The problem consists of three parts: 1. Graph the inequality $3x + 4y < 24$ and shade the unwanted region.

1. Problem Description

The problem consists of three parts:

1. Graph the inequality $3x + 4y < 24$ and shade the unwanted region.

2. Given the function $f(x) = 3x - 12$, determine $f(5)$ and solve $f(x) = 3$.

3. Determine the equation of the line with slope $m = \frac{3}{5}$ passing through the point $(5, -2)$.

2. Solution Steps

1. To graph $3x + 4y < 24$, first consider the line $3x + 4y = 24$. Find the x and y intercepts. When $y=0$, $3x = 24$, so $x = 8$. When $x=0$, $4y = 24$, so $y = 6$. Thus, the line passes through $(8,0)$ and $(0,6)$. Since the inequality is $<$, we draw a dashed line through these points. To determine which region to shade, we can test the point $(0,0)$. $3(0) + 4(0) = 0 < 24$, so $(0,0)$ is in the solution region. Therefore, shade the region that does not contain the origin.

2. i. $f(x) = 3x - 12$. To find $f(5)$, substitute $x=5$ into the expression for $f(x)$:

$f(5) = 3(5) - 12 = 15 - 12 = 3$.

ii. To solve $f(x) = 3$, set $3x - 12 = 3$. Then $3x = 15$, so $x = 5$.

3. The equation of a line is given by $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. We are given that $m = \frac{3}{5}$ and the line passes through $(5, -2)$. Plug these values into the equation $y = mx + b$:

$-2 = \frac{3}{5}(5) + b$

$-2 = 3 + b$

$b = -5$

Therefore, the equation of the line is $y = \frac{3}{5}x - 5$.

Alternatively, we can use the point-slope form of a line: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line. Using $(5, -2)$ and $m = \frac{3}{5}$:

$y - (-2) = \frac{3}{5}(x - 5)$

$y + 2 = \frac{3}{5}x - 3$

$y = \frac{3}{5}x - 5$.

3. Final Answer

1. The inequality $3x + 4y < 24$ is graphed with a dashed line through (8,0) and (0,6). The region above the line is shaded.

2. i. $f(5) = 3$

ii. $x = 5$

3. $y = \frac{3}{5}x - 5$