systems of equations in three variables.

**quikwerk** · December 16th 2010, 04:13 PM

I have a question here that I have only partially worked.

Question:
3c+4y+z=7
2y+z=3
-5+3y+8z=-31

Working:
Z=3-2y, given because 2y+z=3.

$3c+4y+z=7 \rightarrow 3c+4y+3-2y=7 \rightarrow 3c+2y+3=7$ so $3x+2y+3-3=7-3 \rightarrow 3x+2y=4 \therefore y=2-\frac{3}{2}x$

My calculations break down here, where I try to solve the third equation for x...
I can't seem to finish the operation.
Here is the attempted working:
[Math]-5c+3y+8z=37 \rightarrow -5+3(2-\frac{3}{2}x)+8z=-31 \rightarrow -5x+6-\frac{9}{2}x+8z=-31 \rightarrow 9\frac{1}{2}C+6+8z=-31[/tex]so $-9\frac{1}{2}C+6-6+8z=-31-6 \rightarrow -9\frac{1}{2}x+8z=-37$
This is the point at which my calculations fall apart.

Thank you!

**dwsmith** · December 16th 2010, 04:14 PM

You can set up the coefficient matrix and do the reduced row echelon form which would be more efficient and probably a little easier.

$\displaystyle \begin{bmatrix} 3 & 4 & 1 & 7\\ 0 & 2 & 1 & 3\\ -5 & 3 & 8 & -31 \end{bmatrix}\Rightarrow rref=\begin{bmatrix} 1 & 0 & 0 & -2\\ 0 & 1 & 0 & 5\\ 0 & 0 & 1 & -7 \end{bmatrix}$

**Wilmer** · December 16th 2010, 04:57 PM

Originally Posted by quikwerk

3c+4y+z=7
2y+z=3
-5+3y+8z=-31

Rewrite:
4y + z = 7 - 3c [1]
2y + z = 3 [2]
3y + 8z = -26 [3]

[1] - [2]: 2y = 4 - 3c [4]

[2]*8: 16y + 8z = 24 [5]

[5] - [3] : y = 50/13

OK? Take over...

**HallsofIvy** · December 17th 2010, 12:31 AM

Originally Posted by quikwerk

I have a question here that I have only partially worked.

Question:
3c+4y+z=7
2y+z=3
-5+3y+8z=-31

First, I recommend you take much more care in copying the problem- this is probably NOT the system you intended. First it would seem peculiar to me to have variables "c", "y", and "z" rather than "x", "y", and "z". More importantly you use both "x" and "c" in your calculations (not to mention "C"). Also I doubt that "-5" in the third equation. I suspect that the system you really meant was
3x+4y+z=7
2y+z=3
-5x+3y+8z=-31

Since the second equation does not involve x, eliminate it from the other two equations:
multipy the first equation by 5: 15x+ 20y+ 5z= 35 and the third equation by 3: -15x+ 9y+ 24z= -93. Now add those two equations: 29y+ 29z= -58. Divide the equation by 29: y+ z= -2.

Now you have y+ z= -2 and 2y+ z= 3.

Working:
Z=3-2y, given because 2y+z=3.

$3c+4y+z=7 \rightarrow 3c+4y+3-2y=7 \rightarrow 3c+2y+3=7$ so $3x+2y+3-3=7-3 \rightarrow 3x+2y=4 \therefore y=2-\frac{3}{2}x$

My calculations break down here, where I try to solve the third equation for x...
I can't seem to finish the operation.
Here is the attempted working:
[Math]-5c+3y+8z=37 \rightarrow -5+3(2-\frac{3}{2}x)+8z=-31 \rightarrow -5x+6-\frac{9}{2}x+8z=-31 \rightarrow 9\frac{1}{2}C+6+8z=-31[/tex]so $-9\frac{1}{2}C+6-6+8z=-31-6 \rightarrow -9\frac{1}{2}x+8z=-37$
This is the point at which my calculations fall apart.

Thank you!

**Soroban** · December 17th 2010, 05:32 AM

Hello, quikwerk!

I agree with HallsofIvy . . .You have a couple of typos.
And it looks like you're using Substitution.

$\begin{array}{cccc}\;3x+4y+z&=&7 & {\bf[1]} \\ \qquad \;2y+z &=& 3 & {\bf[2]} \\ \text{-}5x+3y+8z &=&\text{-}31 & {\bf[3]} \end{array}$

From [2], we have: . $z \:=\:3-2y$ .[4]

Substitute into [1]: . $3x + 4y + (3-2y) \:=\:7 \quad\Rightarrow\quad 3x + 2y \:=\:4$ .[5]

Substitute into [3]: . $\text{-}5x + 3y + 8(3-2y) \:=\:\text{-}31 \quad\Rightarrow\quad \text{-}5x - 13y \:=\:\text{-}55$ .[6]

From [5], we have: . $y \:=\:\dfrac{4-3x}{2}$ .[7]

Substitute into [6]: . $\text{-}5x - 13\left(\dfrac{4-3x}{2}\right) \:=\:\text{-}55 \quad\Rightarrow\quad x \:=\:\text{-}2$

Substitute into [7]: . $y \:=\:\dfrac{4-3(\text{-}2)}{2} \quad\Rightarrow\quad y \:=\:5$

Substitute into [4]: . $z \:=\:3 - 2(5) \quad\Rightarrow\quad z \:=\:\text{-}7$

Therefore: . $\begin{Bmatrix}x &=& \text{-}2 \\ y &=& 5 \\ z &=& \text{-}7 \end{Bmatrix}$

**quikwerk** · December 17th 2010, 11:27 AM

Stupid touch keypad...I should have proof read my post. Sorry guys.
Thank you very much for the help!

