Find the area (in square units) bounded by $x=6, y=0, 3x-2y=0$, and $x-4y+10=0$.

We need to find the area enclosed by four boundary lines.

Let's rewrite each line in slope-intercept form:

$x = 6$ is a vertical line

$y = 0$ is the x-axis

$3x - 2y = 0$ gives us $y = \dfrac{3x}{2}$

$x - 4y + 10 = 0$ gives us $y = \dfrac{x + 10}{4}$

---

Find where each pair of lines intersects:

Intersection of $y = 0$ and $y = \dfrac{3x}{2}$:

Setting equal: $0 = \dfrac{3x}{2}$ gives us $x = 0$

Point: $(0, 0)$

Intersection of $y = 0$ and $x = 6$:

Point: $(6, 0)$

Intersection of $x = 6$ and $y = \dfrac{x + 10}{4}$:

Substituting: $y = \dfrac{6 + 10}{4} = \dfrac{16}{4} = 4$

Point: $(6, 4)$

Intersection of $y = \dfrac{3x}{2}$ and $y = \dfrac{x + 10}{4}$:

Setting equal: $\dfrac{3x}{2} = \dfrac{x + 10}{4}$

Cross multiply: $12x = 2x + 20$

Solving: $10x = 20$, so $x = 2$

Then $y = \dfrac{3(2)}{2} = 3$

Point: $(2, 3)$

---

Split the quadrilateral into two triangles:

Triangle 1: $(0,0)$, $(2,3)$, $(6,0)$

Triangle 2: $(2,3)$, $(6,4)$, $(6,0)$

Using the triangle area formula:

Area = $\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

Triangle 1 area:

$\dfrac{1}{2}|0(3-0) + 2(0-0) + 6(0-3)| = \dfrac{1}{2}|0 + 0 - 18| = 9$

Triangle 2 area:

$\dfrac{1}{2}|2(4-0) + 6(0-3) + 6(3-4)| = \dfrac{1}{2}|8 - 18 - 6| = \dfrac{1}{2} \times 16 = 8$

Total Area = $9 + 8 = 17$ square units