Question

Alguém me ajuda nessa 10? Pfv

gab: 10) k= 61/6

Answer 1

Resposta:

$k=\frac{61}{6}$

Explicação passo-a-passo:

$\left \{ {{kx-6y=k-1&~eq(1)} \atop {2x+3y=11~eq(2)}} \right.$

Multiplicando a (eq2) por dois:

$\left \{ {{kx-6y=k-1&~eq(1)} \atop {4x+6y=22~eq(2)}} \right$

Somando as equações:

$(4+k)x+0y=k+21\Rightarrow x=\frac{k+21}{k+4}$

Substituindo x na eq(2)

$2x+3y=11 \Rightarrow 2.(\frac{k+21}{k+4})+3y=11 \Rightarrow 3y=11-2.(\frac{k+21}{k+4})\\\\y=\frac{1}{3}.(11-2.(\frac{k+21}{k+4}))$

No problema é informado que x=y, logo:

$x=y\\\\\frac{k+21}{k+4}=\frac{1}{3}.(11-2.(\frac{k+21}{k+4})\\\\3(k+21)=(k+4)(11-2.(\frac{k+21}{k+4}))\\\\3k+63=11(k+4)-2(k+21)=11k+44-2k-42=9k+2\\\\6k=61\\\\k=\frac{61}{6}$