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OliviaxRaex3
Group Title
Solve the system by elimination.
x+5y4z=10
2xy+5z=9
2x10y5z=0
a.) (5, –1, 0)
b.) (–5, 1, 0)
c.) (–5, –1, 0)
d.) (–5, –1, –2)
 one year ago
 one year ago
OliviaxRaex3 Group Title
Solve the system by elimination. x+5y4z=10 2xy+5z=9 2x10y5z=0 a.) (5, –1, 0) b.) (–5, 1, 0) c.) (–5, –1, 0) d.) (–5, –1, –2)
 one year ago
 one year ago

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OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
@amorfide Helppp!):
 one year ago

OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 ??
 one year ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
x+5y4z=10 equation 1 2xy+5z=9 equation 2 2x10y5z=0 equation 3 okay so you want to rearrange an equation lets rearrange the first one and make x the subject x+5y4z=10 equation 1 x=105y+4z now we want to substitute this into the other two equations we have equation 2 and 3 2xy+5z=9 equation 2 2x10y5z=0 equation 3 these become 2(105y+4z)y+5z=9 equation 2 2(105y+4z)10y5z=0 equation 3 simplify these two get simultaneous equations with two unknowns 2010y+8zy+5z=9 11y+13z=11 equation 2 2010y+8z10y5z=0 20y+3z=20 equation 3 put these side by side 11y+13z=11 equation 2 20y+3z=20 equation 3 now rearrange one of these to get y the subject or z the subject 11y=1113z y=(1113z)/11 substitute this into equation 3 20(1113z)/11 + 3z=20 simplify (220+260z)/11 +3z=20 multiply by 11 to get rid of the fraction 220+260z 33z = 220 157z=0 z=0 now substitute z=0 into 11y+13z=11 equation 2 11y+13(0)=11 11y=11 y=1 substitute z=0 and y=1 into original equation x+5y4z=10 equation 1 x+5(1)4(0)=10 x5=10 x=5 just to make sure substitute x=5 y=1 z=0 into the other two original equations 2xy+5z=9 equation 2 2(5)(1)+5(0)=9 10+1=9 9=9 2x10y5z=0 2(5)10(1)5(0)=0 10+10=0 0=0 x=5 y=1 z=0
 one year ago

OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much!!! Would you be able to help me with one more?
 one year ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
does this help/ does this make sense?
 one year ago

OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
Solve the system by substitution. 2xy+z=4 z=5 2x+3yz=10 a.) (8, 7, 5) b.) (–8, –7, 5) c.) (8, –7, 5) d.)(–8, –7, –5)
 one year ago

OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
Can you help me with this one too? /:
 one year ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
you can do this... you are given the value of Z substitute it into the other 2 equations, then you will have two equations with two unknowns then you will rearrange one of the equations to make y the subject or x the subject lets say you make y the subject substitute y into the other equation you have then you will get a value for x then substitute x into one of the two equations you have to get y
 one year ago

OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
Okk, let me try it. Then will you check my answers?
 one year ago

OliviaxRaex3 Group TitleBest ResponseYou've already chosen the best response.0
Ok.... I'm confused..
 one year ago

amorfide Group TitleBest ResponseYou've already chosen the best response.0
confused with what
 one year ago
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