7x - 0 \:=\:14\quad\Rightarrow\quad\boxed{x \,=\,2}$

Substitute into ${\color{blue}[2]}$ : . $3(2) + y - 0 \:=\:5\quad\Rightarrow\quad\boxed{y \,=\,\text{-}1}$

Answer: . $(x,\,y,\,z) \:=\:(2,\,\text{-}1,\,0)$

**derekv** · August 1st 2007, 05:19 AM

You make it so much easier to understand, thank you both.

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