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anonymous
 4 years ago
can someone check my answer:
2x+51<6
add 1
2x+5<7
2x+5>7
subtract 5
2x>12
anonymous
 4 years ago
can someone check my answer: 2x+51<6 add 1 2x+5<7 2x+5>7 subtract 5 2x>12

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precal
 4 years ago
Best ResponseYou've already chosen the best response.0when you drop the asolute value it should be \[7<2x+5<7\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does that mean my answer is wrong?

precal
 4 years ago
Best ResponseYou've already chosen the best response.0yes 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

precal
 4 years ago
Best ResponseYou've already chosen the best response.0\[12<2x<2\] \[6<x<1\] Your solution is from 6 to 1 but not including 6 or 1 just the numbers in between

precal
 4 years ago
Best ResponseYou've already chosen the best response.0or add 5 same thing and it is because you want to move it to the other side

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do i subtract if from the 1 or the 6?

precal
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left 2x+5 \right1<6\] 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 \[6<x<1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if i were to graph it on a number line what would it look like? as in wouldi t point to the left?

precal
 4 years ago
Best ResponseYou've already chosen the best response.0put 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327885941882:dwok so i put the circles on 6 then shade to the right and i put circles on 1 and shade to the left?

Directrix
 4 years ago
Best ResponseYou've already chosen the best response.1Study up on how to remove absolute value notation in a linear "less than" inequality. This pattern for "less than" absolutevalue 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 20002011 All Rights Reserved Solve  2x + 3  < 6. Since this is a "less than" absolutevalue 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" absolutevalue 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 20002011 All Rights Reserved Solve  2x + 3  < 6. Since this is a "less than" absolutevalue 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
 4 years ago
Best ResponseYou've already chosen the best response.0shade 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327886150381:dw the squigly lines are the shading lol

precal
 4 years ago
Best ResponseYou've already chosen the best response.0yes but put hollow circles on 6 and 1 because they are not solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hollow as in shaded circles?

precal
 4 years ago
Best ResponseYou've already chosen the best response.0no 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)
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