Question 189411
First mistake: mixed up notation.
Please keep your equations and the application of them separate. Your understanding will be clearer and your teacher will be happier. Since  equations are abstract, the only things that belong in them are abstract symbols. Use lower-case variables in algebra. C does not equal c.

This is the way to set up the problem:

{{{c}}} = number of bottles of apple juice
{{{d}}} = number of bottles of orange juice

{{{c + d = 12}}}
{{{1c + 1.5d = 15}}} or {{{1x + (3d/2) = 15}}}

Then it is solved like this: ({{{-d}}} doesn't equal zero)

{{{c + d - d = 12 - d}}}
{{{c = 12 - d}}}

Replace {{{c}}} in the other equation with your result (zero times anything equals zero). It is solved like this:
{{{1(12 - d) + 1.5d = 15}}}
{{{12 - d + 1.5d = 15}}}
{{{12 + .5d = 15}}}
{{{12 + .5d - 12 = 15 - 12}}}
{{{.5d = 3}}}
{{{(.5d)/.5 = 3/.5}}}
{{{d = 6}}}

Second mistake: you didn't answer the question. This is related to the first mistake. To solve a word problem, translate the information in it to algebraic notation first, so make it abstract. Then you have to make it concrete again. So translate your result to its application or meaning. When you are at this point in the problem, go back to the question and ask, "What do I have to find?" Then you look at your result and see what it says about that and formulate your answer. Sometimes more than one thing is required, like here. We have found there were 6 bottles of orange juice bought. This equation {{{c + d = 12}}} lets you find the number of apple juice bottles bought.

{{{c + 6 = 12}}}
{{{c + 6 - 6 = 12 - 6}}}
{{{c = 12}}}

Your answer is, "I bought 6 bottles each of apple and orange juice."

I've seen a lot of these mistakes before, so I think they are common, probably caused by a (very unfortunate) lack of rigor and logic in schools.