[MOVED] PRE- CALC absolute value inequalities

[lahoku808]

I honestly have tried to work this problem for about 2 hours now and I am either just completely in left field or was sick when this problem was explained. Any help is greatly appreciated!

Describe the interval -5 < x < 7 with an inequality of the form |x - x₀| < d

In most cases, the problem usually starts with an inequality, such as |5 - x| < 2, and is put into interval form. This problem is asking for the opposite. Is there a formula for this?

Thank you to anyone who is kind enough to help!!!

-Elizabeth

 

[stapel]

When you had |5 - x| < 2, you did the following:

. . . . .|5 - x| < 2

. . . . .-2 < 5 - x < 2

. . . . .-2 - 5 < 5 - 5 - x < 2 - 5

. . . . .-7 < -x < -3

. . . . .7 > x > 3

In this case, work backwards by finding something that will allow you to have the same value (ignoring the signs for the moment) on either end:

. . . . .-5 < x < 7

. . . . .7 - (-5) = 7 + 5 = 12

. . . . .Since the end values are 12 units apart, and since
. . . . .-6 and 6 are symmetrically 12 units apart, is
. . . . .there a way to convert the above to having end
. . . . .values of -6 and 6?

. . . . .Yes:

. . . . .-5 - 1 < x - 1 < 7 - 1

Can you take it from there? :wink:

Eliz.