# Question 114fc

Nov 11, 2017

$\left(y \ge \left\mid x + 1 \right\mid - 3\right) \cap \left(2 y < x + 5\right)$

#### Explanation:

For reference lets call the "upper" line that crosses the Y-axis at $2.5$ Line A
and the "lower" line that crosses the Y-axis at $\left(- 2\right)$ Line B

For Line A
We note the following points are on the line:
$\textcolor{w h i t e}{\text{XXX}} \left(x , y\right) \in \left\{\begin{matrix}0 & \frac{5}{2} \\ - 5 & 0\end{matrix}\right\}$
Since this is a straight line, the slope is constant for all pairs of points:
$\textcolor{w h i t e}{\text{XXX}} \frac{y - 0}{x - 5} = \frac{\frac{5}{2} - 0}{0 - 5} = \frac{1}{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 2 y = x - 5$

The identified area is below this line and the hollow "bubbles" indicate the points on the line are not to be included.

Therefore Line A provides the constraint
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{2 y < x - 5}$

For Line B
The V shape of Line B indicates that the equation for this line involves the absolute value of a linear expression of $x$.
Since the "turning point" occurs at $x - 1 \rightarrow x + 1 = 0$,
the absolute expression is $\left\mid x + 1 \right\mid$

Since the "turning point" is at $\left(x , y\right) = \left(- 1 , - 3\right)$
we note that relative to $y = \left\mid x + 1 \right\mid$, all the $y$ values have been reduced by $3$;
that is the equation for Line B $y = \left\mid x + 1 \right\mid - 3$

The identified area is above (and includes, since there are no "bubbles") Line B.

Therefore Line B provides the constraint
color(white)("XXX")color(blue)(y >= abs(x+1)-3

Combining the Areas
The identified area meets both Line A and Line B constraints.

Therefore the identified area is constrained by
color(white)("XXX")color(red)(2y < x-5) and color(blue)(y >= abs(x+1)-3
or (in set notation)
color(white)("XXX")color(red)((2y < x-5))nncolor(blue)((y >= abs(x+1)-3)#