please, please help....cannot figure this out...keep getting negative numbers. Solve by the system of elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5
-2x+2y+3z=0 <--- let's try this one -2x-y+z=-3 <--- and this one 2x+3y+3z=5 \(\begin{array}{llll} -2x+2y+3z=0&\implies &-2x+2y+3z=0\\ -2x-y+z=-3&\times -3\implies &6x+3y-3z=9\\ \hline\\ &&4x+5y+0=9\\ \quad \\ \color{blue}{4x+5y=9} \end{array}\) now let's try -2x+2y+3z=0 <---- this one -2x-y+z=-3 2x+3y+3z=5 <---- and this one \(\bf \begin{array}{llll} -2x+2y+3z=0&\implies &-2x+2y+3z=0\\ 2x+3y+3z=5&\times -1\implies &-2x-3y-3z=-5\\ \hline\\ &&0-y+0=9\\ \quad \\ \color{blue}{y =9} \end{array} \)
hmmm one sec... on the 2nd one
\( \begin{array}{llll} -2x+2y+3z=0&\implies &-2x+2y+3z=0\\ 2x+3y+3z=5&\times -1\implies &-2x-3y-3z=-5\\ \hline\\ &&-4x-y+0=9\\ \quad \\ \color{blue}{-4x-y =9} \end{array}\)
so now we have a system of equations of 2 variables, 4x + 5y = 9 -4x-y = 9 let's us solve that by elimination, we can add them straight up \(\bf \begin{array}{llll} 4x + 5y = 9\\ -4x-y = 9\\ \hline\\ 0x+4y=18\implies y = \cfrac{18}{4}\implies \color{green}{{\bf y = \cfrac{9}{2}}}\\ \quad \\ \textit{let us use our found "y" in say... }\quad 4x + 5y = 9\\ \quad \\ 4x+5\left(\cfrac{9}{2}\right)=9\implies 4x=9-\cfrac{45}{2}\implies 4x = -\cfrac{27}{2}\\ \quad \\ \color{green}{{\bf x = -\cfrac{27}{8}}} \end{array}\)
so now that we have our "x" and "y", let use use it on either... say... hmm \(\bf -2x-y+z=-3\) \(\bf \implies -2\left(-\cfrac{27}{8}\right)-\left(\cfrac{9}{2}\right)+z=-3\implies \cfrac{27}{4}-\cfrac{9}{2}+z=-3\\ \quad \\ z=-3-\cfrac{27}{4}+\cfrac{9}{2}\implies z = -\cfrac{12}{4}-\cfrac{27}{4}+\cfrac{18}{4}\implies \color{green}{z = -\cfrac{21}{4}}\)
so 2 ended up negative... and 1 is positve, if that's what you got, it was ok
Thank you so much.....I kept coming up with those negative numbers and thought I was doing something wrong. This helps to understand it better. Thanks again.
yw
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