Answer
$a. b=2a,\: a≠−3\\b. b≠2a\\c. a=−3,\:b=−6$
Work Step by Step
$\left(a\right).\:For\:a\:system\:to\:have\:no\:solution,\:there\:must\:be\:an\:inconsistency\:in\:the\:equations.\\Multiplying\:the\:first\:equation\:by\:-a\:and\:adding\:it\:to\:the\:second\:equation:\\\left(b-2a\right)y=-12\\If\:b-2a=0,\:we\:have\:the\:statement\:0=-12\:which\:is\:an\:inconsistency.\\Therefore,\:if\:b=2a\:and\:a≠−3,\:there\:will\:be\:no\:solution.\\$$\\\left(b\right).\:We\:saw\:that\:if\:there\:is\:no\:solution,\:b=2a.\:If\:b\ne 2a,\:we\:can\:solve\:for\:y\:from\:the\:equation:\\y=-\frac{12}{b-2a}\\Plugging\:this\:into\:the\:first\:equation,\:we\:can\:solve\:for\:x.\:Therefore,\:there\:will\:be\:only\:one\:solution.\\$ $\\\left(c\right).\:For\:there\:to\:be\:an\:infinite\:number\:of\:solutions,\:one\:equation\:must\:be\:a\:multiple\:of\:the\:other.\:Therefore,\:looking\:at\:the\:constants,\:we\:know\:that\:the\:second\:equation\:is\:-3\:times\:the\:first\:equation.\:Our\:solutions\:for\:a\:and\:b\:are:\\a= -3\\b=-6$
