We are given a system of two linear equations with two variables $x$ and $y$, and a parameter $m$: $(m+1)x - my = 4$ $3x - 5y = m$ We want to find the values of $m$ for which the system has a solution $(x, y)$ such that $x - y < 2$.

1. Problem Description

We are given a system of two linear equations with two variables $x$ and $y$, and a parameter $m$:

$(m+1)x - my = 4$

$3x - 5y = m$

We want to find the values of $m$ for which the system has a solution $(x, y)$ such that $x - y < 2$.

2. Solution Steps

First, we solve the system of equations for $x$ and $y$ in terms of $m$.

From the second equation, we can express $x$ in terms of $y$ and $m$:

$3x = 5y + m$, so $x = \frac{5y + m}{3}$.

Substitute this expression for $x$ into the first equation:

$(m+1)(\frac{5y + m}{3}) - my = 4$

$\frac{(m+1)(5y + m)}{3} - my = 4$

$(m+1)(5y + m) - 3my = 12$

$5my + m^2 + 5y + m - 3my = 12$

$2my + 5y + m^2 + m - 12 = 0$

$y(2m + 5) = 12 - m - m^2$

If $2m + 5 \neq 0$, then $y = \frac{12 - m - m^2}{2m + 5}$.

Substituting this into the expression for $x$:

$x = \frac{5y + m}{3} = \frac{5(\frac{12 - m - m^2}{2m + 5}) + m}{3} = \frac{5(12 - m - m^2) + m(2m + 5)}{3(2m + 5)} = \frac{60 - 5m - 5m^2 + 2m^2 + 5m}{3(2m + 5)} = \frac{60 - 3m^2}{3(2m + 5)} = \frac{20 - m^2}{2m + 5}$.

Now, we have $x = \frac{20 - m^2}{2m + 5}$ and $y = \frac{12 - m - m^2}{2m + 5}$.

We want $x - y < 2$, so $\frac{20 - m^2}{2m + 5} - \frac{12 - m - m^2}{2m + 5} < 2$.

$\frac{20 - m^2 - (12 - m - m^2)}{2m + 5} < 2$

$\frac{20 - m^2 - 12 + m + m^2}{2m + 5} < 2$

$\frac{8 + m}{2m + 5} < 2$

$\frac{8 + m}{2m + 5} - 2 < 0$

$\frac{8 + m - 2(2m + 5)}{2m + 5} < 0$

$\frac{8 + m - 4m - 10}{2m + 5} < 0$

$\frac{-3m - 2}{2m + 5} < 0$

$\frac{3m + 2}{2m + 5} > 0$

The critical points are $m = -\frac{2}{3}$ and $m = -\frac{5}{2}$.

We test intervals:

1. $m < -\frac{5}{2}$: $3m + 2 < 0$ and $2m + 5 < 0$, so $\frac{3m + 2}{2m + 5} > 0$.

2. $-\frac{5}{2} < m < -\frac{2}{3}$: $3m + 2 < 0$ and $2m + 5 > 0$, so $\frac{3m + 2}{2m + 5} < 0$.

3. $m > -\frac{2}{3}$: $3m + 2 > 0$ and $2m + 5 > 0$, so $\frac{3m + 2}{2m + 5} > 0$.

Thus, the solution is $m < -\frac{5}{2}$ or $m > -\frac{2}{3}$.

Also, we assumed $2m + 5 \neq 0$, so $m \neq -\frac{5}{2}$.

3. Final Answer

$m \in (-\infty, -\frac{5}{2}) \cup (-\frac{2}{3}, \infty)$