Solve: 3x+2y+2z=-3
2x+3y+3z=-2
-3x-5y+z=-9

Question

Solve: 3x+2y+2z=-3
           2x+3y+3z=-2
           -3x-5y+z=-9

anonymous · Answer

Can you use matrices?

anonymous · Answer

Yes

anonymous · Answer

Well set it up like a matrix first like this

anonymous · Answer

|dw:1358731665356:dw|

anonymous · Answer

then reduce echelon form

anonymous · Answer

How would I do that on a TI-83 plus graphing calculator?

anonymous · Answer

You don't you do it by hand!!

anonymous · Answer

You won't be able to use your calculator in tests so you need to be able to do it by hand

anonymous · Answer

Ok but my son's math teacher is allowing calculator use but we don't know how to do it with this model of calculator. Do you know?

anonymous · Answer

Sorry I don't know how to do it with a calculator :/

anonymous · Answer

Thank you anyway

whpalmer4 · Answer

Have you looked at the manual?  I'm 100% certain that if the calculator has a function for doing this built in, it will describe how to do it.

anonymous · Answer

we found it online thanks! but we still need help on how to do it by hand :)

whpalmer4 · Answer

Okay, now you're talking :-)

$$3x+2y+2z=-3$$$$2x+3y+3z=-2$$$$-3x-5y+z=-9$$
The idea is you're going to add and subtract various combinations of multiples of those equations to end up with some equations in fewer variables, eventually ending up with 1 equation in 1 variable, which after solving, you can back propagate the value through the equations until you have all of them.

anonymous · Answer

ok sounds good

whpalmer4 · Answer

For example, adding the 1st and 3rd equations together would cause the x terms to disappear, giving you an equation in just y and z.

anonymous · Answer

thats is true...i did not think to do that!

whpalmer4 · Answer

Multiplying the second equation by 3 and the third equation by 2 then adding them would similarly give you another equation in y and z.

whpalmer4 · Answer

Now you have two equations in two unknowns, and I suspect that is not unfamiliar territory.

anonymous · Answer

no i can solve it from there...thank you so much for your help :)

whpalmer4 · Answer

The matrix bit that RyanL mentioned is the way you want to do this for big systems, but in essence, what I described is what it does.

anonymous · Answer

thanks again!!

whpalmer4 · Answer

Here's a brief write-up, if you're interested. Probably Khan Academy has a video on this as well. http://people.richland.edu/james/lecture/m116/matrices/matrices.html

anonymous · Answer

i appreciate it!