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New Member
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Nov 11, 2009, 05:26 PM
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Algerba II
Can someone help me with my homeowrk in Algerba II.
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Uber Member
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Nov 11, 2009, 05:35 PM
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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.
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New Member
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Nov 11, 2009, 05:43 PM
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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
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Uber Member
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Nov 11, 2009, 06:04 PM
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 Originally Posted by sweetpea82093
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.
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New Member
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Nov 11, 2009, 06:10 PM
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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.
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Uber Member
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Nov 11, 2009, 06:12 PM
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 Originally Posted by sweetpea82093
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."
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New Member
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Nov 11, 2009, 06:17 PM
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Okay so I solved it but got the wrong answers
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Uber Member
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Nov 11, 2009, 06:34 PM
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 Originally Posted by sweetpea82093
okay so i solved it but got the wrong answers
How do you know you got the wrong answers? Show me what you did and I can steer you in the right direction.
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New Member
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Nov 11, 2009, 06:39 PM
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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
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Uber Member
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Nov 11, 2009, 06:50 PM
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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.
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Uber Member
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Nov 11, 2009, 06:50 PM
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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".
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New Member
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Nov 11, 2009, 06:53 PM
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how did you get this?
14x+6y-8z=28
6x-10y+8z=-24
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New Member
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Nov 11, 2009, 06:57 PM
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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
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Uber Member
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Nov 11, 2009, 07:14 PM
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 Originally Posted by sweetpea82093
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
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New Member
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Nov 11, 2009, 07:18 PM
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6x-10y+8z=-24
so where did you get that
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Uber Member
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Nov 11, 2009, 08:20 PM
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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)
..
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New Member
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Nov 12, 2009, 03:00 PM
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Thank you.
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