Question

Find a and b. If |x-3|<7, then a 12 years ago

Answer 1

-7<x-3<7
-7+3<x<7+3
-4<x<10
add4 in each inequality and get the value of a and b.

Answer 2

If you were to substitute in -4 and10 in to a<x+4<b, wouldn't you subtract 4 from each side to get -8<x<6?

Answer 3

Are we doing piece wise functions here?

Answer 4

I do not understand? Piece wise functions?

Answer 5

You haven't learned that? Then I got my answer.

Answer 6

LOL! Probably not, still in the beginning of College Algebra

Answer 7

This is not college algebra.

Answer 8

This is Algebra 2.

Answer 9

That makes senses, but it is called College Algebra at my school, and towards end of semester it gets in to Trig

Answer 10

That is algebra two. And this is piece wise functions. I am going to go offline for a while be back in about ten or fifteen minutes.

Answer 11

Thank you

Answer 12

\(\bf |x-3|<7 \implies
\begin{cases}
+(x-3)<7\\ \quad \\
\bf -(x-3)<7
\end{cases}\)


if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?

Answer 13

I am so lost

Answer 14

hmm ok, what part?

Answer 15

When I work those problems, which I am probably not doing correctly I get:\[+(x−3)<7 = x<10\] \[−(x−3)<7 = x>4\] but the answer is -4, so I believe I am doing this wrong

Answer 16

So would I have multiplied both sides by -1 and then added three to both sides?

Answer 17

those answers are correct

Answer 18

now to make each of the inequalities look like " x + 4 "
just add 4 to each side :)

Answer 19

hmm... the 2nd one would be -4l

Answer 20

other than that, is ok... so \(\large {|x-3|<7 \implies
\begin{cases}
+(x-3)<7 \implies x < 10\\ \quad \\
-(x-3)<7 \implies x > -4
\end{cases}\\ \quad \\
x < 10 \implies x +4 < 14\\ \quad \\
x > -4 \implies x+4 > 0}\)

Answer 21

so you can see what "a" and "b" are

Answer 22

So I did end up with -4<x<10, so if I supplement those in a<x+4<b = -4<x+4<10. Even though it says find a and b, would I go on to say -8<x<6 ?

Answer 23

\(\large {\begin{cases}
x < 10 \implies x +4 < 14\\ \quad \\
x > -4 \implies x+4 > 0
\end{cases}\implies \bf 0 < x +4 < 14}\)

Answer 24

So I would add 4 to both sides and not subtract?

Answer 25

well, they want \(\large x \color{red}{+}4\)

so you'd need to add, yes

Answer 26

-4<x+4<10, I thought you would work it like a normal equation and to carry over the 4, you would have to subtract from all sides? -4+-4<x+4-4<10-4 = -8<x<6, how would it cancel the +4 unless you subtract?

Answer 27

yeap

when yo have only left and right sides, you add/divide/multiply to BOTH sides

when you have left, middle, and right sides, you add/divide/multiply to ALL sides

Answer 28

-4<x<10
-4+4<x+4<10+4
0<x+4<14
hence a=0,b=14

Answer 29

\[or we can say a \le 0,b \ge 14 .\]