Question

|x+7|+13≥4

Answer 1

oops sorry hold on
x+7+13=4
x+7=-9
x=-16

Answer 2

\[X+7\ge\]

Answer 3

so we are going to subtract 4 from 13?

Answer 4

lol woops

Answer 5

yes
x+7+13=4
x+7+13-13=4-13

Answer 6

\[X+7\ge-9\]

Answer 7

you're subtracting 13 from 4

Answer 8

not the other way around. because you're solving for X

Answer 9

ugh sorry. my bad XD. just out of it XD. but anyway, I showed u the subtraction part

Answer 10

and those bars around X+7 just mean absolute value. and since X+7 is positive you don't need to worry bout that. so then subtract 7 from both sides.

Answer 11

friggin equation typer

Answer 12

we are not going to mind the sign of this one? |x+7|

Answer 13

\[X \ge-16\]

Answer 14

yeah. don't worry about the absolute value. whenever its positive inside the absolute value, it will always be postive

Answer 15

yeah, you don't worry about it. if it was |X-7| then it would equal X+7 because 7 is the value of the used number. so absolute value basically means the total of the equation, but positive

Answer 16

so the answer is X≥−16?

Answer 17

yes

Answer 18

That is only the first part of the answer. There is a second part to it.

Answer 19

lol teamwork!

Answer 20

lolwut

Answer 21

second part?

Answer 22

and the solution will be

x+7+13=4
x+7+13-13=4-13
x+7=-9
x=-9-7
x=-16 is this right?

Answer 23

plz explain. cuz it seems to me like that's it.. unless i forgot something about algebra in which case i'd love for you to remind me of it because i have a psat next wednesday

Answer 24

it is right. idk y there will be a second part

Answer 25

X is going to be greater than or equal to -16

Answer 26

Yes. The part above was solved assuming |x + 7| was positive.
Next assume it was |-(x + 7)| inside the absolute or modulus operation.
|-(x + 7)| will also = |x + 7|.
So solve -x - 7 >= -9

Answer 27

For example, solve |x| = 7.
The answer is x = 7 and x = -7

Answer 28

ur over thinking this. unless it states a negative outside the bars for absolute value it will always be positive...

Answer 29

so the second answer will be x>= -2?

Answer 30

yeah. I don't think there is a need to find the other solution

Answer 31

@Aero+Exis No, not really. Look at the example I gave earlier.
What is the solution for |x| = 2 ?

Answer 32

lol. that's encompassed by \[X \ge-16\]

Answer 33

im getting confused xD

Answer 34

the answer is -16. if the problem asked u to do more, then yeah doing the other method will help but don't confuse urself

Answer 35

The only instruction was SOLVE,

Answer 36

@ranga can you show me your solution on the 2nd part and @mew55 can you show me yours please?

Answer 37

Hold on one sec. Attending to another matter at home. Will be back soon.

Answer 38

ok..

Answer 39

lx+7l+13>4

the absolute value of x+7 is x+7, always positive.
x+7+13>4

add 7 and 13 together

x+20>4

subtract 20 to both sides

x+20-20>4-20

x>-16

sorry still didn't get the greater than equal to symbol XD

Answer 40

\[x \ge -16? \]

Answer 41

yes

Answer 42

so in the second one i assume that |x+7| will be -(x+7) ?

Answer 43

-(x+7) >= -9
-x-7 >= -9
-x >= -2
Multiply both sides by -1 (but flip the sign cuz multiplying by a negative)
x <= 2
-16 <= x <= 2 I think is the solution.

Answer 44

r u getting the problem out of a book MysteryOwl?

Answer 45

oh snap. ranga's right...

Answer 46

really?

Answer 47

yep

Answer 48

dis guy...

Answer 49

gj ranga

Answer 50

huh. maybe XD

Answer 51

:O

Answer 52

@ranga can please make it clear?

Answer 53

... omg now i'm confused... it can't be right. because there is no - sign outside the bars. which means that the value will always be above -16! so then my original is correct! i'm sorry @MysteryOwl we're not helping, we're just confusing you. over such a simple problem too

Answer 54

yeah I agree with Aero_Exis. we r sorry owl.

Answer 55

the answer must be \[X \ge-16\]
no if ands or buts.

Answer 56

i typed in on wolfram. same answer

Answer 57

its ok :D i believe you

Answer 58

so disregard renga, he was just overthinking the problem, he was correct in a certain situation, but it can not fit this one. sorry for all the trouble mystery :)

Answer 59

its ok.. so the solution would be like this?
lx+7l+13>4
the absolute value of x+7 is x+7, always positive.
x+7+13>4 add 7 and 13 together
x+20>4
subtract 20 to both sides x+20-20>4-20
x>-16

Answer 60

yep

Answer 61

thanks :D

Answer 62

ur welcome. sorry for the confusion

Answer 63

yes :) remember that's greater than or equal too though. so it's gotta have the dash under the >. like this. \[X \ge-16 :)\]

Answer 64

now go forth with your newfound knowledge. and conquer my friend. XD

Answer 65

LOL xD

Answer 66

hahaha XD. btw, does anybody know the Euler Method?

