Question: I'm currently working on a decision problem, and for some reason I am struggling with a system of equations, which should be the easiest part of the problem.

The correct answers are $\left(\frac{2}{11},\frac{1}{11} \right)$ in $(p,q)$

I have managed to compute the following utility functions (they are correct)

$U(R)= p-3q-1$

$U(B)= 1-11p-q$

$U(N)= -4p-4q$

I want to find unique values of $p$ and $q$ which would result in indifference between the utility functions.
I have tried for ages yet I keep getting negative values for q (which make no sense when I look at the context of the question).

However, If I apply $p=\frac{2}{11}$ and $q=\frac{1}{11}$ to the system of equations, the result is $U(R) \equiv U(B) \equiv U(N)$

The solutions I have reached only imply indifference between $U(R)$ and $U(B)$.

I would appreciate any help.

Answer: 
Thanks for the replies everyone, I was just being silly.
I managed to get the value of $q$ through the gauss algorithm but I have realised a much simpler solution.

$p-3q-1=1-11p-q=-4p-4q$

$$12p-1=2q$$

$$p=\frac{1+q}{6}$$

$$1-11\left(\frac{1+q}{6} \right )-q=-4 \left(\frac{1+q}{6} \right)-4q$$

$$\frac{1}{6}=\frac{11}{6}q$$

$$(p,q)=\left(\frac{2}{11},\frac{1}{11} \right)$$
you have to solve $p-3q-1=a$,$1-11p-q=a$,$-4p-4q=a$ you can use the Gauss algorithm to do this. Do need a hint?
rewriting your system in the form  

$p-3q-a=1$ (1)  

$-11p-q-a=-1$ (2)  

$-4p-4q-a=0$ (3)
then we get $11(1)+(2)=-34q-12a=10$ and $4(1)+(3)=-16q-5a=4$
now we have  

$-34q-12a=10$(I)  

$-16q-5a=4$(II)
now we compute $5I-6II=11q=1$ and we have $q=\frac{1}{11}$