Show linear independence of vector addition Let $\{v_1, v_2, v_3\}$ be a linearly independent set of vectors in a vector space $V$. Show that the set $\{v_1 + v_2, v_2 + v_3, v_1 + v_3\}$ is also a linearly independent set.

I put the equations in the form:
$c_1(v_1 + v_2) + c_2(v_2 + v_3) + c_3(v_1 + v_3) = 0$, to see if it implies that $c_1 = c_2 = c_3 = 0$.

Then I distributed to change it to the form:
$v_1(c_1 + c_3) + v_2(c_1 + c_2) + v_3(c_2 + c_3) = 0$

I'm not sure what to do from here. I need to show the linear independence of:
$c1 + c3 = 0$,
$c1 + c2 = 0$, and
$c2 + c3 = 0$

Do I just put this in reduced row echelon form?
The set $\{v_1,v_2,v_3\}$ is linearly independent.   

Since you've reduced it to the form $$(c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3) = 0,$$ you know that this is only true if all $c_i = 0$, since we assumed linear independence i.e. let $d_1 = (c_1+c_3)$, $d_2 = (c_1+c_2)$, $d_3=(c_2+c_3)$. Then we have $$d_1v_1 + d_2v_2 + d_3v_3$$ which is only possible if the $d_i =0$. Ergo, your second set is linearly independent, since they may only sum to $0$ if the coefficients are all $0$. Is this clear?
So once you're at the step
$$(c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3)v_3=0$$
because you know the set $\{v_1,v_2,v_3\}$ is linearly independent, we must have all these coefficients equal to zero, i.e.
$$c_1+c_3=c_1+c_2=c_2+c_3=0$$

From the first equality we can say we must have $c_2=c_3$ (by subtracting $c_1$ from both). Then
$$0=c_2+c_3=c_3+c_3=2c_3\implies c_3=0$$
and since $c_2=c_3$, we must have $c_2=0$. Then $0=c_1+c_3=c_1$, so all the $c_i$ are zero and the set is linearly independent.