**quikwerk** · December 21st 2010, 01:08 PM

In the interest of understanding the way to work this question completely, I decided to dissect the working that Soroban gave me:

Originally Posted by Soroban

Hello, quikwerk!

I agree with HallsofIvy . . .You have a couple of typos.
And it looks like you're using Substitution.

From [2], we have: . $z \:=\:3-2y$ .[4]

Substitute into [1]: . $3x + 4y + (3-2y) \:=\:7 \quad\Rightarrow\quad 3x + 2y \:=\:4$ .[5]

Substitute into [3]: . $\text{-}5x + 3y + 8(3-2y) \:=\:\text{-}31 \quad\Rightarrow\quad \text{-}5x - 13y \:=\:\text{-}55$ .[6]

From [5], we have: . $y \:=\:\dfrac{4-3x}{2}$ .[7]

Substitute into [6]: . $\text{-}5x - 13\left(\dfrac{4-3x}{2}\right) \:=\:\text{-}55 \quad\Rightarrow\quad x \:=\:\text{-}2$

Substitute into [7]: . $y \:=\:\dfrac{4-3(\text{-}2)}{2} \quad\Rightarrow\quad y \:=\:5$

Substitute into [4]: . $z \:=\:3 - 2(5) \quad\Rightarrow\quad z \:=\:\text{-}7$

Therefore: . $\begin{Bmatrix}x &=& \text{-}2 \\ y &=& 5 \\ z &=& \text{-}7 \end{Bmatrix}$

In the process I have found that I make some kind of mistake at this step:
Substitute into [6]: . $\text{-}5x - 13\left(\dfrac{4-3x}{2}\right) \:=\:\text{-}55 \quad\Rightarrow\quad x \:=\:\text{-}2$

When I expand it into its smallest steps:
Since $y=\frac{4-3x}{2}$ ,then: $-5x-13y=-55 \rightarrow -5x-13(\frac{4-3x}{2})=-55 \rightarrow -5x-\frac{52-39x}{2}=-55 \rightarrow -5x-26-19.5x=-55 \rightarrow -24.5x-26=-55$ .
I know that the expression $-24.5x-26=-55$ is incorrect if x=-2 and I know that the reason is that I must add 19x not subtract it, since -5x=10.
However, I do not understand just where I am going wrong....

**quikwerk** · December 21st 2010, 01:09 PM

In the interest of understanding the way to work this question completely, I decided to dissect the working that Soroban gave me:

Originally Posted by Soroban

From [2], we have: . $z \:=\:3-2y$ .[4]

Substitute into [1]: . $3x + 4y + (3-2y) \:=\:7 \quad\Rightarrow\quad 3x + 2y \:=\:4$ .[5]

Substitute into [3]: . $\text{-}5x + 3y + 8(3-2y) \:=\:\text{-}31 \quad\Rightarrow\quad \text{-}5x - 13y \:=\:\text{-}55$ .[6]

From [5], we have: . $y \:=\:\dfrac{4-3x}{2}$ .[7]

Substitute into [6]: . $\text{-}5x - 13\left(\dfrac{4-3x}{2}\right) \:=\:\text{-}55 \quad\Rightarrow\quad x \:=\:\text{-}2$

Substitute into [7]: . $y \:=\:\dfrac{4-3(\text{-}2)}{2} \quad\Rightarrow\quad y \:=\:5$

Substitute into [4]: . $z \:=\:3 - 2(5) \quad\Rightarrow\quad z \:=\:\text{-}7$

Therefore: . $\begin{Bmatrix}x &=& \text{-}2 \\ y &=& 5 \\ z &=& \text{-}7 \end{Bmatrix}$

In the process I have found that I make some kind of mistake at this step:
Substitute into [6]: . $\text{-}5x - 13\left(\dfrac{4-3x}{2}\right) \:=\:\text{-}55 \quad\Rightarrow\quad x \:=\:\text{-}2$

When I expand it into its smallest steps:
Since $y=\frac{4-3x}{2}$ ,then: $-5x-13y=-55 \rightarrow -5x-13(\frac{4-3x}{2})=-55 \rightarrow -5x-\frac{52-39x}{2}=-55 \rightarrow -5x-26-19.5x=-55 \rightarrow -24.5x-26=-55$ .
I know that the expression $-24.5x-26=-55$ is incorrect if x=-2 and I know that the reason is that I must add 19.5x not subtract it, since -5x=10.
However, I do not understand just where I am going wrong....

**Archie Meade** · December 21st 2010, 01:31 PM

Originally Posted by quikwerk

In the interest of understanding the way to work this question completely, I decided to dissect the working that Soroban gave me:

In the process I have found that I make some kind of mistake at this step:
Substitute into [6]: . $\text{-}5x - 13\left(\dfrac{4-3x}{2}\right) \:=\:\text{-}55 \quad\Rightarrow\quad x \:=\:\text{-}2$

When I expand it into its smallest steps:
Since $y=\frac{4-3x}{2}$ ,then: $-5x-13y=-55 \rightarrow -5x-13(\frac{4-3x}{2})=-55 \rightarrow -5x-\frac{52-39x}{2}=-55 \rightarrow -5x-26-19.5x=-55 \rightarrow -24.5x-26=-55$ .
I know that the expression $-24.5x-26=-55$ is incorrect if x=-2 and I know that the reason is that I must add 19.5x not subtract it, since -5x=10.
However, I do not understand just where I am going wrong....

$\displaystyle-\left[\frac{52-39x}{2}\right]=-\frac{52}{2}-\left[-\frac{39x}{2}\right]=-\frac{52}{2}+\frac{39x}{2}$

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