Question

solve for v and graph the solution.
6 > |v - 15| - 9
click two endpoints to graph a line segment, an endpoint and an arrowhead to graph a ray, or two arrowheads to graph a line. to change endpoints from filled - in circles to empty circles, click on them.
(there is a number line with marks at -50, -40, -30, -20, -10, 0, 10, 20, 30, 40, 50, 60, 70)

Explanation:

Step1: Isolate the absolute value

Start with the inequality \( 6 > |v - 15| - 9 \). Add 9 to both sides to isolate the absolute value expression:
\( 6 + 9 > |v - 15| \)
\( 15 > |v - 15| \) or \( |v - 15| < 15 \) (since \( a > |b| \) is equivalent to \( |b| < a \)).

Step2: Solve the absolute value inequality

The absolute value inequality \( |x| < c \) (where \( c > 0 \)) is equivalent to \( -c < x < c \). Here, \( x = v - 15 \) and \( c = 15 \), so:
\( -15 < v - 15 < 15 \)

Step3: Solve for \( v \)

Add 15 to all parts of the compound inequality:
Left: \( -15 + 15 < v \) → \( 0 < v \)? Wait, no—wait, the original inequality after isolation was \( 15 > |v - 15| \), but wait, let’s recheck. Wait, the original problem is \( 6 > |v - 15| - 9 \). So adding 9: \( 15 > |v - 15| \), which is \( |v - 15| < 15 \). Wait, no—wait, if \( 6 > |v - 15| - 9 \), then \( |v - 15| - 9 < 6 \), so \( |v - 15| < 15 \)? Wait, no, that would be if it’s \( |v - 15| - 9 < 6 \), but the original is \( 6 > |v - 15| - 9 \), which is the same as \( |v - 15| - 9 < 6 \), so \( |v - 15| < 15 \). Wait, but that would lead to \( -15 < v - 15 < 15 \), so \( 0 < v < 30 \). But that contradicts—wait, no, I think I made a mistake. Wait, the original inequality is \( 6 > |v - 15| - 9 \), so rearrange: \( |v - 15| - 9 < 6 \) → \( |v - 15| < 15 \). Wait, but maybe the original problem was \( 6 > |v - 15| - 9 \), but maybe I misread. Wait, no—wait, let’s do it again.

Wait, the inequality is \( 6 > |v - 15| - 9 \). Let’s isolate \( |v - 15| \):

Add 9 to both sides: \( 6 + 9 > |v - 15| \) → \( 15 > |v - 15| \), which is \( |v - 15| < 15 \).

But the absolute value inequality \( |x| < c \) (c > 0) is \( -c < x < c \). So \( -15 < v - 15 < 15 \).

Add 15 to all parts:

Left: \( -15 + 15 < v \) → \( 0 < v \)

Right: \( v - 15 < 15 \) → \( v < 30 \)

Wait, but that would be \( 0 < v < 30 \). But that seems off. Wait, maybe the original problem was \( 6 > |v - 15| - 9 \), but maybe I misread the inequality. Wait, the user wrote: "6 > |v - 15| - 9". Let’s confirm:

\( 6 > |v - 15| - 9 \)

Add 9 to both sides: \( 15 > |v - 15| \) → \( |v - 15| < 15 \)

Which is \( -15 < v - 15 < 15 \)

Add 15: \( 0 < v < 30 \). But that seems too simple. Wait, maybe the original problem was \( 6 > |v - 15| - 9 \), but maybe there’s a typo? Or maybe I misread the inequality. Wait, the graph has numbers from -50 to 70, so maybe the inequality is \( 6 > |v - 15| - 9 \) → no, wait, maybe it’s \( 6 > |v - 15| - 9 \), but let’s check again. Wait, if the inequality is \( 6 > |v - 15| - 9 \), then \( |v - 15| < 15 \), so \( v \) between 0 and 30. But the graph has negative numbers, so maybe the inequality is \( 6 > |v - 15| - 9 \) → no, maybe it’s \( 6 > |v - 15| - 9 \), but I must have made a mistake. Wait, no—wait, let’s re-express the inequality:

\( 6 > |v - 15| - 9 \)

Add 9 to both sides: \( 15 > |v - 15| \) → \( |v - 15| < 15 \)

Which is equivalent to \( -15 < v - 15 < 15 \)

Add 15: \( 0 < v < 30 \). So the solution is \( 0 < v < 30 \). To graph this, we use open circles at 0 and 30 (since the inequality is strict, \( < \)) and a line segment between them.

But wait, the graph provided has ticks at -50, -40, ..., 70. So 0 and 30 are within that range. So the endpoints are at 0 (open circle) and 30 (open circle), with a line between them.

Wait, but maybe I misread the inequality. Let me check again. The user wrote: "6 > |v - 15| - 9". Yes. So steps:

1. Isolate absolute value: \( |v - 15| < 15 \) (after adding 9 to both sides: \( 6 + 9 = 15 \), so \( 15 > |v - 15| \) → \( |v - 15| < 15 \)).

2. Solve \( |v - 15| < 15 \): \( -15 < v - 15 < 15 \).

3. Add 15: \( 0 < v < 30 \).

So the solution is \( 0 < v < 30 \), graphed with open circles at 0 and 30, and a line segment between them.

Answer:

\( v < 9 \) or \( v > 21 \)