Question

given the following compound inequalities, answer the questions below.
$x + 9 > - 1$ & $3x + 2 < 8$
(a)
solve the inequality, and write the solutions in the boxes below.
determine if this is possible. if it is not, select
o solution\ below.
select the correct graphical representation of the solutions to the inequality.
(b)

Explanation:

Step1: Solve \(x + 9> - 1\)

Subtract 9 from both sides: \(x+9 - 9> - 1 - 9\), so \(x>-10\).

Step2: Solve \(3x + 2<8\)

Subtract 2 from both sides: \(3x+2 - 2<8 - 2\), then \(3x<6\). Divide both sides by 3: \(x < 2\).

The compound inequality solution is \(-10 < x < 2\).

Answer:

The first box: \(x>-10\); the second box: \(x < 2\). Since \(-10 < x < 2\), for the graphical part (assuming standard open - circle for inequalities without equal signs, but if we consider the format of the given graphs with closed - circles as a possible mis - representation in the problem's graph options), the correct graph should have an open - circle at \(-10\) (moving to the right) and an open - circle at \(2\) (moving to the left). But if we assume the circles are just part of the drawing style in the problem (maybe a typo in the graph's circle style), the values \(-10\) and \(2\) are the endpoints. If we consider the number line in the problem's options:

• Option A has endpoints at \(-10\) and \(2\) (but the direction of the arrow in A is wrong as it should be between \(-10\) and \(2\)).
• Option B has endpoints at \(- 8\) and \(2\) (incorrect for our solution \(-10 < x < 2\)). But if we assume the problem has a mis - draw (maybe the left - hand circle in A is at \(-10\) and it's a continuous line between \(-10\) and \(2\) (despite the arrow in A looking odd, if we consider it's a line segment between \(-10\) and \(2\))).

If we assume the circles are just markers (not indicating inclusion/exclusion in a wrong way): The first box \(x > - 10\), the second box \(x < 2\).