system with parameter

Question

$$\left\{\begin{matrix} 2(x+y)&+ \ mxy &=5 \\ (m-1)(x+y) &+ \ xy &=1 \\ 3(x+y) &-\ xy &=m+1 \end{matrix}\right.$$

I need to find the number of $m$ values such that this system has solutions $(x,y)\in \mathbb R \times \mathbb R$.

After I noted $x+y=s$ and $xy=p$ I obtained a system in $s$ and $p$.From the last 2 equations I get $s=1$ and $p=2-m$ then I replace these values in first equation and I get $p=3/m$.Then I get $3/m=2-m$ with no real solutions.

Where's my mistake?There's another method to approach this system ?The right answer is $1$.

Answer 1

Summing of the second and the third equations gives: $$(m+2)(x+y)=m+2.$$ 1. $m=-2$.

We have, $$2(x+y)-2xy=5$$ and $$3(x+y)-xy=-1,$$ which gives $$xy=-\frac{17}{4}<0,$$ which says that our system has solutions.

2. $m\neq-2$.

Thus, $$x+y=1,$$ $$xy=\frac{3}{m}$$ and $$xy=2-m,$$ which gives $$\frac{3}{m}=2-m$$ or $$m^2-2m+3=0,$$ which is impossible.

Id est, there is unique value of $m$, for which our system has solutions.

Answer 2

Hint: Eliminating $$xy$$ so we get for $$x+y$$ two equations $$(x+y)(2-m^2+m)=5-m$$ and $$(x+y)(m+2)=m+2$$ Can you proceed?