How do you solve and graph #abs((3h+1)/2)

Question

How do you solve and graph #abs((3h+1)/2)<8#?

Isaiah Anthony · Accepted Answer

See a solution process below:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent. #-8 < (3h + 1)/2 < 8# First, multiply each segment of the system of inequalities by #color(red)(2)# to eliminate the fraction while keeping the system balanced: #color(red)(2) xx -8 < color(red)(2) xx (3h + 1)/2 < color(red)(2) xx 8# #-16 < cancel(color(red)(2)) xx (3h + 1)/color(red)(cancel(color(black)(2))) < 16# #-16 < 3h + 1 < 16# Next, subtract #color(red)(1)# from each segment to isolate the #h# term while keeping the system balanced: #-16 - color(red)(1) < 3h + 1 - color(red)(1) < 16 - color(red)(1)# #-17 < 3h + 0 < 15# #-17 < 3h < 15# Now, divide each segment by #color(red)(3)# to solve for #h# while keeping the system balanced: #-17/color(red)(3) < (3h)/color(red)(3) < 15/color(red)(3)# #-17/3 < (color(red)(cancel(color(black)(3)))h)/cancel(color(red)(3)) < 5# #-17/3 < h < 5# Or #h > -17/3# and #h < 5# Or, in interval notation: #(-17/3, 5)#

Nathan Axel · Answer

undefined To solve and graph the inequality $ \frac{|3h + 1|}{2} < 8 $:

1. **Isolate the absolute value expression**:
$$ \frac{|3h + 1|}{2} < 8 $$
$$ |3h + 1| < 16 $$

2. **Consider both cases**:
   - Case 1: $3h + 1$ is positive:
     $$ 3h + 1 < 16 $$
     $$ 3h < 15 $$
     $$ h < 5 $$

- Case 2: $3h + 1$ is negative:
     $$ -(3h + 1) < 16 $$
     $$ -3h - 1 < 16 $$
     $$ -3h < 17 $$
     $$ h > -\frac{17}{3} $$

3. **Combine the solutions**:
   $$ -\frac{17}{3} < h < 5 $$

4. **Graph the solution**:
   - Draw a number line.
   - Mark the points $h = -\frac{17}{3}$ and $h = 5$.
   - Shade the region between these two points, excluding the endpoints, since $h$ cannot equal these values.

How do you solve and graph #abs((3h+1)/2)&lt;8#?

How do you solve and graph #abs((3h+1)/2)<8#?