Targeted Review Solutions

Solve: $\frac{4 x + 1}{6} = \frac{3 x - 4}{8}$

$\begin{array}{rcl} 6 (3 x - 4) & = & 8 (4 x + 1) \\ 18 x - 24 & = & 32 x + 8 \\ - 32 & = & 14 x \end{array}$

$x = - \frac{16}{7}$

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

$\begin{array}{rcl} (\frac{4}{3}) \frac{3}{4} (3 x + \frac{1}{2}) & = & - 1 (\frac{4}{3}) \\ 3 x + \frac{1}{2} & = & - \frac{4}{3} \\ (6) (3 x + \frac{1}{2}) & = & - \frac{4}{3} (6) \\ 18 x + 3 & = & - 8 \\ 18 x & = & - 11 \end{array}$

$x = - \frac{11}{18}$

Name the domain and range.

x y

–2 1

–1 2

0 4

1 8

2 16

$Domain : \{- 2, - 1, 0, 1, 2\} Range : \{1, 2, 4, 8, 16\}$

Note

The domain is the set of x-values. The range is the set of y-values.

$Domain : ℝ Range : y \leq 4$

Note

The shorthand for all real numbers is ℝ.

Solve: $\frac{x - 8}{5} = \frac{2 x + 5}{10}$

$\begin{array}{rcl} 5 (2 x + 5) & = & 10 (x - 8) \\ 10 x + 25 & = & 10 x - 80 \\ 25 & = & - 80 \end{array}$

no solution

Factor: $x^{3} + 64$

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

Note

This is the sum of cubes.

Divide.

$(3 x^{3} + x^{2} + 2 x + 10) {(x^{2} - 2 x + 1)}^{- 1}$

$x^{2} - 2 x + 1 3 x + 7 3 x^{2} + x^{2} + 2 x + 10 - (3 x^{2} - 6 x^{2} + 3 x) 7 x^{2} - x + 10 - (7 x^{2} - 14 x + 7) 13 x + 3 + \frac{13 x + 3}{x^{2} - 2 x + 1}$

$3 x + 7 + \frac{13 x + 3}{x^{2} - 2 x + 1}$

Note

Long division must be used because the divisor is a quadratic term.

$(4 x^{3} - 14 x^{2} - 2) \div (2 x - 1)$

$\begin{array}{rcl} \frac{(4 x^{3} - 14 x^{2} - 2) \div 2}{(2 x - 1) \div 2} & = & \frac{2 x^{3} - 7 x^{2} - 1}{x - \frac{1}{2}} \\ x - \frac{1}{2} & = & 0 \\ x & = & \frac{1}{2} \end{array}$

$2 x^{2} - 6 x - 3 - \frac{\frac{5}{2}}{x - \frac{1}{2}} \frac{\frac{5}{2} (2)}{(x - \frac{1}{2}) (2)} = \frac{5}{2 x - 1}$

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

Multiple Choice

Describe the transformation of the parabola $y = {(x - 2)}^{2} + 5$ as compared to the parent function.

The parabola shifts 2 spaces right and 5 spaces up.
The parabola shifts 2 spaces left and 5 spaces up.
The parabola shifts 2 spaces right and 5 spaces down.
The parabola shifts 2 spaces left and 5 spaces down.

$y = {(x - h)}^{2} + k$
(h, k) is the vertex.
$(x - h)$ shifts right, (x + h) shifts left, +k moves up, $- k$ moves down

Note

B, D) (x – h) shifts the graph right h-spaces, not left.
C, D) + k shifts the graph up k-spaces, not down.

Solve: $\frac{2}{3} n - 5 = \frac{5}{4}$

$\frac{8}{75}$
$\frac{75}{8}$
$- \frac{45}{8}$
$\frac{55}{8}$

$(12) (\frac{2}{3} n - 5 = \frac{5}{4}) 8 n - 60 = 15 8 n = 75 n = \frac{75}{8}$

Note

This option is the reciprocal of the answer.
This option is the answer when the constant 60 is subtracted from 15.
This option occurs when the fractions are not all written with a common denominator.

Use the graph to answer problems 11–12.

Select all vertices of the system:
$y \leq \frac{4}{5} x + 4 y \geq 4 x - 12 y \geq - \frac{4}{5} x + 4$

(4, 0)
(3, 0)
(0, 4)
(5, 8)
(8, 5)

Note

(4, 0) and (8, 5) are not vertices on the graph.

Using the graph from problem 11, select the minimum and maximum values using the objective function $f (x, y) = x - 2 y .$

4

$f (4, 0) = 4 - 2 (0) = 4$
3

$f (3, 0) = (3) - 2 (0) = 3$ maximum
–8

$f (0, 4) = (0) - 2 (4) = - 8$
–11

$f (5, 8) = (5) - 2 (8) = - 11$ minimum
–2

$f (8, 5) = (8) - 2 (5) = - 2$

Note

4, –8, –2 do not represent the minimum or maximum value.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	A1	A1	A1	A1	A1	L03	L05	L06	A1	A1	L01	L01

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

x	y
–2	1
–1	2
0	4
1	8
2	16