Test 4 (Lessons 7–8): Rational Expressions Solutions

Name the restrictions for the denominator.

$\frac{1}{2 x^{2} + 7 x - 30}$

$(x + 6) (2 x - 5) x + 6 = 0, 2 x - 5 = 0$

$x \neq - 6, \frac{5}{2}$

$\frac{1}{x^{2} - 5 x + 6}$

$(x - 2) (x - 3) x - 2 = 0, x - 3 = 0$

$x \neq 2, 3$

Simplify. State the restrictions on the denominator.

$\frac{\frac{1}{x} + \frac{4}{x^{2}}}{1 + \frac{2}{x} - \frac{8}{x^{2}}}$

$Numerator LCD : x^{2}, x \neq 0 \frac{1 (x)}{x (x)} + \frac{4}{x^{2}} = \frac{x + 4}{x^{2}} Denominator LCD : x^{2}, x \neq 0 \frac{(1) x^{2}}{x^{2}} + \frac{2 (x)}{x (x)} - \frac{8}{x^{2}} = \frac{x^{2} + 2 x - 8}{x^{2}}$

$\begin{array}{rcl} Combined : \\ \frac{x + 4}{x^{2}} \div \frac{x^{2} + 2 x - 8}{x^{2}} \\ \frac{x + 4}{x^{2}} \cdot \frac{x^{2}}{(x + 4) (x - 2)} \\ \frac{x + 4}{x^{2}} \cdot \frac{x^{2}}{(x + 4) (x - 2)} \end{array}$

$\frac{1}{x - 2}, x \neq - 4, 0, 2$

Simplify. State the restrictions on the denominator.

$\frac{6 x^{2} + 7 x + 2}{2 x^{2} - 5 x - 12} \div \frac{3 x + 2}{3 x - 12}$

$\frac{(3 x + 2) (2 x + 1)}{(x - 4) (2 x + 3)} \div \frac{3 x + 2}{3 (x - 4)} \frac{(3 x + 2) (2 x + 1)}{(x - 4) (2 x + 3)} \cdot \frac{3 (x - 4)}{3 x + 2}, x \neq - \frac{3}{2}, - \frac{2}{3}, 4 \frac{(3 x + 2) (2 x + 1)}{(x - 4) (2 x + 3)} \cdot \frac{3 (x - 4)}{3 x + 2}$

Note

The restrictions are needed for the numerator and denominator of the expression after the division symbol.

$\frac{3 (2 x + 1)}{2 x + 3}, x \neq - \frac{3}{2}, - \frac{2}{3}, 4$

Determine the probability of landing on the triangle.

$P (triangle) = \frac{area of triangle}{total area of rectangle (T A)} length of rectangle : (x + 4) + (x + 1) = 2 x + 5 A_{∆} = \frac{1}{2} (x + 4) (x + 4) T A = (x + 4) (2 x + 5) P (triangle) = \frac{\frac{1}{2} (x + 4) (x + 4)}{(x + 4) (2 x + 5)} P (triangle) = \frac{\frac{1}{2} (x + 4) (x + 4)}{(x + 4) (2 x + 5)}$

$P (triangle) = \frac{\frac{1}{2} x + 2}{2 x + 5}, x \neq - 4, - \frac{5}{2} or P (triangle) = \frac{x + 4}{4 x + 10}, x \neq - 4, - \frac{5}{2}$

Note

The triangle has 2 equal legs, use this to determine the total length of the side of the rectangle.

Simplify. Name the restrictions on the domain.

