Practice 2 Solutions

Solve. Check your solutions for extraneous values.

$\frac{4 x}{x - 2} = \frac{3 x}{x + 3}$

$\begin{array}{rcl} x \neq - 3, 2 \\ 4 x (x + 3) & = & 3 x (x - 2) \\ 4 x^{2} + 12 x & = & 3 x^{2} - 6 x \\ x^{2} + 18 x & = & 0 \\ x (x + 18) & = & 0 \end{array}$

$x = - 18, 0$

Note

Q: When is it possible to have zero as a solution to a rational equation?
A: Any time, as long as it does not make the denominator equal to zero.

$\frac{2 h - 7}{h^{2} - h - 2} + \frac{3 h}{h + 1} = 4$

$\frac{2 h - 7}{(h - 2) (h + 1)} + \frac{3 h}{h + 1} = 4 h \neq - 1, 2 LCD : (h - 2) (h + 1)$
$\begin{array}{rcl} \frac{2 h - 7}{(h - 2) (h + 1)} + \frac{3 h (h - 2)}{(h - 2) (h + 1)} & = & \frac{4 (h - 2) (h - 1)}{(h - 2) (h - 1)} \\ 2 h - 7 + 3 h (h - 2) & = & 4 (h - 2) (h + 1) \\ 2 h - 7 + 3 h^{2} - 6 h & = & 4 (h^{2} - h - 2) \\ 3 h^{2} - 4 h - 7 & = & 4 h^{2} - 4 h - 8 \\ 0 & = & h^{2} - 1 \\ 0 & = & (h + 1) (h - 1) \\ h & = & - 1, 1 \end{array}$

$h = 1$

Note

The value h = –1 is extraneous because it is a restriction.

$\frac{2}{b + 2} - \frac{4}{b} = \frac{1}{b^{2} + 2 b}$

$\frac{2}{b + 2} - \frac{4}{b} = \frac{1}{b (b + 2)} LCD : b (b + 2) b \neq - 2, 0 \frac{2 b}{b (b + 2)} - \frac{4 (b + 2)}{b (b + 2)} = \frac{1}{b (b + 2)} 2 b - 4 (b + 2) = 1 2 b - 4 b - 8 = 1 - 2 b = 9$

$b = - \frac{9}{2}$

$\frac{5}{x - 3} - \frac{2}{4 - x} = \frac{7 x}{x^{2} - 7 x + 12}$

$\frac{5}{x - 3} - \frac{2}{- (x - 4)} = \frac{7 x}{(x - 3) (x - 4)} \frac{5}{x - 3} + \frac{2}{x - 4} = \frac{7 x}{(x - 3) (x - 4)} LCD : (x - 3) (x - 4) x \neq 3, 4$
$\begin{array}{rcl} \frac{5 (x - 4)}{(x - 3) (x - 4)} + \frac{2 (x - 3)}{(x - 3) (x - 4)} & = & \frac{7 x}{(x - 3) (x - 4)} \\ 5 (x - 4) + 2 (x - 3) & = & 7 x \\ 5 x - 20 + 2 x - 6 & = & 7 x \\ 7 x - 26 & = & 7 x \\ - 26 & = & 0 \end{array}$

No solution

$\frac{100}{x^{2} - 25} - \frac{2 x}{x + 5} = \frac{- x}{x - 5}$

$\frac{100}{(x + 5) (x - 5)} - \frac{2 x}{x + 5} = \frac{- x}{x - 5} LCD : (x + 5) (x - 5) x \neq \pm 5$

$\begin{array}{rcl} \frac{100}{(x + 5) (x - 5)} - \frac{2 x (x - 5)}{(x + 5) (x - 5)} & = & \frac{- x (x + 5)}{(x + 5) (x - 5)} \\ 100 - 2 x (x - 5) & = & - x (x + 5) \\ 100 - 2 x^{2} + 10 x & = & - x^{2} - 5 x \\ 0 & = & x^{2} - 15 x - 100 \\ 0 & = & (x + 5) (x - 20) \\ x & = & - 5, 20 \end{array}$

$x = 20$

Note

The value x = –5 is extraneous because it is a restriction.

$\frac{5}{y + 5} + 1 = \frac{7 y + 2}{y^{2} + y - 20}$

$\frac{5}{y + 5} + 1 = \frac{7 y + 2}{(y + 5) (y - 4)} LCD : (y + 5) (y - 4) y \neq - 5, 4$
$\begin{array}{rcl} \frac{5 (y - 4)}{(y + 5) (y - 4)} + \frac{(y + 5) (y - 4)}{(y + 5) (y - 4)} & = & \frac{7 y + 2}{(y + 5) (y - 4)} \\ 5 (y - 4) + (y + 5) (y - 4) & = & 7 y + 2 \\ 5 y - 20 + y^{2} + y - 20 & = & 7 y + 2 \\ y^{2} + 6 y - 40 & = & 7 y + 2 \\ y^{2} - y - 42 & = & 0 \\ (y + 6) (y - 7) & = & 0 \end{array}$

$y = - 6, 7$

$\frac{4}{x} + \frac{10}{x + 5} = \frac{9}{2 x - 5}$

$LCD : x (x + 5) (2 x - 5) x \neq - 5, 0, \frac{5}{2}$

$\begin{array}{rcl} \frac{4 (x + 5) (2 x - 5)}{x (x + 5) (2 x - 5)} + \frac{10 x (2 x - 5)}{x (x + 5) (2 x - 5)} & = & \frac{9 x (x + 5)}{x (x + 5) (2 x - 5)} \\ 4 (x + 5) (2 x - 5) + 10 x (2 x - 5) & = & 9 x (x + 5) \\ 4 (2 x^{2} + 5 x - 25) + 20 x^{2} - 50 x & = & 9 x^{2} + 45 x \\ 8 x^{2} + 20 x - 100 + 20 x^{2} - 50 x & = & 9 x^{2} + 45 x \\ 28 x^{2} - 30 x - 100 & = & 9 x^{2} + 45 x \\ 19 x^{2} - 75 x - 100 & = & 0 \\ (19 x + 20) (x - 5) & = & 0 \end{array}$

