The problem is to solve the system of inequalities: $y - x - 1 < 0$ $3x + y - 6 > 0$

1. Problem Description

The problem is to solve the system of inequalities:

$y - x - 1 < 0$

$3x + y - 6 > 0$

2. Solution Steps

We rewrite the inequalities to isolate $y$.

The first inequality is:

$y - x - 1 < 0$

$y < x + 1$

The second inequality is:

$3x + y - 6 > 0$

$y > -3x + 6$

Thus we have

$y < x + 1$

$y > -3x + 6$

The solution is the region where both inequalities are satisfied, i.e., the region bounded by the lines $y = x + 1$ and $y = -3x + 6$.

We can find the intersection of the two lines by setting $x + 1 = -3x + 6$, which gives $4x = 5$, so $x = \frac{5}{4}$. Then $y = \frac{5}{4} + 1 = \frac{9}{4}$.

So the intersection point of the two lines is $(\frac{5}{4}, \frac{9}{4})$.

The solution region is the area between the two lines, such that $-3x + 6 < y < x + 1$.

3. Final Answer

The solution to the system of inequalities is the region defined by $-3x + 6 < y < x + 1$. The lines $y = -3x+6$ and $y = x+1$ intersect at $(\frac{5}{4}, \frac{9}{4})$.