can someone check my answer:
|2x+5|-1<6
add -1
|2x+5|<7
2x+5>-7
subtract 5
2x>-12

- anonymous

can someone check my answer:
|2x+5|-1<6
add -1
|2x+5|<7
2x+5>-7
subtract 5
2x>-12

- jamiebookeater

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- precal

when you drop the asolute value it should be \[-7<2x+5<7\]

- anonymous

does that mean my answer is wrong?

- precal

yes because you never switch the inequality from less than to greater than unless you multiply by a negative number or if you divide by a negative number.
subtract 5 to all sides and then divide by 2

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## More answers

- precal

\[-12<2x<2\]
\[-6

- anonymous

why do i subract 5?

- precal

or add -5 same thing and it is because you want to move it to the other side

- anonymous

do i subtract if from the -1 or the 6?

- precal

\[\left| 2x+5 \right|-1<6\]
add 1 to both sides
\[\left| 2x+5 \right|<7\]
drop absolute value
\[-7<2x+5<7\]
subtract 5 to all 3 sides
\[-12<2x<2\]
divide by 2 to all 3 sides
\[-6

- precal

hope that helps

- anonymous

if i were to graph it on a number line what would it look like? as in wouldi t point to the left?

- precal

put hollow circles on -6 and 1 that shows that they are not included and shade in between those two numbers.
-5, -4, -3, -2, -1, 0 are solutions but so are -5.999 to .99999

- anonymous

|dw:1327885941882:dw|ok so i put the circles on -6 then shade to the right and i put circles on 1 and shade to the left?

- Directrix

Study up on how to remove absolute value notation in a linear "less than" inequality.
This pattern for "less than" absolute-value inequalities always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the exercises get more complicated, the pattern still holds. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
Solve | 2x + 3 | < 6.
Since this is a "less than" absolute-value inequality, the first step is to clear the absolute value according to the pattern. Then I'll solve the linear inequality.
| 2x + 3 | < 6
–6 < 2x + 3 < 6 [this is the pattern for "less than"]
–6 – 3 < 2x + 3 – 3 < 6 – 3
–9 < 2x < 3
–9/2 < x < 3/2
Then the solution to | 2x + 3 | < 6 is the interval –9/2 < x < 3/2.
This pattern for "less than" absolute-value inequalities always holds: Given the inequality | x | < a, the solution is always of the form –a < x < a. Even when the exercises get more complicated, the pattern still holds. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
Solve | 2x + 3 | < 6.
Since this is a "less than" absolute-value inequality, the first step is to clear the absolute value according to the pattern. Then I'll solve the linear inequality.
| 2x + 3 | < 6
–6 < 2x + 3 < 6 [this is the pattern for "less than"]
–6 – 3 < 2x + 3 – 3 < 6 – 3
–9 < 2x < 3
–9/2 < x < 3/2
Then the solution to | 2x + 3 | < 6 is the interval –9/2 < x < 3/2.
http://www.purplemath.com/modules/absineq.htm

- precal

shade all of it from -6 to 1, think of this as all possible solutions.
Any number from -6 to 1 are solutions except for -6 and 1.
-5.99 is a solution to .999 is a solution.

- anonymous

|dw:1327886150381:dw|
the squigly lines are the shading lol

- precal

yes but put hollow circles on -6 and 1 because they are not solutions.

- anonymous

hollow as in shaded circles?

- precal

no shaded circles are solid points and represent that the point is a solution. You do not have an equal sign under your inequality therefore it is a hollow circle (not shaded circle)

- anonymous

thank!

- precal

You are welcome.

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