anonymous one year ago Solve and check for extraneous solutions 16=3(x-1)-(x-7)

1. anonymous

@satellite73

2. anonymous

not sure what "extraneous" solutions means in this context it is a linear equation right? no radials, no denominators

3. anonymous

distribute first on the right both the 3 and the minus sign

4. anonymous

let me know what you get

5. anonymous

Ok, so I got 2x+6=16

6. anonymous

think there is a mistake there the $$2x$$ is right, but where did the $$6$$ come from?

7. anonymous

-1+7

8. anonymous

$3(x-1)-(x-7)=3x-3-x+7$ is a start

9. anonymous

oops forgot the distributive law didn't you?

10. anonymous

$3(x-1)=3x-3\times 1$

11. anonymous

Oh okay! So 2x+4?

12. anonymous

right $2x+4=16$ takes only 2 steps to solve

13. anonymous

a) subtract 4 b) divide by 2

14. anonymous

2x=10 x=2

15. anonymous

???

16. anonymous

first off $$16-4\neq 10$$ and secondly $$\frac{10}{2}\neq 2$$ so both steps are wrong

17. anonymous

what is $$16-4$$?

18. anonymous

16-4=12

19. anonymous

ok so $2x+4=16\\ 2x=12$ is a the first step

20. anonymous

x=6! So sorry, it's late where I am lol so I have a sleepy brain I suppose

21. anonymous

second step is $$x=\frac{12}{2}$$

22. anonymous

x=6

23. anonymous

right

24. anonymous

How do we check for extraneous solutions?

25. anonymous

there are no extraneous solutions with linear equations, nothing to check

26. anonymous

you can replace x by 6 in the original equation and see that it works if you like but it is right, nothing is extraneous here

27. anonymous

ok:) Can I ask another? I'll be better haha

28. anonymous

sure

29. anonymous

x/3=x/2-2 Check for extraneous solutions

30. anonymous

is this $\frac{x}{3}=\frac{x}{2}-2$ like that ?

31. anonymous

yup

32. anonymous

you have a choice you can work with the fractions or you can get rid of them

33. anonymous

you pick

34. anonymous

get rid of them.

35. anonymous

Actually, which ever is simpler

36. anonymous

then since the least common multiple of 2 and 3 is 6, if you multiply both sides by 6 they will be gone most people i think like getting rid of fractions

37. anonymous

$6\times \frac{x}{3}=6\times (\frac{x}{2}-2)$

38. anonymous

I think I did it wrong bc I got 2x/6-3x/6

39. anonymous

since $$\frac{6}{2}=3$$ and $$\frac{x}{3}=2$$ you end up with $2x=3x-18$

40. anonymous

you forgot to get rid of the 6 when you cancelled

41. anonymous

oh okay:)

42. anonymous

$6\times \frac{x}{3}=2x$ not $$\frac{2x}{6}$$

43. anonymous

so now you have $2x=3x-18$ which takes 2 steps to solve you see where the $$-18$$ came from?

44. anonymous

Yes!

45. anonymous

x=18

46. anonymous

yes

47. anonymous

So it wouldnt be extraneous?

48. anonymous

no it would not