## anonymous one year ago solve the equation. check for extraneous solutions. will fan and medal

1. anonymous

$9\left| 9-8x \right|=2x+3$

2. SolomonZelman

well, you will get two results: $$9\left| 9-8x \right|=2x+3$$ $$9(9-8x )=2x+3$$ OR $$9(9-8x )=-2x-3$$

3. SolomonZelman

solve each case individually.

4. anonymous

i dont know where to start. will you walk me through it?

5. SolomonZelman

well, expand the right side in case 1

6. SolomonZelman

I mean the left side (not the right side)

7. anonymous

would you move the x to the left side first?

8. SolomonZelman

$$9(9-8x )=2x+3$$ $$9\cdot9-9\cdot8x =2x+3$$ $$81-72x =2x+3$$

9. SolomonZelman

So in case two you would have the same, but the right side would be -2x-3

10. SolomonZelman

Solve for each: $$81-72x =2x+3$$ $$81-72x =-2x-3$$

11. anonymous

x would go on the left side?

12. SolomonZelman

you can add 72x to both sides in case 2 and case 1

13. anonymous

leaving us with 81=74x+3

14. SolomonZelman

two cases: $$81-72x =2x+3$$ $$81-72x+72x =2x+72x+3$$ $$81=74x+3$$ right!

15. SolomonZelman

then subtract 3 from both sides.

16. anonymous

78=74x?

17. SolomonZelman

yes, so x=?

18. anonymous

74/78?

19. SolomonZelman

yes

20. anonymous

what about in the second case?

21. SolomonZelman

Now in case 2: $$81-72x =-2x-3$$

22. SolomonZelman

you also add 72x to both sides, THEN, add 3 to both sides

23. anonymous

x=70/84?

24. SolomonZelman

oh the first one is incorrect, and this one too

25. anonymous

noooo

26. SolomonZelman

when in case one you have: 78=74x you divide by 74 on both sides, and thus x=78/74 (not 74/78)

27. SolomonZelman

And in case two you made the same mistake

28. anonymous

so it would be 84/70?

29. SolomonZelman

Yes, for case 2.

30. SolomonZelman

And they are not extraneous.

31. SolomonZelman

let me say correctly what would have been extraneous:

32. SolomonZelman

Any x-solution, in this case, that is equal to $$C$$ that satisfies: $$C<-2/3$$

33. SolomonZelman

because if x is less than -2/3 then 2x+3 is negative, and absolute value can NOT e equal to negative (since absolute value is really a distance)