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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)
 2 years ago
 2 years 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)
 2 years ago
 2 years ago

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

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

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

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

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

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

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