need help with a systems of equations question I have these systems of equations: 

$$\frac{3}{4}x-\frac{1}{2}y=12$$

$$ax-by=9$$

If $a$ and $b$ are constants, what is $\frac{a}{b}$?

I know that no solutions means that LHS $=0$ basically. so i tried making the RHS equal to 36 by multiplying the top equation by 3 and bottom by 4. that way when i subtract the two, i have the LHS $=0$. however this didn't work. any other way i can do this? the answer is $\frac{3}{2}$
Hint: Plugging $$y=\frac{3}{2}x-24$$in the second equation we get
$$ax-b\left(\frac{3}{2}x-24\right)=9$$
Can you finish?
Unless you've left something out, this isn't solvable. For example, letting $y=\frac32 x-24$ and subbing in to the second equation, we get $$ax-\frac32 bx+24b=9,$$ so $$\left(a-\frac32b\right)x=9-24b.$$ From this, we can conclude either that $a-\frac32b=0$ and $9-24b=0$ (so that $a=\frac32b$ and $b=\frac38,$ so that $a=\frac9{16}$ and, indeed, $\frac ab=\frac32$) or that $$x=\frac{9-24b}{a-\frac32}=\frac{18-48b}{2a-3}.$$ In the latter case, we can then conclude that $$y=\frac{27-72b}{2a-3}-24=\frac{99-72b-48a}{2a-3}.$$ However, this tells us nothing about $a$ and $b$! As long as $a\ne-\frac32$ and $b\ne 0,$ $\frac{a}{b}$ could be literally anything. In fact, since $b$ can be $0,$ then $\frac{a}{b}$ need not even be defined!!

If you know (for example) that both equations are supposed to be for the same line, then you can solve it. The easiest way to do so is to note that the lines should have the same slope, since $\frac ab$ will be the slope; all you have to do, then, is convert them both to slope-intercept form, and you're done.