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mathcalculus
Group Title
HELP! how to solve system of equations
x2y+z=1
y+2z=4
x+y+3z=5
what do we do when there is a missing variable??? :(
 one year ago
 one year ago
mathcalculus Group Title
HELP! how to solve system of equations x2y+z=1 y+2z=4 x+y+3z=5 what do we do when there is a missing variable??? :(
 one year ago
 one year ago

This Question is Closed

woody84 Group TitleBest ResponseYou've already chosen the best response.0
what are you solving for??
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
x,y, and z
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure what to do since the x is missing.
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
Write the system as: x  2y +z =1 0x +y+ 2z=4 x + y + 3z = 5 In the first and third equations, eliminate the variable x. Then, you will have a system of 2 by 2 equations in the unknowns y and z. Of those two equations, eliminate a variable and solve for the other. If you are allowed to use technology, take a look at the online 3 x 3 linear equation system solver at: http://math.bd.psu.edu/~jpp4/finitemath/3x3solver.html @mathcalculus
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
Question?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
thank you
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im trying to solve it step by step
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
x  2y +z =1 0x +y+ 2z=4 x + y + 3z = 5 x  2y +z =1 x  2y +z =1 x + y + 3z = 5 multiply through by 1 > x  y  3z = 5  3y  2z = 4
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got a little lost. now i know why
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got that, so what about the second equation?
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
@mathcalculus Your Task: Solve this system  Just add them. 3y  2z = 4 y+ 2z=4 
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
it bothers me that the 0x is there.
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
Don't look at it. When I eliminated the "x" in equations #1 and #3, the 0x was not an issue.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got this: 2y=0
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
but it can't be. it's undefined.
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
Note that here: Solve this system  Just add them. 3y  2z = 4 y+ 2z=4  I excluded the 0x term. It is just a placeholder of sorts.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yeah i know.
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
2y = 0 y= 0/2 y = 0> You have division by 0 and division into 0 mixed up.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so y is 0?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay i got that step right.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
it's wrong... :(
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
It is wrong. But, unless you show your steps, I cannot help you find the error.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ok hold on.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got it right! x=1 y=0 z=2 ! =]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
thanks so much! i really aprreciate it!
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.1
You are welcome.
 one year ago
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