anonymous
  • anonymous
can someone check my answer: |2x+5|-1<6 add -1 |2x+5|<7 2x+5>-7 subtract 5 2x>-12
Mathematics
jamiebookeater
  • jamiebookeater
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precal
  • precal
when you drop the asolute value it should be \[-7<2x+5<7\]
anonymous
  • anonymous
does that mean my answer is wrong?
precal
  • 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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precal
  • precal
\[-12<2x<2\] \[-6
anonymous
  • anonymous
why do i subract 5?
precal
  • precal
or add -5 same thing and it is because you want to move it to the other side
anonymous
  • anonymous
do i subtract if from the -1 or the 6?
precal
  • 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
  • precal
hope that helps
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
|dw:1327886150381:dw| the squigly lines are the shading lol
precal
  • precal
yes but put hollow circles on -6 and 1 because they are not solutions.
anonymous
  • anonymous
hollow as in shaded circles?
precal
  • 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
  • anonymous
thank!
precal
  • precal
You are welcome.

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