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|x-8|<|2x+1|

Help would be greatly appreciated.

- Thread starter Brady
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- #1

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|x-8|<|2x+1|

Help would be greatly appreciated.

- #2

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In addition, write |-5x^5y^2| two ways.

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Tide

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- #4

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- #5

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please help me

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Tide

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Did you make a sketch?

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- #8

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Brady said:

|x-8| - |2x+1| < 0Brady said:

|x-8|<|2x+1|

Help would be greatly appreciated.

This solution is based on the fact that if a continuous function changes its sign, it might happen only in this function zeroes.

1.

Mark zeroes for each |expression| on the number line: x=8 and x=-1/2.

This two points break the whole number line into three intervals:

I. {-inf, -1/2}: -inf < x < 8

II. [-1/2, 8}: -1/2 <= x < 8

III. [8, +inf}: 8<= x < +inf.

Each expression "keeps" its sign unchanged on each interval.

Draw the number line and put the expressions' signs on the intervals

x-8: .............-..........|........-.........|.......+.........

2x+1: ...........-..........|........+.........|.......+.........

Solve inequality on each region separately following the definition

|a| = -a, if a<0

|a| = a, if a>=0

I.

x-8<0 and 2x+1<0 therefore

-(x-8) - (-(2x+1)) < 0

-x + 8 + 2x + 1 < 0

x < -9 - agrees with -inf < x < 8 condition.

II.

x-8<0 and 2x+1>=0 therefore

-(x-8) - (2x+1) < 0

-x + 8 - 2x - 1 < 0

3x > 7

x > 7/3 - considering -1/2 <= x < 8 restriction

the answer on this interval

7/3 < x < 8

III.

x-8>=0 and 2x+1>=0 therefore

x - 8 - (2x+1) < 0

x - 8 - 2x - 1 < 0

x > - 9 - considering x>=8 restriction

the answer on this interval

x >= 8.

Combininig all the answers,

x < -9 or x > 7/3.

Faster way - just like Tide recommended.

Draw two graphs:

f(x) = |x-8| and g(x) = |2x+1|

(do you need help with plotting them?).

|x-8|<|2x+1| is equvalent to

graph f(x) is below graph g(x).

These two graphs intersect at x=-9 and x=7/3.

Graph f(x) is below graph g(x) left of -9 and right of 7/3.

The same answer.

- #9

HallsofIvy

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Tide said:

Tide just did!Brady said:please help me

Tide said:Did you make a sketch?

Yes, you did. You didBrady said:

In fact, although you say you tried, you have shown us NOTHING of what you tried.

That is, after all, a requirement for homework help on this forum.

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