$\frac{x}{5} + \frac{x - 6}{x^{2} + 6 x + 5} + \frac{2 x + 5}{x + 5}$

$\frac{x}{5} + \frac{x - 6}{(x + 5) (x + 1)} + \frac{2 x + 5}{x + 5} LCD : 5 (x + 5) (x + 1) x \neq - 5, - 1 \frac{x (x + 5) (x + 1)}{5 (x + 5) (x + 1)} + \frac{5 (x - 6)}{5 (x + 5) (x + 1)} + \frac{5 (2 x + 5) (x + 1)}{5 (x + 5) (x + 1)} x (x^{2} + 6 x + 5) + 5 (x - 6) + 5 (2 x^{2} + 7 x + 5) x^{3} + 6 x^{2} + 5 x + 5 x - 30 + 10 x^{2} + 35 x + 25 x^{3} + 16 x^{2} + 45 x - 5$

$\frac{x^{3} + 16 x^{2} + 45 x - 5}{5 (x + 1) (x + 5)} x \neq - 5, - 1$

$\frac{7 x + 8}{x + 8} - \frac{x + 3}{3 x - 12}$

$\frac{7 x + 8}{x + 8} - \frac{x + 3}{3 (x - 4)} LCD : 3 (x + 8) (x - 4) x \neq - 8, 4 \frac{3 (7 x + 8) (x - 4)}{3 (x + 8) (x - 4)} - \frac{(x + 3) (x + 8)}{3 (x + 8) (x - 4)} 3 (7 x^{2} - 20 x - 32) - (x^{2} + 11 x + 24) 21 x^{2} - 60 x - 96 - x^{2} - 11 x - 24 20 x^{2} - 71 x - 120$

$\frac{20 x^{2} - 71 x - 120}{3 (x + 8) (x - 4)} x \neq - 8, 4$

$\frac{\frac{x}{x - 1} + \frac{3}{x + 1} + \frac{6}{x^{2} - 1}}{\frac{1}{x - 1}}$

$\frac{\frac{x}{x - 1} + \frac{3}{x + 1} + \frac{6}{(x + 1) (x - 1)}}{\frac{1}{x - 1}} Numerator LCD : (x + 1) (x - 1) x \neq \pm 1 \frac{x (x + 1)}{(x + 1) (x - 1)} + \frac{3 (x - 1)}{(x + 1) (x - 1)} + \frac{6}{(x + 1) (x - 1)} x (x + 1) + 3 (x - 1) + 6 x^{2} + x + 3 x - 3 + 6 x^{2} + 4 x + 3 (x + 1) (x + 3) \frac{(x + 1) (x + 3)}{(x + 1) (x - 1)} \frac{(x + 1) (x + 3)}{(x + 1) (x - 1)} \frac{x + 3}{x - 1}$

$Combined : \frac{x + 3}{x - 1} \div \frac{1}{x - 1} \frac{x + 3}{x - 1} \cdot \frac{x - 1}{1} \frac{x + 3}{x - 1} \cdot \frac{x - 1}{1}$

$x + 3 x \neq \pm 1$

Calculate the efficiency ratio of the total surface area of a cone to volume of a cone.
$S A = π r^{2} + π r l$ $V = \frac{1}{3} π r^{2} h$

$\begin{array}{rcl} \frac{S A}{V} & = & \frac{π r^{2} + π r l}{\frac{1}{3} π r^{2} h} \\ = & \frac{π r (r + l)}{π r (\frac{1}{3} r h)} \\ = & \frac{π r (r + l)}{π r (\frac{1}{3} r h)} \\ = & \frac{r + l}{\frac{1}{3} r h} (\frac{3}{3}) \end{array}$

$\frac{S A}{V} = \frac{3 (r + l)}{r h}$

If a lower efficiency rating is desired, which is the best option for the cone?
Use the simplified ratio from problem 9.

P Q

radius, r 7 8

lateral height, l 10 12.3

height, h 11.5 10

Option P
$P = \frac{3 (7 + 10)}{7 (11.5)} P = 0.634$

Option Q
$Q = \frac{3 (8 + 12.3)}{8 (10)} Q = 0.761$

Option P is more efficient because the ratio is lower.

	P	Q
radius, r	7	8
lateral height, l	10	12.3
height, h	11.5	10