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anonymous
 5 years ago
these are hard to put on here but what is the solution to the system of equations
{ 3x3y+2z=7
<{ z=1
{ 2x3y+z=6
anonymous
 5 years ago
these are hard to put on here but what is the solution to the system of equations { 3x3y+2z=7 <{ z=1 { 2x3y+z=6

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first plug z = 1 into equns. 1 and 3 to give 2 equns. in x and y which u can solve by elimination

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03x 3y +2 = 7 2x 3y +1 = 6 subtractng the above eqautions : x +1 = 1 x = 2 Plug x=2 in second equn: 2(2) 3y + 1 = 6 3y = 6 1 +4 = 3 y = 1 Solution is x=2, y=1 and z =1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[3x4y+2=7\] \[2x3y+1=6\] first is \[3x4y=9\] second is \[2x3y=7\] first is \[3x+4y=9\] second is \[2x3y=7\] multiply first by 2 and second by 3 \[6x+8y=18\] \[6x9y=21\] add tp get \[y=3\] or \[y=3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops maybe i am wrong sorry . let me check

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0jimmy rep is right and i am wrong. i will see why

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh because i wrote 4y and it is 3y. my mistake. sorry
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