3x – 5y + 2z = 7 4x + 9y – z = 2 x – 3y + 3z = 7
i dont see any that are scalars; so parallel "planes" is off the table; which means there should be a solution to find
This is the question it asked.. If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and second equations, then the second and third equations.
i could find a det. for it, but I got no idea how I would solve for the numerators using a cramer
elimination is simple enough :)
I don't know how to do these lol. These are the answers i have to choose from though.. A.) x – 3y = 7 3x – 5y = 7 B.) x – 3y = 7 13x + 24y = 13 C.) 11x + 13y = 11 3x – 5y = 7 D.) 11x + 13y = 11 13x + 24y = 13
3x – 5y + 2z = 7 4x + 9y – z = 2 ; *2 x – 3y + 3z = 7 3x – 5y +2z = 7 8x + 18y - 2z =4 --------------- 11x +13y = 11 ; /11 x +13y/11 = 1 4x + 9y – z = 2 ; *3 x – 3y + 3z = 7 12x + 27y –3z = 6 x – 3y + 3z = 7 ------------------ 13x + 24y = 13 ; /13 x + 24y/13 = 1 ------------------------------ x +13y/11 = 1 x + 24y/13 = 1
oh, they want a method and not an answer .... ok
\[\left[\begin{matrix}3 &-5 & 2 \\ 4 & 9&-1\\1&-3&3\end{matrix}\right] \left(\begin{matrix}x\\y \\ z\end{matrix}\right)= \left(\begin{matrix}7\\2\\ 7\end{matrix}\right)\]
@UnkleRhaukus, can you solve the other one i posted, too?
choose the new system of equations that would result: after the variable z is eliminated in the 1st and 2nd the 2nd and 3rd equations become .....
..What?
if you cant help, dont interrupt :)
:( okay. lol
im trying to make sense of the directions
I gotcha :)
if we eliminate z from the first and second equations; the the 3rd equation is unaffected as far as i can tell .... unless im reading the directions wrong
im thinking the answer is "d" but i got no idea what the directions are actually suggesting ....
Well whatever, might as well take your guess over mine (: Thank you anyway lol
:) good luck
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