Answer 67

never heard of that

Answer 68

errrr can't say i do. i'll research it for you though if ya want. i got nothing better to do :P

Answer 69

its this problem. y'=y(3-ty) y(0)=0.5 h=0.5.

Answer 70

dat stuff went right over my head. wikipedia doesn't explain things in laymen's terms.

Answer 71

Now I am getting confused.
Start with the original problem:
|x + 7| + 13 >= 4
|x + 7| >= -9

It does not matter what value x takes the left hand side will ALWAYS be positive and will always be greater than -9!!!

Where is the error?

Answer 72

shhhh...

Answer 73

xD

Answer 74

kk. so the real question is. who gets the medal O.o

Answer 75

irdk xD

Answer 76

nosebleed..

Answer 77

Man, such a simple algebra problem is stumping me.

But one thing though. If I have an inequality say:
| 2x + 3 | < 6
The correct way to solve the problem is not to just solve 2x + 3 < 6 but
-6 < 2x + 3 < 6
-9/2 < x < 3/2

And to solve | 2x – 3 | > 5
2x – 3 < –5 OR 2x – 3 > 5
2x < –2 or 2x > 8
x < –1 or x > 4

Answer 78

Going back to the original problem:
|x + 7| + 13 >= 4

|x + 7| can never be negative no matter whether x is negative, zero or positive.
And 13 will always be > 4.
So the left hand side will always be greater than the right hand side

So the answer is ALL values of x satisfies the inequality.

Answer 79

so what ur saying is \[X \ge0\]

Answer 80

ALL values of x: x <= 0, x = 0, x >= 0.

Try picking any number for x and put it in the inequality and the inequality will always be valid.

Answer 81

guys? are you quarreling? or debating?

Answer 82

Stumped by an unusual problem and stumbling into a solution.

Answer 83

yeah... but X can't be less than zero... so then X>= 0 is the right answer i suppose...

Answer 84

The problem doesn't state x can't be negative.

In the example I gave earlier, | 2x – 3 | > 5
the solution is x < -1 or x > 4
So if I pick x = -8, the LHS will be |-16-3| = 19 which is greater than 5 and so it is a valid solution.
But if we pick any number between -1 and 4 the inequality will not be valid
Say x = 0, then LHS = |0-3| = 3 which is not > 5 and so it os not a solution.

Answer 85

Later this week I am going to post this riddle of a problem and ask some math wiz to answer why this problem gives different answers when the real answer seems to be all values of x is good.

Answer 86

:O

Answer 87

... i don't like this world anymore. hahaha oh algebra

Answer 88

Finally figured out the error in the logic.

For solving an inequality with ABSOLUTE values, the rule says there are two parts to the solution.

If the inequality is |x| < a, then the solution is: -a < x AND x < a

If the inequality is |x| > a, then the solution is: x < -a OR x > a

Example: Solve |x| < 3. Solution: -3 < x < 3

Example: Solve |x| > 5. Solution: x < -5 or x > 5

In BOTH cases, the right hand side value of the inequality ("a") MUST BE POSITIVE. ONLY then you can proceed to solve the problem using the two part solution mentioned above.

Here the problem was: |x+7| +13 ≥ 4
We subtract 13 from both sides to isolate the absolute value.
|x+7| >= -9

The right hand side is negative and therefore we cannot use the two part solution here. That is the mistake all of us made earlier. I solved the two parts. Some of you solved one part. But both are wrong.

When you have an absolute value on the left side of the inequality and you have a negative value on the right hand side of the inequality then we will have to treat them as special cases.

So for |x+7| >= -9 we can see that the left hand side, being an absolute value, will ALWAYS be greater than the right hand side which is a negative number. So the solution is ALL values of x are good.

If the original problem had a <= sign instead of a >= sign then the problem will be:
|x+7| <= -9. Since the right hand side is negative this is also a special case and we can easily see that the inequality is absurd because an absolute value can never be less than a negative number. Therefore, there are NO solutions to this inequality.
So in the first case, all values of x are good.
In the second case, no value of x is good.
These are the special cases.

So the moral of the story is: In all other cases of inequality involving absolute values we will have to remember that there are two parts to the solution. But we should not use the two part solution when the left hand side is an absolute value and the right hand side is a negative value.

Answer 89

When I say the solution to this problem is all values of x I mean all REAL values of x.