How do you solve 7|x+2|+5>4 ?

Question

The | are absolute value signs, no typos. In this type of inequality equation there should be 2 answers but I can't seem to get it.

I said there are NO TYPOS. I did this question myself and ended up with a strange interval set, which is why I decided to ask other people. I thought it should've been two disjoint sets but I was right after all.

Thanks everyone

Answer 1

7∣x + 2∣ + 5 > 4

7x + 14 + 5 > 4

7x + 19 > 4

7x + 19 - 19 > 4 - 19

7x > - 15

7x / 7 > - 15 / 7

x > - 15 / 7

x = - 2 1/7

- - - - - - - - -s-

Answer 2

Are you *sure* there are no typos? This particular equation has the whole of R as a solution set - note that |x+2| >= 0 for all x, so 7|x+2| >= 0 for all x, and hence 7|x+2| + 5 >= 5 for all x.

To solve it formally:
7|x+2| + 5 > 4
<=> |x+2| > -1/7
<=> x + 2 > -1/7 or x + 2 < 1/7
<=> x > -15/7 or x < -13/7
<=> x ∈ (-∞, -13/7) ∪ (-15/7, ∞)
<=> x ∈ (-&infin, &infin).

Normally you'd get two disjoint intervals, but in this case they overlap.

Answer 3

7|x + 2| + 5 > 4

Normally, for absolute value inequalities, there will be two cases. For some positive constant c
1) |y| > c

In this case, this translates to an OR. This translates into
(y > c) OR (y < -c).

2) |y| < c

In this case, this translates to an AND. This translates into

(y < c) AND (y > -c)

This case presents a rather unusual one though, and doesn't qualify for #1 or #2 ...

7|x + 2| + 5 > 4
7|x + 2| > -1
|x + 2| > (-1/7)

The function f(x) = |x + 2| has a range of 0 to infinity; in other words, it is either 0 or positive. What the above inequality is asking is when |x + 2| > (-1/7) is a true statement. But it's *always* true, because |x + 2| is always 0 or positive.

That makes the solution all real numbers. It's similar to the inequality x^2 > -2; it will always be true because x^2 will always be greater than or equal to 0.

Answer 4

7 |x+2| + 5 > 4

7 |x+2| + 5 - 5 > 4 - 5 => 7 |x+2| > -1
(7 |x+2|) / 7 > -1 / 7
|x + 2| > -1/7

this means that either
1) x + 2 > -1/7
or
2) - (x + 2) > -1/7
is true.

from 1)
x + 2 > -1/7 => x > -15/7
from 2)
- (x + 2) > -1/7 => x + 2 < 1 / 7 => x < -13 / 7

this means that
x > -15 / 7 or x < -13 / 7

note that the union of these solution sets comprise the real number system. which means that x can be any real number.

(this is easy to see since |x+2| is always positive or zero and multiplying this to 7 will only yield non-negative answer. adding 5 to a non-negative number will surely result to a number greater than 4)

Answer 5

2 + 7/x = 4/x^2 2x^2 + 7x=4 ---- Multiplying each and every term with the help of with the help of x^2 2x^2 + 7x - 4 = 0 2x^2 + 8x - 1x - 4 = 0 2x(x + 4) -a million(x + 4) = 0 (x + 4)(2x -a million) = 0 x + 4 = 0 -------> x = -4 2x -a million = 0 -------> x = a million/2

Answer 6

You'll treat with the absolute sign one time by positive sign and other time with negative sign.
So the soultion has 2 parts:
First part :
7(x+2)+5 > 4
7x+14+5 > 4
7x > 4-19
7x > -15 (Both sides / 7)
x > (-15/7)
x is belongs to ]-15/7 , infinity [ -------1
second part :
7(-x-2)+5 > 4
-7x-14+5 > 4
-7x > 4+9
-7x > 13 (Both sides / -7)
x > (-13/7)
x is belongs to ]-13/7 , infinity [ -------2

From 1 and 2 so, X is belongs to ]-15/7 , infinity [

Mohamed Hesham
Egypt

Answer 7

|x+2|>0 for any x. So 7|x+2|>0 for any x too. Then 7|x+2|+5>5 for any x. And 5>4 so it's true for any x.
I think there is an error. Maybe it's 7|x+2|-5>4, or 7|x+2|+5>14

Answer 8

This is a question related to solving equations involving Absolute Values.

I could give you the full worked solution to the question you have given, however, it's really in your own interest to figure out how to do this on your own! I suggest that you do a little reading of your own, and you'll be able to solve this yourself! Being able to do this on your own will make it much easier for you during exam time!

A good tutorial that can help you solve this question can be found at:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut21_abseq.htm

Answer 9

there will be 2 answers because lx+2l will be solved by (x+2) and -(x+2)

first : x+2

7x+14+5>4
7x+19>4
7x>-15
x>-(15/7)

second: -(x+2)

-7x-14+5>4
-7x-9>4
-7x>13
x>-(13/7)

that's your 2 answers
hope that's right

Answer 10

When |x+2| is positive,

7x+14+5>4
7x+19>4
7x>-15
x>-2.14

When |x+2| is negative,

-7x-14+5>4
-7x-9>4
-7x>13
x<13/7