Solve the following inequalities and represent the solution on a graph. a) 1/2x > 4 b) 15x + 8 < 48. Find the inequality represented by each graph.

Multiply both sides of the inequality by 2:

Draw a number line. Place an open circle at $x = 8$ (because the inequality is strict, $x$ cannot be equal to 8). Draw an arrow extending to the right from the open circle, indicating all values greater than 8.

The solution is $\boxed{x > 8}$.

b) $15x + 8 < 48$

Subtract 8 from both sides of the inequality:

Divide both sides by 15:

Draw a number line. Place an open circle at $x = \frac{8}{3}$ (approximately 2.67, because the inequality is strict). Draw an arrow extending to the left from the open circle, indicating all values less than $\frac{8}{3}$.

The solution is $\boxed{x < \frac{8}{3}}$.

13. Find the inequality represented by each graph.

a) Graph 1

The line is a dashed line, which means the inequality does not include "equal to" ($<$ or $>$).

The line passes through points like $(0, 0)$, $(1, 1)$, $(2, 2)$, etc.
The slope $m = \frac{1-0}{1-0} = 1$.
The y-intercept is $c = 0$.
So, the equation of the line is $y = x$.

The region above the dashed line $y = x$ is shaded.
Choose a test point in the shaded region, for example, $(0, 1)$.
Substitute into $y$ and $x$: 1 and 0.
Since the shaded region is above the line, $y$ values are greater than $x$ values.
So, the inequality is $y > x$.

The inequality represented by Graph 1 is $\boxed{y > x}$.

b) Graph 2

The line is a solid line, which means the inequality includes "equal to" ($\le$ or $\ge$).

The line passes through points like $(0, 2)$, $(1, 3)$, $(-1, 1)$, etc.
The y-intercept is $c = 2$.
The slope $m = \frac{3-2}{1-0} = 1$.
So, the equation of the line is $y = x + 2$.

The region below the solid line $y = x + 2$ is shaded.
Choose a test point in the shaded region, for example, $(0, 0)$.
Substitute into $y$ and $x$: 0 and 0.
0 compared to $0 + 2$ is $0 \le 2$.
Since the shaded region is below the line, $y$ values are less than or equal to $x+2$.
So, the inequality is $y \le x + 2$.

The inequality represented by Graph 2 is $\boxed{y \le x + 2}$.