Resolving by the substitution method:
[tex] \left \{ {{y=2x-6\:(I)} \atop {4x-2y=14\:(II)}} \right. [/tex]
Replace the value of "Y" in the second equation:
[tex]4x-2y=14\:(II)[/tex]
[tex]4x - 2*(2x-6) = 14[/tex]
[tex]4x - 4x + 12 = 14[/tex]
[tex]0x = 14 - 12[/tex]
[tex]\boxed{\boxed{0x = 2 \: (\nexists)}}\end{array}}\qquad\quad\checkmark[/tex]
or
[tex] \left \{ {{y=2x-6\:(I)} \atop {4x-2y=14\:(II)}} \right. [/tex]
[tex]\left \{ {{-2x+y=-6\:*(2)} \atop {4x-2y=14\:\:\:\:\:\:}} \right.
[/tex]
[tex]\left \{ {{-\diagup\!\!\!\!4x+\diagup\!\!\!\!2y=-12} \atop {\diagup\!\!\!\!4x-\diagup\!\!\!\!2y=14}} \right.[/tex]
[tex]\boxed{\boxed{0 = 2 \: (\nexists)}}\end{array}}\qquad\quad\checkmark[/tex]
Answer:
[tex]\underline{No\:solutions (Undetermined)}[/tex]
