Solving Absolute Value Equations?

1)
|2x - 6| = 12
2x - 6 = ±12

2x - 6 = 12
2x = 12 + 6
2x = 18
x = 18/2
x = 9

2x - 6 = -12
2x = -12 + 6
2x = -6
x = -6/2
x = -3

∴ x = -3 , 9

= = = = = = = =

2)
|4n + 7| = 1
4n + 7 = ±1

4n + 7 = 1
4n = 1 - 7
4n = -6
n = -6/4
n = -3/2 (-1.5)

4n + 7 = -1
4n = -1 - 7
4n = -8
n = -8/4
n = -2

∴ n = -2 , -3/2 (-1.5)

= = = = = = = =

3)
|1/2x - 3| = 4
1/2x - 3 = ±4

1/2x - 3 = 4
x/2 = 4 + 3
x = 2(7)
x = 14

1/2x - 3 = -4
x/2 = -4 + 3
x = 2(-1)
x = -2

∴ x = -2 , 14

= = = = = = = =

4)
|7 + 2x| = 9
7 + 2x = ±9

7 + 2x = 9
2x = 9 - 7
x = 2/2
x = 1

7 + 2x = -9
2x = -9 - 7
x = -16/2
x = -8

∴ x = -8 , 1

= = = = = = = =

5)
|1/3x - 3| = 9
1/3x - 3 = ±9

1/3x - 3 = 9
x/3 = 9 + 3
x = 3(12)
x = 36

1/3x - 3 = -9
x/3 = -9 + 3
x = 3(-6)
x = -18

∴ x = -18 , 36

= = = = = = = =

6)
|8 - 3x| = 11
8 - 3x = ±11

8 - 3x = 11
3x = 8 - 11
x = -3/3
x = -1

8 - 3x = -11
3x = 8 - (-11)
x = 19/3

∴ x = -1 , 19/3

attempt to isolate truthfully the cost section and sparkling up by potential of utilising the two situations. -8|7x - 4| = -19 |7x - 4| = 19/8 case a million is the place the respond is effective, so: 7x - 4 = 19/8 7x = 19/8 + 4 x = fifty one/fifty six case 2 is the place the respond is unfavorable, so: 7x - 4 = -19/8 7x = -19/8 + 4 x = thirteen/fifty six hence, x = fifty one/fifty six or x = thirteen/fifty six

14n+71 = 1
14n = 1-71
= -70
n = -70/14
n = -5
If you are still having confusion then you may check by putting n value -5
14x-5 +71 = 1

ok so basically you should think of an 'absolute value' as a number's 'distance from zero.' therefore, whether a number is positive of negative is irrelevant, and as such, |-x| = x.
this raises the problem that if you are asked to solve for an absolute value, there is going to be two possible results. you may recall from your earlier work that the major difficulty is deciding if the two answers are seperated by an 'AND' or an 'OR'. In any case, such as those above, where there is an equals (=) sign, rather than a greater-than/ less-than symbol, it will always be an 'OR' sum. But enough explaining.

1. |2x - 6| = 12
therfore, either 2x - 6 = 12
2x = 18
x = 9

OR 2x - 6 = -12
2x = -6
x = -3
This should be read as "x is equal to 9 OR -3".

1. I2x - 6I = 12
2x - 6 = +- 12
2x = 6 +- 12
2x = 6+12, 6-12
2x = 18, -6
x = 18/2, -6/2
x = 9, -3

Solve the others likewise. Let me know if you need any further help. My contact info are there in my profile.

For these questions you can think:
|f(x)| = a
-> f(x) = a and f(x) = -a
So there will be 2 solutions

I'll solve the first one, rest is same
|2x - 6| = 12
2x - 6 = 12 & 2x - 6 = -12
x = 9 & x = -3