# Thread: Systems of equations

1. ## Systems of equations

I have a test tomorrow, and I can do all the problems on the review sheet except these two:

$\displaystyle x+y-3z=9$
$\displaystyle y+4z=6$
$\displaystyle y-4z= 12$

I don't even ever recall him giving us a lesson on three variable equations.

And:

$\displaystyle 2/x-5/y=4$
$\displaystyle 3/x+4/y= 10$

I know for this one I do this:

$\displaystyle A=1/x$
$\displaystyle B=1/y$

$\displaystyle 2A-5B=4$
$\displaystyle 3A+4B=10$

I know you can multiply it from their, but I always get these huge fraction that don't plug in right.

2. Originally Posted by B-Ren
I have a test tomorrow, and I can do all the problems on the review sheet except these two:

$\displaystyle x+y-3z=9$
$\displaystyle y+4z=6$
$\displaystyle y-4z= 12$

I don't even ever recall him giving us a lesson on three variable equations.

And:

$\displaystyle 2/x-5/y=4$
$\displaystyle 3/x+4/y= 10$

I know for this one I do this:

$\displaystyle A=1/x$
$\displaystyle B=1/y$

$\displaystyle 2A-5B=4$
$\displaystyle 3A+4B=10$

I know you can multiply it from their, but I always get these huge fraction that don't plug in right.

For the first question, the 3rd row needs to be changed. To do this, Subtract the 2nd row from it.

Then you'll be able to solve for z, and back substitute to find y and x.

3. Originally Posted by B-Ren
I have a test tomorrow, and I can do all the problems on the review sheet except these two:

$\displaystyle x+y-3z=9$
$\displaystyle y+4z=6$
$\displaystyle y-4z= 12$

add the last two equations ...

2y = 18 ... you should be able to get x and z from here

And:

$\displaystyle 2/x-5/y=4$
$\displaystyle 3/x+4/y= 10$

I know for this one I do this:

$\displaystyle A=1/x$
$\displaystyle B=1/y$

$\displaystyle 2A-5B=4$
$\displaystyle 3A+4B=10$

multiply the first equation by -3, the second by 2 , then sum ...

-6A + 15B = -12
6A + 8B = 20
---------------
23B = 8

B = 1/y = 8/23 ... y = 23/8

2A - 5(8/23) = 4

46A - 40 = 92

46A = 132

A = 1/x = 132/46 = 66/23 ... x = 23/66

.

4. Originally Posted by B-Ren
$\displaystyle x+y-3z=9$
$\displaystyle y+4z=6$
$\displaystyle y-4z= 12$

I don't even ever recall him giving us a lesson on three variable equations.
Ouch!

To make up the deficiency, you might want to try studying some online lessons, such as this one. In general, though, you'll be using the same techniques for three-variable, three-equation systems as you did for the simpler case covered in class.

Warning: As you add more variables and equations, it becomes that much easier to make silly arithmetic errors. (At least, that's the biggest problem I have! *blush*) Do your work very clearly and carefully!