## anonymous 4 years ago can someone check my answer: |2x+5|-1<6 add -1 |2x+5|<7 2x+5>-7 subtract 5 2x>-12

1. precal

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

2. anonymous

does that mean my answer is wrong?

3. 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

4. precal

$-12<2x<2$ $-6<x<1$ Your solution is from -6 to 1 but not including -6 or 1 just the numbers in between

5. anonymous

why do i subract 5?

6. precal

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

7. anonymous

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

8. 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 $-6<x<1$

9. precal

hope that helps

10. anonymous

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

11. 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

12. 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?

13. 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

14. 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.

15. anonymous

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

16. precal

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

17. anonymous

18. 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)

19. anonymous

thank!

20. precal

You are welcome.

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