Question

1- Determine duas soluçoes de cada uma das equaçoes

a) -x + $\frac{1}{2}$ y = 0

b) x + y - 2z = 0

2- Calcule a , de modo que (-1, a +1,2 ) nao seja soluçao da equaçao 2x + 4y + z = 0

Quero resposta correta, com resoluçao passo a passo...sem brincadeiras se nao vou denunciar...grata

Answer 1

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1)

$\tt a)~\sf -x+\dfrac{y}{2}=0\\\sf multiplicando~igualdade~por~2~temos:\\\sf2\cdot(-x)+2\cdot\dfrac{1}{2}y=2\cdot0\\\sf -2x+y=0\implies y=2x\\\sf se~x=0\implies y=2\cdot0=0\\\sf (0,0)~\acute e~soluc_{\!\!,}\tilde ao\\\sf se~x=\dfrac{1}{2}\implies y=\diagup\!\!\!2\cdot\dfrac{1}{\diagup\!\!\!2}=1\\\sf (1,0)~tamb\acute em~\acute e~soluc_{\!\!,}\tilde ao\\\tt\checkmark~ perceba~que~a~equac_{\!\!,}\tilde ao~admite~infinitas~soluc_{\!\!,}\tilde oes\\\sf\checkmark por\acute em~a~soluc_{\!\!,}\tilde ao~geral~\acute e~da~forma~(\beta,2\beta)=\beta\cdot(1,2)\\\sf~todas~as~possíveis~soluc_{\!\!,}\tilde oes~são~combinac_{\!\!,}\tilde oes~lineares~deste~vetor~se~assim\\\sf desejar~interpretar.$

$\tt b)~\sf x+y-2z=0\\\sf 2z=x+y\implies z=\dfrac{x+y}{2}\\\rm aqui,a~ soluc_{\!\!,}\tilde ao~vai~depender~dos~valores~de~x~e~y.\\\sf se~x=1~e~y=3~ent\tilde ao~ z=\dfrac{1+3}{2}=\dfrac{4}{2}=2\\\sf logo~o~terno~(1,3,2)~\acute e~soluc_{\!\!,}\tilde ao. \\\sf se~x=-1~y=7~ent\tilde ao~ z=\dfrac{-1+7}{2}=\dfrac{6}{2}=3~e~o~terno~(-1,7,3)~\acute e~soluc_{\!\!,}\tilde ao$

2)

$\sf2x+4y+z=0\\\sf 4y=-z-2x\\\sf y=\dfrac{-z-2x}{4}\\\sf a+1=\dfrac{-2-2\cdot(-1)}{4}\\\sf a+1=\dfrac{-\diagup\!\!\!2+\diagup\!\!\!2}{4}\\\sf a+1=0\\\sf a=-1\\\sf se~a\ne-1~o~terno (-1,a+1,2)~deixa~de~ser~soluc_{\!\!,}\tilde ao.$