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How do you graph and solve #3|8-x| + 2 <7- 2|x-8|#?

Answer 1

Eva Adamson

#7<=x<=9#

The graph is all the area between and including x =7 ; x=9

Given: #3|8-x|+2<7-2|x-8|#

Collecting like terms

#3|8-x|+2|x-8|<7-2#

But #|8-x| = |x-8| -> test->|8-2|=|2-8 | -> 6=6#

using only #|8-x|# for both

Factoring out gives:

#|8-x|(3+2)<5#

#|8-x|<1#

Absolute is always 'not negative' so we need:

#0<=|8-x|<1#

If #x = 9" then "|8-x|=1# If #x>9" then "|8-x|>1# If #x=7" then "|8-x|=1# If #x<7" then "|8-x|>1#

So #color(white)(.)7<=x<=9#

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Answer 2

Savannah Anderson

To graph and solve the inequality (3|8-x| + 2 < 7 - 2|x-8|), follow these steps:

Identify the critical points where the expressions within the absolute value signs change sign. For (3|8-x|), the critical point is (x = 8), and for (2|x-8|), the critical point is also (x = 8).
Split the number line into intervals based on the critical points.
Test a value from each interval into the original inequality to determine the solution set.
Graph the solution set on the number line.

After solving, you'll find the solution set and graphically represent it on the number line.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you graph and solve #3|8-x| + 2 &lt;7- 2|x-8|#?

How do you graph and solve #3|8-x| + 2 <7- 2|x-8|#?