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sweetpea82093
Nov 11, 2009, 05:26 PM
Can someone help me with my homeowrk in Algerba II.

s_cianci
Nov 11, 2009, 05:35 PM
We'll help you but we won't do it for you. Do you have any specific questions you need help with? Post them here.

sweetpea82093
Nov 11, 2009, 05:43 PM
I am willing to do the work.
It's a three variable problem
7x-3y+4z=-14
8x+2y-24z=18
6x-10y+8z=-24

s_cianci
Nov 11, 2009, 06:04 PM
I am willing to do the work.
Its a three variable problem
7x-3y+4z=-14
8x+2y-24z=18
6x-10y+8z=-24 For starters, I'd recommend you combine your 1st and 3rd equations and eliminate the z. This will leave you with an equation containing x and y. Then combine your 1st. And 2nd. Equations and again eliminate the z. This will leave you with another equation containing x and y. Then combine those and eliminate the x or y ; your choice. Finally, solve for the remaining variable. Then back-substitute that into one of your 2-variable equations to solve for the 2nd variable. Finally, back-substitute those into one of your original 3-variable equations to solve for the 3rd. And remaining variable.

sweetpea82093
Nov 11, 2009, 06:10 PM
Finally, solve for the remaining variable. Then back-substitute that into one of your 2-variable equations to solve for the 2nd variable. Finally, back-substitute those into one of your original 3-variable equations to solve for the 3rd. And remaining variable.

I don't know how to do from there on.

s_cianci
Nov 11, 2009, 06:12 PM
Finally, solve for the remaining variable. Then back-substitute that into one of your 2-variable equations to solve for the 2nd variable. Finally, back-substitute those into one of your original 3-variable equations to solve for the 3rd. and remaining variable.

i don't know how to do from there on.But that's the last step ; there is no "from there on."

sweetpea82093
Nov 11, 2009, 06:17 PM
Okay so I solved it but got the wrong answers

s_cianci
Nov 11, 2009, 06:34 PM
okay so i solved it but got the wrong answersHow do you know you got the wrong answers? Show me what you did and I can steer you in the right direction.

sweetpea82093
Nov 11, 2009, 06:39 PM
my text book shows me the right answer and that's not what I got, this is what did.

7x-3y+4z=-14
6x-10y+8z=-24
-14x+6y-8z=-38
-8-4y=-62

7x-3y+4z=-14
8x+2y-24z=18
42x-18y+24z=-84
50x-16y=-46

I then combined my two answers and got
32+16y=248
y=13.5

x=3.4
z=2.7/4

KISS
Nov 11, 2009, 06:50 PM
Na, something is wrongs with your logic:

7x-3y+4z=-14
6x-10y+8z=-24

Multiply the 1st equation by -2

14x+6y-8z=28
6x-10y+8z=-24

Now add.

Try not to subtract. Always multiply by a negative number and ADD.

s_cianci
Nov 11, 2009, 06:50 PM
Mistake #1; What's 2(-14)? It's not -38. I'm also lost when you say "I then combined my two answers and got
32+16y=248".

sweetpea82093
Nov 11, 2009, 06:53 PM
how did you get this?


14x+6y-8z=28
6x-10y+8z=-24

sweetpea82093
Nov 11, 2009, 06:57 PM
s_cianci

I combined 50x-16y=-46 and -8x-4y=14
and got y is equal to -5.5
after I changed the mistake you told me about and its still not right

KISS
Nov 11, 2009, 07:14 PM
how did you get this?


14x+6y-8z=28 = (7x-3y+4z=-14)*-2; oops sign problems
6x-10y+8z=-24

New eqn:
-14x+6y-8z=28
6x-10y+8z=-24

sweetpea82093
Nov 11, 2009, 07:18 PM
6x-10y+8z=-24
so where did you get that

KISS
Nov 11, 2009, 08:20 PM
I am willing to do the work.
It's a three variable problem
7x-3y+4z=-14 (The unmodified eqn)
8x+2y-24z=18 (Use later)
6x-10y+8z=-24 (what you asked about)

..

KISS
Nov 11, 2009, 08:26 PM
Take a look here: System of linear equations - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/System_of_linear_equations)

The easiest way to solve is to use Cramer's rule, but your not there yet.

sweetpea82093
Nov 12, 2009, 03:00 PM
Thank you.