Explicație pas cu pas:
a)
[tex]\left \{ {{ \frac{x}{6} = \frac{y}{3} } \atop { \frac{x}{10} = \frac{y}{5} }} \right. \iff \left \{ {{ x = 2y} \atop { x = 2y }} \right. \\ [/tex]
[tex] \implies \left \{ {{x \in \mathbb{R}; y \in \mathbb{R}} \atop {x = 2y}} \right. \\ [/tex]
b)
[tex]\left \{ {{ \frac{6 + x}{5} = \frac{y + 12}{10} } \atop { \frac{y - 1}{3} = - \frac{x + 7}{6} }} \right. \iff \left \{ { {2(6 + x) = y + 12 } \atop { 2(y - 1) = -(x + 7)}} \right. \\ [/tex]
[tex]\left \{ { {12 + 2x = y + 12 } \atop { 2y - 2 = - x - 7}} \right.\iff \left \{ { {2x - y = 0} \atop { x + 2y = - 5}} \right. \\ [/tex]
[tex]\iff \left \{ { {2x - y = 0} \atop { - 2x - 4y =10}} \right. \\ [/tex]
[tex] - 5y = 10 \implies y = - 2 \\ [/tex]
[tex]2x - ( - 2) = 0\iff 2x = - 2 \\ \implies x = - 1 \\ [/tex]
c)
[tex]\left \{ {{ \frac{x}{2} + 2 = \frac{y + 5}{2}} \atop {x + \frac{y}{2} = \frac{9}{2} }} \right. \iff \left \{ {{ x + 4 = y + 5} \atop {2x + y = 9}} \right. \\[/tex]
[tex]\left \{ {{ x - y = 1} \atop {2x + y = 9}} \right.\\ [/tex]
[tex]3x = 10 \implies x = \frac{10}{3} \\ [/tex]
[tex]\frac{10}{3} - y = 1 \iff y = \frac{10}{3} - 1 \\ \implies y = \frac{7}{3} \\ [/tex]
d)
[tex]\left \{ {{ \frac{x + 1}{2} + 2y = \frac{ - 3x + y}{4}} \atop { \frac{x - 1}{3} + \frac{y + 1}{2} = 0}} \right. \iff \left \{ {{2(x + 1) + 8y = - 3x + y} \atop {2(x - 1) + 3(y + 1) = 0}} \right. \\ [/tex]
[tex]\left \{ {{2x + 2 + 8y = - 3x + y} \atop {2x - 2 + 3y + 3 = 0}} \right. \iff \left \{ {{5x + 7y = - 2} \atop {2x + 3y = - 1}} \right. \\ [/tex]
[tex]\left \{ {{10x + 14y = - 4} \atop { - 10x - 15y = 5}} \right. \\ [/tex]
[tex] - y = 1 \implies y = - 1[/tex]
[tex]2x + 3( - 1) = - 1\iff 2x - 3 = - 1 \\ \implies x = 1[/tex]
