Question

money did you have initially? write an inequality that rep
a) $d - 22 > 28$;\t\t\tc) $d + 22 \geq 28$;
b) $d + 22 \leq 28$;\t\t\t\\(\bigcirc\\)d) $d - 22 \geq 28$;

1. solve for $x$: \\t $-6x + 8 < 20$

\bigcirc a) $x > -2$ \\t b) $x < -2$ \\t c) $x > 2$ \\t d) $x < 2$

1. solve for $g$: \\t $3 + 6g \geq 5(g + 3)$

a) $g \geq 12$ \\t b) $g \geq -12$ \\t c) $g \geq 15$ \\t d) $g \geq 18$

1. which graph represents the inequality $y < x - 4$

(graphs a, b, c, d with coordinate planes and shaded regions)

1. which equation represents the line shown in the graph?

a. $y = \frac{4}{5}x + 3$
b. $y = \frac{-4}{5}x + 3$
c. $y = \frac{-5}{4}x + 3$
d. $y = \frac{5}{4}x + 3$
(graph with a line on coordinate plane)

60 students went on a field trip. five buses ($b$) were equally filled with students and 15 students traveled in cars. which of the following equations represents the field trip?
) $b + 15 = 160$ \\t (b) $5b - 15 = 160$ \\t (c) $5b + 15 = 160$ \\t (d) $-15b + 5 = 160$

which ordered pair is a solution of the equation $y = -2x + 5$?
(2, 0) \\t \bigcirc b) (0, 5) \\t c) (1, 2) \\t d) (5, 0)

fall semester multiple choice review c (2023-2024)\\t\\t\\t\\t pg. 6

Explanation:

Question 18: Solve for \( x \): \( -6x + 8 < 20 \)

Step 1: Subtract 8 from both sides

Subtract 8 from each side of the inequality to isolate the term with \( x \).
\( -6x + 8 - 8 < 20 - 8 \)
Simplifies to: \( -6x < 12 \)

Step 2: Divide by -6 (reverse inequality)

Divide both sides by -6. Remember, dividing by a negative number reverses the inequality sign.
\( \frac{-6x}{-6} > \frac{12}{-6} \)
Simplifies to: \( x > -2 \)

Step 1: Expand the right side

Distribute the 5 on the right side: \( 3 + 6g \geq 5g + 15 \)

Step 2: Subtract \( 5g \) from both sides

Subtract \( 5g \) to get: \( 3 + g \geq 15 \)

Step 3: Subtract 3 from both sides

Subtract 3: \( g \geq 15 - 3 \)
Simplifies to: \( g \geq 12 \)? Wait, no—wait, 15 - 3 is 12? Wait, no: \( 3 + g \geq 15 \) → \( g \geq 15 - 3 \) → \( g \geq 12 \)? Wait, no, let's recheck:

Wait, original expansion: \( 3 + 6g \geq 5g + 15 \)
Subtract \( 5g \): \( 3 + g \geq 15 \)
Subtract 3: \( g \geq 12 \). Wait, but option A is \( g \geq 12 \), but let's check again.

Wait, \( 3 + 6g \geq 5g + 15 \)
Subtract \( 5g \): \( 3 + g \geq 15 \)
Subtract 3: \( g \geq 12 \). So the answer is A) \( g \geq 12 \)? Wait, no—wait, maybe I made a mistake. Wait, \( 5(g + 3) = 5g + 15 \). Then \( 3 + 6g \geq 5g + 15 \). Subtract \( 5g \): \( 3 + g \geq 15 \). Subtract 3: \( g \geq 12 \). So yes, A) \( g \geq 12 \).

(corrected):

Step 1: Expand \( 5(g + 3) \)

\( 3 + 6g \geq 5g + 15 \)

Step 2: Subtract \( 5g \)

\( 3 + g \geq 15 \)

Step 3: Subtract 3

\( g \geq 12 \)

Answer:

A) \( x > -2 \)

Question 19: Solve for \( g \): \( 3 + 6g \geq 5(g + 3) \)