Practice 1 Solutions

Solve. Check your solutions for extraneous values.

$\frac{x + 4}{- x - 2} = \frac{3 x}{3 x - 10}$

$\begin{array}{rcl} x \neq - 2, \frac{10}{3} \\ (x + 4) (3 x - 10) & = & 3 x (- x - 2) \\ 3 x^{2} - 10 x + 12 x - 40 & = & - 3 x^{2} - 6 x \\ 6 x^{2} + 8 x - 40 & = & 0 \\ 2 (3 x^{2} + 4 x - 20) & = & 0 \\ 2 (3 x + 10) (x - 2) & = & 0 \end{array}$

$x = - \frac{10}{3}, 2$

Note

Q: Why does this equation need to be factored to find the values of x?
A: Because the equation is degree 2.

$\frac{5 x - 9}{2 x + 1} = \frac{3 x - 7}{x + 2}$

$\begin{array}{rcl} x \neq - 2, - \frac{1}{2} \\ (5 x - 9) (x + 2) & = & (2 x + 1) (3 x - 7) \\ 5 x^{2} + x - 18 & = & 6 x^{2} - 11 x - 7 \\ x^{2} - 12 x + 11 & = & 0 \\ (x - 1) (x - 11) & = & 0 \end{array}$

$x = 1, 11$

$\frac{4}{a^{2} - 2 a - 8} = \frac{12}{a^{2} - a - 6}$

$\begin{array}{rcl} \frac{4}{(a + 2) (a - 4)} & = & \frac{12}{(a + 2) (a - 3)} \\ a \neq - 2, 3, 4 \\ 4 (a^{2} - a - 6) & = & 12 (a^{2} - 2 a - 8) \\ 4 a^{2} - 4 a - 24 & = & 12 a^{2} - 24 a - 96 \\ 8 a^{2} - 20 a - 72 & = & 0 \\ 4 (2 a^{2} - 5 a - 18) & = & 0 \\ 4 (2 a - 9) (a + 2) & = & 0 \\ a & = & - 2, \frac{9}{2} \end{array}$

$a = \frac{9}{2}$

Note

The value $a = - 2$ is extraneous because it makes the denominator zero.

$\frac{7}{4 x} - \frac{1}{3} = \frac{2}{9 x}$

$\begin{array}{rcl} LCD : 36 x \\ x \neq 0 \\ \frac{7 (9)}{(4 x) (9)} - \frac{1 (12 x)}{(3) (12 x)} & = & \frac{2 (4)}{(9 x) (4)} \\ 7 (9) - 1 (12 x) & = & 2 (4) \\ 63 - 12 x & = & 8 \\ - 12 x & = & - 55 \end{array}$

$x = \frac{55}{12}$

Note

Remember to find the LCD and multiply each expression by the missing factor or factors only, as shown in the examples.

$\frac{5}{g + 6} + \frac{4}{g - 3} = \frac{12 g}{g^{2} + 3 g - 18}$

$\frac{5}{g + 6} + \frac{4}{g - 3} = \frac{12 g}{(g + 6) (g - 3)} LCD : (g + 6) (g - 3) g \neq - 6, 3$

$\begin{array}{rcl} \frac{5 (g - 3)}{(g + 6) (g - 3)} + \frac{4 (g + 6)}{(g + 6) (g - 3)} & = & \frac{12 g}{(g + 6) (g - 3)} \\ 5 (g - 3) + 4 (g + 6) & = & 12 g \\ 5 g - 15 + 4 g + 24 & = & 12 g \\ 9 g + 9 & = & 12 g \\ 9 & = & 3 g \\ g & = & 3 \end{array}$

No solution

Note

$g \neq 3$ is a restriction.

Q: Why is it important to check the denominators for restrictions?
A: Because a restriction cannot be a solution to the equation.
This is a case where there is no solution because the only value you solved for is already a restriction for the domain.

$\frac{5}{x} - \frac{2}{x + 1} = \frac{3}{x + 6}$

$LCD : x (x + 1) (x + 6) x \neq - 6, - 1, 0$

$\begin{array}{rcl} \frac{5 (x + 1) (x + 6)}{x (x + 1) (x + 6)} - \frac{2 x (x + 6)}{x (x + 1) (x + 6)} & = & \frac{3 x (x + 1)}{x (x + 1) (x + 6)} \\ 5 (x + 1) (x + 6) - 2 x (x + 6) & = & 3 x (x + 1) \\ 5 (x^{2} + 7 x + 6) - 2 x^{2} - 12 x & = & 3 x^{2} + 3 x \\ 5 x^{2} + 35 x + 30 - 2 x^{2} - 12 x & = & 3 x^{2} + 3 x \\ 3 x^{2} + 23 x + 30 & = & 3 x^{2} + 3 x \\ 20 x + 30 & = & 0 \\ 20 x & = & - 30 \end{array}$

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

Note

Remember to combine like terms on each side of the equation and then combine like terms across the equal sign. This may help if you need to break the problem into small pieces.

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

$\frac{x}{x + 4} - \frac{4}{- (x - 4)} = \frac{x^{2} + 16}{(x + 4) (x - 4)}$

Factor out −1 from the second denominator so terms are in standard form.

$\frac{x}{x + 4} + \frac{4}{x - 4} = \frac{x^{2} + 16}{(x + 4) (x - 4)} LCD : (x + 4) (x - 4) x \neq \pm 4$

$\begin{array}{rcl} \frac{x (x - 4)}{(x + 4) (x - 4)} + \frac{4 (x + 4)}{(x + 4) (x - 4)} & = & \frac{x^{2} + 16}{(x + 4) (x - 4)} \\ x (x - 4) + 4 (x + 4) & = & x^{2} + 16 \\ x^{2} - 4 x + 4 x + 16 & = & x^{2} + 16 \\ x^{2} + 16 & = & x^{2} + 16 \end{array}$

$ℝ, x \neq \pm 4$

Note

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

Q: What does it mean when the solution is all real numbers?
A: Any number, fraction, decimal, negative, etc, can be substituted for the value of the variable and the equation will be true.

