Targeted Review Solutions

Solve.

$m + \frac{m}{m - 5} = \frac{5}{m - 5}$

$\begin{array}{rcl} LCD : m - 5 \\ m & \neq & 5 \\ m (\frac{m - 5}{m - 5}) + \frac{m}{m - 5} & = & \frac{5}{m - 5} \\ m (m - 5) + m & = & 5 \\ m^{2} - 5 m + m & = & 5 \\ m^{2} - 4 m - 5 & = & 0 \\ (m + 1) (m - 5) & = & 0 \\ m & = & - 1, 5 \end{array}$

$m = - 1$

$2 (x - 3) = \frac{1}{2} (4 x + 5)$

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

no solution

$5 p - 3 > \frac{1}{3} (6 p + 27)$

$\begin{array}{rcl} 5 p - 3 & > & 2 p + 9 \\ 3 p & > & 12 \end{array}$

$p > 4$

Graph the solution to problem 3 on a number line.

Graph the linear function on a coordinate plane.
$y - 3 = \frac{2}{3} (x - 1)$

$m = \frac{2}{3}, (1, 3)$

Name the domain and range of a linear function in set-builder notation.
domain $\{x | x \in ℝ\}$
range $\{y | y \in ℝ\}$

Note

The domain and range of a linear function are both the set of real numbers. In interval notation this is $(- \infty, \infty)$ .

Sketch a graph of the rational function on the coordinate plane.
$f (x) = \frac{1}{x + 4} - 3$

$a = 1, h = - 4, k = - 3$

Name the domain and range for problem 7 in set-builder notation.

domain $\{x | x \in ℝ, x \neq - 4\}$
range $\{y | y \in ℝ, y \neq - 3\}$

Ron and Will volunteer to pick up trash along the road. If Ron worked alone, it would take him 5 hours. Working together, it would take them 3 hours. How long would it take Will if he worked alone?

1.875 hours
7.5 hours
8 hours
15 hours

Note

This option would happen if students had the equation: $\frac{1}{5} + \frac{1}{3} = \frac{1}{t}$ .

This option adds the total time and Ron’s time.
This option multiplies the total time and Ron’s time.

	Time in hours	Rate of work
Ron	5	$\frac{1}{5}$
Will	t	$\frac{1}{t}$
Together	3	$\frac{1}{3}$

$\frac{1}{5} + \frac{1}{t} = \frac{1}{3} LCD : 15 t 15 t (\frac{1}{5} + \frac{1}{t} = \frac{1}{3}) 3 t + 15 = 5 t 15 = 2 t t = 7.5$

It would take Will 7.5 hours to pick up trash alone.

Determine the expression that will form a polynomial identity with ${(2 x + 1)}^{2} .$

$(5 x^{2} - 6 x + 10) - (x^{2} + 10 x - 9)$
$(5 x^{2} - 6 x + 10) - (x^{2} + 6 x + 9)$
$(5 x^{2} - 6 x + 10) - (x^{2} - 10 x + 9)$
no identity given

$4 x^{2} + 4 x + 1$

Note

If the –1 is not distributed across all terms in the second expression, you may have picked this choice.
This option would be true if the binomial squared was equal to the expression $4 x^{2} + 1$ .

Determine the restrictions for the denominator of the rational function.
$y = \frac{4 (x + 5)}{x^{2} + 3 x - 10}$

$x \neq - 5$
$x \neq 2$
$x \neq - 5, 2$
$x \neq - 2, 5$

$\begin{array}{rcl} x^{2} + 3 x - 10 & = & 0 \\ (x + 5) (x - 2) & = & 0 \\ x & = & - 5, 2 \end{array}$

Note

A–B) These options only include one of the possible restrictions.

D) These options are the values in the factored answer before solving for x.

Select the number line that best represents the solution to $| x - 6 | < 2$ .

$\begin{matrix} x - 6 < 2 & and & - (x - 6) < 2 \\ x < 8 & x - 6 > - 2 \\ x > 4 \end{matrix} 4 < x < 8$

Note

This option shows case 1 but is missing case 2 for the absolute value inequality.
This option has an incorrect sign on the lower number.
This option is the graph of an OR problem rather than an AND problem.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L9	A1	A1	A1	A1	A1	L10	L10	L9	L4	L9	A1

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics, FD = Foundational Knowledge