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