Prove that a set of elements in a vector space are linearly dependent

I have this question on my homework and its as follows:

Given $${\{X_1,X_2,X_3}\}$$ which belongs to a vector space $$V$$ and $${Y_1=X_1+X_2+2X_3}$$ and $${Y_2=X_1+X_2-X_3}$$. If $${\{Y_1,Y_2}\}$$ is linearly dependent prove that $${\{X_1,X_2,X_3}\}$$ is linearly dependent.

What I thought to do was first show that $${c_1Y_1+c_2Y_2=0}$$ then substitute in $${Y_1=X_1+X_2+2X_3}$$ and $${Y_2=X_1+X_2-X_3}$$ into the equation such that $${c_1Y_1+c_2Y_2=c_2(X_1+X_2-X_3)+c_1(X_1+X_2+2X_3)=0}$$ then manipulate the equation as follows:

$${X_1(c_1+c_2)+X_2(c_1+c_2)+X_3(2c_1-c_2)=0}$$ and that would show that the set $${\{X_1,X_2,X_3}\}$$ is linearly dependent because they are non zero constants. But now I am considering what if $${c_1=1}$$ and $${c_2=-1}$$ which would make the constants $${(c_1+c_2)=0}$$ for $$X_1$$ and $$X_2$$. Any advice on where to go from here?

thanks!

Note that you only need to prove that $$c_1 + c_2$$ and $$2c_1 - c_2$$ are not both zero to show that $$\{X_1,X_2,X_3\}$$ is linearly dependent.
To see this, suppose that $$c_1 + c_2 = 2c_1 - c_2 = 0$$. This would imply that $$c_1 = c_2 = 0$$. Do you see how this contradicts your assumption that $$\{ Y_1, Y_2\}$$ is linearly dependent?
In this case I think it would be easiest and cleanest to prove the contrapositive, namely, that if $$\{X_1,\dotsc, X_3\}$$ is a linearly independent set of vectors, then so is $$\{Y_1, Y_2\}$$. To this end suppose that $$c_1Y_1+c_2Y_2=0\tag{1}$$ for some $$c_i\in\mathbb{R}$$. Since $$Y_1=X_1+X_2+2X_3$$ and $$Y_2=X_1+X_2-X_3$$ we can rewrite (1) as $$(c_1+c_2)X_1+(c_1+c_2)X_2+(2c_1-c_2)X_3=0$$ whence $$c_1+c_2=0$$ and $$2c_1=c_2$$ by linear independence of the $$X_i$$. In particular $$3c_1=0$$ whence $$c_1=0$$ and so $$c_2=0$$. It follows that the $$Y_i$$ are linearly independent.
You're on the right track: you have a linear combination $$d_1 X_1 + d_2 X_2 + d_3 X_3 = 0$$ where $$d_1 = d_2 = c_1 + c_2$$ and $$d_3 = 2c_1 - c_2$$.
What you need to show is that not all of $$d_1,d_2,d_3$$ are $$0$$. So, suppose $$d_1 = d_2 = 0$$. This means $$c_1 = -c_2$$. In this case, what can you say about $$d_3$$? (If you can conclude it is non-zero, you are done).