Systems with No Solution or Infinitely Many Solutions

The second equation's $x$-coefficient is $3$ times the first, and so is its $y$-coefficient. So the whole second equation must be $3$ times the first: $6x+15y=24$.

Hence $\boxed{c=24}$. For every other $c$ the system has no solution.

Same slope requires $3k=6\cdot4=24$, so $k=8$.

Now confirm the system really has no solution rather than infinitely many. With $k=8$ the system is $3x+6y=12$ and $4x+8y=9$, i.e.\ $x+2y=4$ and $x+2y=\tfrac94$ — contradictory.

Answer: $\boxed{k=8}$.

Divide the second equation by $2$: $y=\tfrac{a}{2}x+7$. The constants already agree, so the slopes must agree too: $\tfrac{a}{2}=4$, giving $\boxed{a=8}$.

Equivalently, doubling the first equation gives $2y=8x+14$, which matches term for term.

The constants tell you the scale factor immediately: $10=2\cdot5$, so the first equation must be $2$ times the second.

Therefore $a=2\cdot3=6$ and $b=2\cdot4=8$, so $a+b=\boxed{14}$.

Divide the first equation by $3$: $4x+5y=8$. The two left sides are now identical, so the lines are parallel no matter what $c$ is.

They coincide when $c=8$ and are distinct otherwise. So no solution occurs for $\boxed{c\neq 8}$, and $c=8$ is the single value that must be excluded.

Rewrite the second equation: $2y=3x-k$, so $y=\tfrac32x-\tfrac{k}{2}$.

Both lines have slope $\tfrac32$ for every $k$, so they are never allowed to cross once. The system has infinitely many solutions when $\tfrac{k}{2}=4$, i.e.\ $k=8$, and no solution otherwise.

Answer: $\boxed{0}$ values of $k$.

Parallel condition: $k(k-1)=3\cdot4=12$, so $k^2-k-12=0$ and $(k-4)(k+3)=0$, giving $k=4$ or $k=-3$.

Test $k=4$: the system is $4x+3y=6$ and $4x+3y=8$ — parallel and distinct, no solution.

Test $k=-3$: the system is $-3x+3y=6$, i.e.\ $-x+y=2$, and $4x-4y=8$, i.e.\ $x-y=2$, i.e.\ $-x+y=-2$ — parallel and distinct, no solution.

Both survive, so the sum is $4+(-3)=\boxed{1}$.

Parallel condition: $1\cdot4-k\cdot k=0$, so $k^2=4$ and $k=2$ or $k=-2$. Now test each.

$k=2$: the system is $x+2y=2$ and $2x+4y=4$. The second is exactly twice the first — infinitely many solutions.

$k=-2$: the system is $x-2y=2$ and $-2x+4y=0$, i.e.\ $x-2y=0$. Same left side, different constant — no solution.

Only one candidate survives: $\boxed{k=2}$.

Equal slopes: $ab=3\cdot2=6$.

That alone would also permit the coincident case, so we must exclude it: coincidence would need $5a=2\cdot6=12$, i.e.\ $a=\tfrac{12}{5}$.

So the conditions are $\boxed{ab=6 \text{ and } a\neq\tfrac{12}{5}}$.