Question

what is the solution of 5|x+1|+6=21

Answer 1

how would we start it?

Answer 2

5|χ+1|=15 => |x+1| = 3.

Answer 3

im sure as long as you just get an answer; you will continue to type in the same kind of problem over and over again :)

Answer 4

assuming x is real:

\[5 \sqrt((x+1)^2)+6 = 21\]

otherwise:

|x+1|=3

Answer 5

2

Answer 6

don't just give it to him.

Answer 7

i don't know how to start it absolute value i just don't understand it

Answer 8

the absolute value is the last thing to work on; how do you get that absolute value all by itself?

Answer 9

5(x+1)+6=21
(5x+5)+6=21
5x+6=16
5x=10
x=2

Answer 10

looking at this question, what is the other solution for x if it was not an absolute value?

Answer 11

i'm seriously terrible in math i really don't know:[

Answer 12

the term "absolute value" means that there is a quantifiable value; something that can be measured. length, height, area, volume, distance are all examples of abslute values

Answer 13

dont jus give him the answer he is probably usin this site for a summer school!!!

Answer 14

in other words; an absolute value doesnt care about what direction the sign takes us in; it only cares about a quantifiable measure

Answer 15

|a| is always positive, you discount the negative answer.

Answer 16

sky, you isolate the absolute value just like you would isolate a variable.

Llort, no you do not 'discount' the negative answer. It is a valid solution.

Answer 17

|-12| has the same absolute value as |12|

Answer 18

as a result; if we are determining any value for inside of the absolute value notation; we have to account for the positive and the negative results that occur within it

Answer 19

So if you have

\[5|x+1| + 6 = 21\]

You want to isolate the term with the absolute value in it first. You do this by subtracting 6 from both sides of the equation.

Answer 20

What do you get when you subtract the 6?

Answer 21

-2?

Answer 22

No,
\(5|x+1| + 6 - 6 = 21 -6\)
\[\implies 5|x+1| + 0 = 15\]\[\implies 5|x+1| = 15\]

Answer 23

So now lets say that |x+1| is something, lets call it k..

So we have
\(5k = 15 \)\[\implies \frac{1}{5}\cdot5k = \frac{1}{5}15\]\[\implies k = 3\]
And since we said k was |x+1| we have \[|x+1| = 3\]So what values can x have and still make this equation true?

Answer 24

I am assuming the other answer is x=-4

Answer 25

x =-4 is one solution yes, because

|-4 + 1 | = |-3| = 3