# Thread: Systems of equations help

1. ## 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.

2. ## if i understand you correctly

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.
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.

3. 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"

4. 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!