$x = - \frac{20}{19}, 5$

$\frac{3 x}{x + 4} - \frac{1}{x - 4} = \frac{3 x^{2} - 13 x - 4}{x^{2} - 16}$

$\frac{3 x}{x + 4} - \frac{1}{x - 4} = \frac{3 x^{2} - 13 x - 4}{(x + 4) (x - 4)} LCD : (x + 4) (x - 4) x \neq \pm 4$

$\begin{array}{rcl} \frac{3 x (x - 4)}{(x + 4) (x - 4)} - \frac{(x + 4)}{(x + 4) (x - 4)} & = & \frac{3 x^{2} - 13 x - 4}{(x + 4) (x - 4)} \\ 3 x (x - 4) - (x + 4) & = & 3 x^{2} - 13 x - 4 \\ 3 x^{2} - 12 x - x - 4 & = & 3 x^{2} - 13 x - 4 \\ 3 x^{2} - 13 x - 4 & = & 3 x^{2} - 13 x - 4 \\ 0 & = & 0 \end{array}$

$ℝ, x \neq \pm 4$

Note

Later you will learn to write this solution using set-builder notation: $\{x | x \in ℝ, x \neq \pm 4\}$

Jacob and Elena need to paint their living room. Alone, Jacob can paint the room in 4 hours while Elena can paint the room by herself in 3 hours. Determine how long it will take if Jacob and Elena paint the room together. Round to the nearest quarter hour.

Jacob Elena Together

time: 4 3 t

rate of work: $\frac{1}{4}$ + $\frac{1}{3}$ = $\frac{1}{t}$

$\frac{1}{4} + \frac{1}{3} = \frac{1}{t} LCD : 12 t \frac{1 (3 t)}{4 (3 t)} + \frac{1 (4 t)}{3 (4 t)} = \frac{1 (12)}{(12) t} 3 t + 4 t = 12 7 t = 12 t = \frac{12}{7} hours \approx 1.71 hours$

Note

Q: About how many minutes are in 0.71 hours?
A: About 43 minutes.

If you said this is about $\frac{3}{4}$ of an hour and round near 45 minutes, this is also a correct estimation.

It will take approximately $1 \frac{3}{4}$ hours for Jacob and Elena to paint the room together.

A bricklayer constructs a retaining wall in 12 hours when working alone. When the bricklayer and an apprentice work together, the retaining wall is constructed in 8 hours. How long would it have taken for the apprentice to do the job alone?

Bricklayer Apprentice Together

time: 12 t 8

rate of work: $\frac{1}{12}$ + $\frac{1}{t}$ = $\frac{1}{8}$

$\frac{1}{12} + \frac{1}{t} = \frac{1}{8} LCD : 24 t \frac{1 (2 t)}{12 (2 t)} + \frac{1 (24)}{(24) t} = \frac{1 (3 t)}{(8) (3 t)} 2 t + 24 = 3 t 24 = t$

It would take the apprentice 24 hours to complete the job alone.

A bicyclist has an initial velocity of 4 m/s before they begin to travel downhill. If the hill is 32 meters long and they accelerate at 2 m/ $s^{2}$ , how fast will the bicyclist be traveling when they reach the bottom of the hill? Use the formula $\frac{2 ∆ x}{v_{f} + v_{i}} = \frac{v_{f} - v_{i}}{a}$ , where $∆ x$ = change in distance, $v_{f}$ = final velocity, $v_{i}$ = initial velocity, a = acceleration.

$\frac{2 (32)}{v_{f} + 4} = \frac{v_{f} - 4}{2} 2 (2) (32) = (v_{f} + 4) (v_{f} - 4) 128 = {v_{f}}^{2} - 16 0 = {v_{f}}^{2} - 144 0 = (v_{f} + 12) (v_{f} - 12) v_{f} = 12 \frac{m}{s}$

Note

The solution of –12 m/s is extraneous because the bicycle did not change direction when traveling downhill.

The bicyclist was traveling 12 m/s by the time they reached the bottom of the hill.

The difference between the reciprocals of two numbers is $\frac{1}{10}$ . The second number is 5 more than the first. Determine the value(s) of each number.

` $\frac{1}{n} - \frac{1}{n + 5} = \frac{1}{10} LCD : 10 n (n + 5) n \neq - 5, 0 Let n = first number, n + 5 = second number$

$\begin{array}{rcl} \frac{1 (10) (n + 5)}{10 n (n + 5)} - \frac{1 (10 n)}{10 n (n + 5)} & = & \frac{1 n (n + 5)}{10 n (n + 5)} \\ 10 (n + 5) - 10 n & = & n (n + 5) \\ 10 n + 50 - 10 n & = & n^{2} + 5 n \\ 0 & = & n^{2} + 5 n - 50 \\ 0 & = & (n + 10) (n - 5) \\ n & = & - 10, 5 \end{array}$

Note

Since there are two possible values for n, find two possible pairs of answers.

If 5 is the first number, the second number is 10.

If –10 is the first number, the second number is –5.

	Jacob		Elena		Together
time:	4		3		t
rate of work:	$\frac{1}{4}$	+	$\frac{1}{3}$	=	$\frac{1}{t}$

	Bricklayer		Apprentice		Together
time:	12		t		8
rate of work:	$\frac{1}{12}$	+	$\frac{1}{t}$	=	$\frac{1}{8}$