Is there a pair of prime $(p,q)$ such that $p >q>7$ and $2q-p=3$ and $p, q$ are successive? I'm failed to find at a least one pair of primes$(p,q)$  $p >q>7$ and $2q-p=3$ with $p, q$ are successive , I think tha's impossible because we do not know more about Gaps between prime and if there is those pair it should be large , then my question how i can find that pair or to disproof it's existence ? 
Edit:  I have added the missed condition wich is $p, q$ are successives 
 A: The key requirement is that the primes $p,q$ be sequential (that is, that there should be no prime between them).
Your assumption implies that $$q=\frac {p+3}2$$
However, a strong form of Bertrand's postulate  tells us that, for $n>25$, there must be a prime between $n$ and $\frac 65 n$.  Applying this to $n=\frac {p+3}2$ shows that, except for very small $p$, there must be a prime between $\frac {p+3}2$ and $p$ and from there a simple finite search shows that your desired primes do not exist.
A: If $p>q>7$ are succcessive primes with $2q-p=3$ then $p=2q-3$. This implies that there are no primes between $q+1$ and $2q-4$. This comes close to contradicting Bertrand's postulate.
In fact it does contradict improvements on Bertrand's postulate; the theorem at the bottom of page 180 of this article states that for every $n\geq9$ there is a prime $r$ such that
$$n\leq r\leq\frac{4}{3}n.$$
If $q>7$ is prime then $q+1\geq9$ so we find that there is a prime $r$ such that
$$q+1\leq r\leq\frac{4}{3}(q+1)\leq 2q-4,$$
and hence $q<r<p$. This shows that $q$ and $2q-3$ are not successive primes. So two successive primes $p>q>7$ never satisfy $2q-p=3$.
A: $(p,q)=(31,17)$ works unless I'm missing something.
A: Since $p, q$ are successive then you can rewrite your equation as $$2p_n-p_{n+1}=3$$
This shows that $p_{n+1}$ is almost double $p_n$ which seems to be forbidden for sufficiently large primes. This is because between $x$ and $2x$ exist many primes.
Write your equation in the form $p_{n+1}=2p_n-3$. This means that the number of primes in the interval $(x, 2x-3]$ is one (for $x=p_n$).
It is proved here that for $x\ge25$ there is a prime between $x$ and $\frac{6}{5}x$.
So, we have one prime between $p_n$ and $\frac{6}{5}p_n$ and one more between $\frac{6}{5}p_n$ and $\frac{6}{5}\cdot \frac{6}{5}p_n=\frac{36}{25}p_n$ if $p_n\ge 25$.
Since you have only one prime between $p_n$ and $2p_n-3$ we must have $2p_n-3<\frac{36}{25}p_n$vwhich gives $14p_n<75$ or $p_n<6$. Hence $p=2, 3$ or $5$.
The prime $5$ is the only good one to satisfy your equation.
