Question

Me ajudem a resolver essa questão... N° 54

A,B,C e D

Preciso do resultado certo e da resolução... Agradeço :)

Answer 1

Segue em anexo respostas com os devidos cálculos!
Bons estudos

Answer 2

Outra maneira seria por determinantes: a) $\left[\begin{array}{ccc}x&y&1\\5&2&1\\-1&3&1\end{array}\right \left\begin{array}{ccc}x&y\\5&2\\-1&3\end{array}\right] = 0; 2x - y + 15 + 2 -3x -5 y = 0;$;- x + 17 = 6 y --> $y = \frac{-x}{6} + \frac{17}{6} ;$ , b)$\left[\begin{array}{ccc}x&y&1\\0&-4&1\\1&6&1\end{array}\right \left\begin{array}{ccc}x&y\\0&-4\\1&6\end{array}\right] = 0; -4x + y+4 - 6x = 0; y = 10x - 4$; ' ' C)$\left[\begin{array}{ccc}x&y&1\\-12&-7&1\\-5&2&1\end{array}\right \left\begin{array}{ccc}x&y\\-12&-7\\-5&2\end{array}\right] = 0; -7x - 5 y - 24 - 35 - 2x + 1 2y = 0;$;--> 7y = 9x + 59 --> $y = \frac{9x}{7} + \frac{59}{7} ;$ , d)$\left[\begin{array}{ccc}x&y&1\\8&-3&1\\-4&-6&1\end{array}\right \left\begin{array}{ccc}x&y\\8&-3\\-4&-6\end{array}\right] = 0;$; $-3x -4y - 48-12+ 6x - 8y= 0 ; 3x - 60 =12y ;x - 20 = 4y ;$;$y = \frac{x}{4} - \frac{20}{4} ;--\ \textgreater \ y = \frac{x}{4} - 5$