How do you solve and write the following in interval notation: #( 2 x - 1 ) / ( x - 3 )

Question

How do you solve and write the following in interval notation: #( 2 x - 1 ) / ( x - 3 ) <1#?

Ethan Adkins · Accepted Answer

#x in (-2, 3)# #(2x-1)/(x-3)=2+5/(x-3) < 1#. So, #5/(x-3) < -1#. And so, #LHS < 0 to x-3 < 0 to x < 3# Cross multiply and reverse the inequality. #5 > -(x-3) to x > -2#. Thus, #-2 < x <3#

Kennedy Adams · Answer

undefined To solve the inequality $ \frac{2x - 1}{x - 3} < 1 $ and write it in interval notation, you first find the critical points by setting the inequality equal to 1 and solving for x. Then, you test the intervals between these critical points to determine where the inequality holds true. 1. Set $ \frac{2x - 1}{x - 3} = 1 $ and solve for x: $$ 2x - 1 = x - 3 $$ $$ x = -2 $$ 2. Test the intervals: a) Test the interval (-∞, -2): Choose a test point, for example, x = -3: $ \frac{2(-3) - 1}{-3 - 3} = \frac{-7}{-6} > 1 $ The inequality does not hold in this interval. b) Test the interval (-2, 3): Choose a test point, for example, x = 0: $ \frac{2(0) - 1}{0 - 3} = \frac{-1}{-3} < 1 $ The inequality holds in this interval. c) Test the interval (3, ∞): Choose a test point, for example, x = 4: $ \frac{2(4) - 1}{4 - 3} = \frac{7}{1} > 1 $ The inequality does not hold in this interval. Therefore, the solution to the inequality $ \frac{2x - 1}{x - 3} < 1 $ in interval notation is (-2, 3).

Julia Adams · Answer

undefined To solve and write the inequality $ \frac{2x - 1}{x - 3} < 1 $ in interval notation, follow these steps: 1. Find the critical points by setting the inequality equal to 1 and solving for $ x $. 2. Determine the sign of the expression $ \frac{2x - 1}{x - 3} $ in each interval created by the critical points. 3. Write the solution in interval notation. Step 1: Find critical points: $ \frac{2x - 1}{x - 3} = 1 $ Solve for $ x $: $ 2x - 1 = x - 3 $ $ x = -2 $ Step 2: Determine the sign of $ \frac{2x - 1}{x - 3} $: Choose test points in the intervals created by the critical point $ x = -2 $. For $ x < -2 $, choose $ x = -3 $. $ \frac{2(-3) - 1}{(-3) - 3} = -\frac{7}{6} $, negative. For $ -2 < x < 3 $, choose $ x = 0 $. $ \frac{2(0) - 1}{0 - 3} = \frac{1}{3} $, positive. For $ x > 3 $, choose $ x = 4 $. $ \frac{2(4) - 1}{4 - 3} = 7 $, positive. Step 3: Write the solution in interval notation: Since the expression is less than 1, it's negative when $ x < -2 $ and positive when $ -2 < x < 3 $ and $ x > 3 $. Therefore, the solution is: $$ (-\infty, -2) \cup (3, \infty) $$

How do you solve and write the following in interval notation: #( 2 x - 1 ) / ( x - 3 ) &lt;1#?

How do you solve and write the following in interval notation: #( 2 x - 1 ) / ( x - 3 ) <1#?