Solve the equation 5+7 | 2x-1 | = -44 the solutions are: ??

i have tried to work out what i need to do to get the solutions that are being asked for but i need help and an explanation of possible! thank you so much!

The easy part is isolating the absolute-value term:

5 + 7 |2x - 1| = -44

7 |2x - 1| = -49

|2x - 1| = -7

Remember that the absolute value function returns a positive number that you can think of as the "size" of that number, or the positive distance between that number and zero. If x is a positive number, its absolute value is the same number, |x| = x. But if x is negative, then the absolute value returns its negative, |x| = -x, which makes it positive. (If x = 0, you can use either result, since -0 is still 0.)

The important thing to take from this is that there are 2 cases to consider: is the expression in the absolute value positive, or is it negative?

• If 2x - 1 > 0, then |2x - 1| = 2x - 1. Then the equation becomes

2x - 1 = -7

2x = -6

x = -3

• If 2x - 1 < 0, then |2x - 1| = - (2x - 1) = 1 - 2x. Then

1 - 2x = -7

-2x = -8

x = 4

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}