# Systems of equations help

• Nov 28th 2009, 07:33 PM
A Beautiful Mind
Systems of equations help
This seems like it requires a lot of intuition because I'm always having trouble figuring the right answers and get different answers...

Like here:

$\displaystyle 3x-5y+5z = 1$
$\displaystyle -5x-2y+3z = 0$
$\displaystyle 7x-y+3z = 0$

My work:

$\displaystyle -5x-2y+3z = 0$
$\displaystyle -7x+y-3z = 0$

$\displaystyle -12x-y = 0$
$\displaystyle 7x +y+3z = 0$
$\displaystyle -5x+3z = 0$
$\displaystyle 5x+2y-3z=0$
$\displaystyle 2y = 0$

Inconsistent. And this isn't how you do it and I don't know why what I'm doing is wrong.
• Nov 28th 2009, 08:12 PM
I<3Kinematics
if i understand you correctly
Quote:

Originally Posted by A Beautiful Mind
This seems like it requires a lot of intuition because I'm always having trouble figuring the right answers and get different answers...

Like here:

$\displaystyle 3x-5y+5z = 1$
$\displaystyle -5x-2y+3z = 0$
$\displaystyle 7x-y+3z = 0$

My work:

$\displaystyle -5x-2y+3z = 0$
$\displaystyle -7x+y-3z = 0$

$\displaystyle -12x-y = 0$
$\displaystyle 7x +y+3z = 0$
$\displaystyle -5x+3z = 0$
$\displaystyle 5x+2y-3z=0$
$\displaystyle 2y = 0$

Inconsistent. And this isn't how you do it and I don't know why what I'm doing is wrong.

(Thinking) if i understand you correctly you are eventually trying to find what each variable equals and the way to do that with a system of equations is to find what each variable equals all with one variable in it such as finding what y equals in terms of x and what z equals in terms of x.

Then insert those equalities into one of the original questions and find what x truly equals.

then plug this number into your equalities for y and z and there you have it.

• Nov 28th 2009, 11:32 PM
aidan
Quote:

Originally Posted by A Beautiful Mind
This seems like it requires a lot of intuition because I'm always having trouble figuring the right answers and get different answers...
Like here:
$\displaystyle 3x-5y+5z = 1$
$\displaystyle -5x-2y+3z = 0$
$\displaystyle 7x-y+3z = 0$

My work:

$\displaystyle -5x-2y+3z = 0$
$\displaystyle -7x+y-3z = 0$ <-- OK
$\displaystyle -12x-y = 0$

$\displaystyle 7x +y+3z = 0$ <-- not consistent

$\displaystyle -5x+3z = 0$
$\displaystyle 5x+2y-3z=0$
$\displaystyle 2y = 0$

Inconsistent. And this isn't how you do it and I don't know why what I'm doing is wrong.

In your "work" you correctly handled the sign for the variables on the Left Hand Side the FIRST TIME.

However, I don't understand why you changed the sign of the y variable and did not change ALL of the Left Hand Side the second time.

It looks as if you understand what you are supposed to do.
Just "watch those signs"
• Nov 28th 2009, 11:34 PM
earboth
Quote:

Originally Posted by A Beautiful Mind
This seems like it requires a lot of intuition <<<< definitely no
because I'm always having trouble figuring the right answers and get different answers...

Like here:

$\displaystyle 3x-5y+5z = 1$
$\displaystyle -5x-2y+3z = 0$
$\displaystyle 7x-y+3z = 0$

...

1. Choose the variable which you want to eliminate in all (3) equations. You started with z. You used the equations #2 and #3. Now do the same with #1 and #2. Multiply the equ.#1 by the leading factor of z of the equ.#2 and multiply the equ.#2 by the leading factor of z of equ.#1:
$\displaystyle \left|\begin{array}{rcl}3x-5y+5z&=&1 / \cdot (3) \\ -5x-2y+3z&=& 0 / \cdot (5)\end{array}\right.$ $\displaystyle \implies$ $\displaystyle \left|\begin{array}{rcl}9x-15y+15z&=&3 \\ -25x-10y+15z&=& 0 \end{array}\right.$
Now subtract columnwise:
$\displaystyle 34x-5y=3$

2. You now have a system of 2 equations in (x,y):

$\displaystyle \left|\begin{array}{rcl}-12x-y &=& 0 \\ 34x-5y&=&3 \end{array}\right.$

Choose the variable you want to eliminate. I guessed that you have chosen y.
Multiply the equ.#1 by the leading factor of y of the equ.#2 and multiply the equ.#2 by the leading factor of y of equ.#1:

$\displaystyle \left|\begin{array}{rcl}-12x-y &=& 0 / \cdot (-5) \\ 34x-5y&=&3 / \cdot (-1) \end{array}\right.$ $\displaystyle \implies$ $\displaystyle \left|\begin{array}{rcl}60x+5y &=& 0 \\ -34x+5y&=&-3 \end{array}\right.$
Subtract columnwise:
$\displaystyle 94x = 3$

3. Solve for x: $\displaystyle 94x = 3~\implies~x=\frac3{94}$

4. Plug in this value into one of the equations at 2. to calculate y. Afterwards plug in the values for x and y into one of the equations at 1. to calculate z.

5. Done!