## jdjrferg4 2 years ago Find a and b. If |x-3|<7, then a<x+4<b ?

1. surjithayer

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

2. jdjrferg4

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?

Are we doing piece wise functions here?

4. jdjrferg4

I do not understand? Piece wise functions?

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

6. jdjrferg4

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

This is not college algebra.

This is Algebra 2.

9. jdjrferg4

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

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.

11. jdjrferg4

Thank you

12. jdoe0001

$$\bf |x-3|<7 \implies \begin{cases} +(x-3)<7\\ \quad \\ \bf -(x-3)<7 \end{cases}$$ if you were to solve those 2 cases, or scenarios, what would be the "values" of "x"?

13. jdjrferg4

I am so lost

14. jdoe0001

hmm ok, what part?

15. jdjrferg4

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

16. jdjrferg4

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

17. jdoe0001

18. jdoe0001

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

19. jdoe0001

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

20. jdoe0001

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

21. jdoe0001

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

22. jdjrferg4

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 ?

23. jdoe0001

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

24. jdjrferg4

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

25. jdoe0001

well, they want $$\large x \color{red}{+}4$$ so you'd need to add, yes

26. jdjrferg4

-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?

27. jdoe0001

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

28. surjithayer

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

29. surjithayer

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