$\frac{3 h}{h + 6} - \frac{h}{h + 5} = - \frac{4}{h^{2} + 11 h + 30}$

$\frac{3 h}{h + 6} - \frac{h}{h + 5} = - \frac{4}{(h + 6) (h + 5)} LCD : (h + 6) (h + 5) h \neq - 6, - 5$

$\begin{array}{rcl} \frac{3 h (h + 5)}{(h + 6) (h + 5)} - \frac{h (h + 6)}{(h + 6) (h + 5)} & = & \frac{- 4}{(h + 6) (h + 5)} \\ 3 h (h + 5) - h (h + 6) & = & - 4 \\ 3 h^{2} + 15 h - h^{2} - 6 h & = & - 4 \\ 2 h^{2} + 9 h + 4 & = & 0 \\ (2 h + 1) (h + 4) & = & 0 \end{array}$

$h = - 4, - \frac{1}{2}$

The community garden can be harvested alone in 10 hours by Farmer Frank. If Garden Gail also helps, it takes only 6 hours to harvest. How long would it take Garden Gail to harvest alone?

Note

Remember that organizing the given information in a chart can help you set up the rational equation correctly. The instructor may need to provide the row and column headings to help you get started.

	Farmer Frank		Garden Gail		Together
time:	10		10		10
rate of work:	$\frac{1}{10}$	+	$\frac{1}{t}$	=	$\frac{1}{6}$

$\frac{1}{10} + \frac{1}{t} = \frac{1}{6} LCD : 60 t \frac{1 (6 t)}{(10) (6 t)} + \frac{1 (60)}{(t) (60)} = \frac{1 (10 t)}{(6) (10 t)} 6 t + 60 = 10 t 60 = 4 t t = 15$

It would take Garden Gail 15 hours to harvest the garden alone.

The sum of the reciprocals of two numbers is equal to $\frac{2}{3}$ . The difference between the two numbers is 4. Find the numbers.

Let x = the first number, and x – 4 be the second number.

$\frac{1}{x} + \frac{1}{x - 4} = \frac{2}{3} LCD : 3 x (x - 4) x \neq 0, 4$

$\begin{array}{rcl} \frac{1 (3) (x - 4)}{3 x (x - 4)} + \frac{1 (3 x)}{3 x (x - 4)} & = & \frac{2 x (x - 4)}{3 x (x - 4)} \\ 3 (x - 4) + 3 x & = & 2 x (x - 4) \\ 3 x - 12 + 3 x & = & 2 x^{2} - 8 x \\ 2 x^{2} - 14 x + 12 & = & 0 \\ 2 (x^{2} - 7 x + 6) & = & 0 \\ 2 (x - 6) (x - 1) & = & 0 \\ x & = & 1, 6 \end{array}$

Note

Remember the word “sum” means to add the terms together. Difference means to subtract, in this case, the number and four.

The reciprocal of x is $\frac{1}{x}$ .

This problem has more than one solution because the degree of the equation is 2, and neither solution is extraneous.

If 1 is the first number, then –3 is the second number.
If 6 is the first number, then 2 is the second number.

Bryson owns a lawn mowing business. If Bryson works alone, he can get all of the lawns mowed for a given day in 8 hours. If he hires his brother, Aaron, he can mow all the lawns in 10 hours. How long would it take to mow all the lawns if they work together? Round to the nearest half hour.

Bryson Aaron Together

time: 8 10 t

rate of work: $\frac{1}{8}$ + $\frac{1}{10}$ = $\frac{1}{t}$

$\begin{array}{rcl} \frac{1}{8} + \frac{1}{10} = \frac{1}{t} LCD (8, 10, t) = 40 t \\ \frac{1}{8} (\frac{5 t}{5 t}) + \frac{1}{10} (\frac{4 t}{4 t}) & = & \frac{1}{t} (\frac{40}{40}) \\ 5 t + 4 t & = & 40 \\ 9 t & = & 40 \\ t & = & \frac{40}{9} \end{array}$

It would take approximately 4.5 hours to mow all the lawns if they work together.

A car, traveling at an initial velocity of 26 m/s, brakes and slows down (decelerates) at a rate of $- 6 m / s^{2} .$
Find the final velocity when the change in distance is 40 meters, using the formula: $\frac{2 ∆ x}{v_{f} + v_{i}} = \frac{v_{f} - v_{i}}{a}$
$∆ x$ = change in distance, $v_{f}$ = final velocity, $v_{i}$ = initial velocity, a = acceleration

$\begin{array}{rcl} \frac{2 (40)}{v_{f} + 26} & = & \frac{v_{f} - 26}{- 6} \\ - 6 (2) (40) & = & (v_{f} + 26) (v_{f} - 26) \\ - 480 & = & {v_{f}}^{2} - 676 \\ {v_{f}}^{2} - 196 & = & 0 \\ (v_{f} + 14) (v_{f} - 14) & = & 0 \end{array}$

$v_{f} = 14 \frac{m}{s}$

Note

Note that in physics we can denote deceleration as acceleration in a negative direction.

The solution of $- 14 \frac{m}{s}$ is extraneous because the car did not change direction as it slowed down.

	Bryson		Aaron		Together
time:	8		10		t
rate of work:	$\frac{1}{8}$	+	$\frac{1}{10}$	=	$\frac{1}{t}$