# How do you graph the system y<= 2x + 3 and x > 1?

Jul 23, 2015

Draw the lines (initially dashed) for $y = 2 x + 3$ and for $x = 1$.
Shade that section that included by inequalities.
Convert the dashed line along this section corresponding to $y = 2 x + 3$ into a solid line.

#### Explanation:

Step 1: draw dashed line for $y = 2 x + 3$
Evaluate $y = 2 x + 3$ for a couple of values of $x$,
for example $x = 0 \rightarrow y = 3$ and $x = 2 \rightarrow y = 7$
Draw the (dashed) line through the corresponding points on the Cartesian plane.

Step 2: draw dashed line for $x = 1$
Draw a vertical (dashed) line through $x = 1$.

Step 3: Identify the section included in both inequalities
Test a few $\left(x , y\right)$ pair values until you identify a pair that is valid for both inequalities.
For example $\left(x , y\right) = \left(2 , 0\right)$ satisfies both inequalities, since
$\textcolor{w h i t e}{\text{XXXX}}$$\left(0\right) < 2 \left(2\right) + 3$ and $\left(2\right) > 1$

Step 4: Shade the section which includes the point that satisfies both inequalities

**Step 5: Convert the dashed line segment bordering the shaded area that corresponds to $y = 2 x + 3$ into a solid line to show that it is included in the relation $y \le 2 x + 3$