Problem with simultaneous equations 
Given that $(5, h)$ is a solution of the simultaneous equations $h(x-y) = x + y -1 = hx^2 - 11y^2$, find
(a) the value of $h$.
(b) the other solution of the equation.(x and y)

I don't really understand the question. What does $(5, h)$ is the solution means? 
Shouldn't simultaneous equations have $1$ solution only (e.g. $5$)?
EDIT: I found that $h = 2$.
Now I'm having problem at (b). Here are my steps.
$h(x-y) = x + y - 1 = hx^2 - 11y^2$
$2(x-y) = 2x - 2y$ -- 1
$x + y - 1$ so $ x = -y + 1$ -- 2
After I substitute 2 into 1, I got $y = -1/2$
I don't think my answer is correct, please help me.
 A: $(5,h)$ is a singular solution, most likely in this case $x = 5, y = h$.
Plugging these in:
$$h(5 - h) = 5 + h - 1 = h+4\implies h^2 - 4h + 4 = 0 \implies h = 2$$
So, $x =5, y = 2$ or '$(5,2)$'
EDIT: In consideration of your edit to the problem:
$x = 5, h = 2$$$h(x-y) = 2(5-y) = 5 + y - 1 = x + y - 1$$
$$2(5-y) = 5+y - 1 \implies 3y = 6$$
So $y = 2 = h$ Note that this should not be a surprise! You know that $(x,y) = (5,h)$ is a solution to the equation. Subbing in $x = 5$ we should find $y = h$ 
A: HINT:
We have $$h(5-h)=5+h-1=h(5^2)-11(h^2)$$
From $h(5-h)=5+h-1$, $$h^2-4h+4=0$$
From $5+h-1=h(5^2)-11(h^2),$  $$11h^2-24h+4=0$$
Find the common roots
A: You have two variables, $x$ and $y$, and one parameter, $h$.  You are told that the solution is $x=5, y=h$ and have two equations because of the two equals signs.  You are expected to solve the two equations $h(x-y)=x=y-1, h(x-y)=hx^2-1y^2$, viewing $h$ as a constant.  You will get a solution where $x$ and $y$ are functions of $h$.  Then if you require $x=5$ you will get and equation for $h$.  Solve it and report the value of $h$ that comes out.
