1. anonymous

I know what the answers are but I cannot seem to get there =(

2. jim_thompson5910

Let's focus on #24 $\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2z = -3\\ z = 2x - 4y\\ \end{cases}$

3. jim_thompson5910

Notice we have a pair of 'z's here $\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2{\LARGE \color{red}{z}} = -3\\ {\LARGE \color{red}{z}} = 2x - 4y\\ \end{cases}$

4. jim_thompson5910

what we can do is replace the 'z' in the second equation with '2x-4y' since z = 2x-4y in the third equation we can then drop the third equation after substitution $\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{z} = -3\\ \color{red}{z} = 2x - 4y\end{cases}$ $\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{(2x-4y)} = -3\end{cases}$

5. anonymous

For 24, which I think I maybe got right, I ended up with $y=3x-13, 4y-3x+2z=-3, z=2x-4y$ and then substituted to end up with $4(3x-13)-3x+2(2x-4(3x-13))=-3$ which simplified to $x=5, y=2, z=2$

6. anonymous

Is that correct (I have all the work but it's so long so I didn't post it sorry)

7. jim_thompson5910

You did the right steps. Nice job

8. anonymous

Awesome, thanks! Um do you think you could help with 25,26,27,28, or 29?

9. jim_thompson5910

Those are definitely harder because one variable isn't already isolated. But we can isolate a variable, say z if we pick on the third equation and solve for z, we get 3x - 2y - z = -9 3x - 2y - z+z = -9+z 3x - 2y = -9+z 3x - 2y + 9 = -9+z+9 3x - 2y + 9 = z z = 3x - 2y + 9

10. jim_thompson5910

agreed so far?

11. anonymous

For which question?

12. jim_thompson5910

#25

13. anonymous

Okay =) Yeah that looks good

14. jim_thompson5910

once we have z isolated, we can replace every copy of z in the first two equations with (3x - 2y + 9)

15. anonymous

Okay, did that =)

16. jim_thompson5910

first equation: x+3y-z = -4 will turn into x+3y-(3x - 2y + 9) = -4 second equation 2x-y+2z=13 will turn into 2x-y+2(3x - 2y + 9) =13

17. jim_thompson5910

at this point, the 'z' has gone away leaving you with a system of 2 equations with 2 unknowns

18. anonymous

That makes sense! I think I tried to overcomplicate it last time lol. So from here, you would just isolate a variable in one of them and then plug whatever that is into the other equation? Or would you solve both for a variable and then solve it like a regular system?

19. jim_thompson5910

you can use a number of methods to solve this new system 1) substitution 2) elimination 3) matrices 4) graphing graphing is probably the fastest way, but it doesn't always guarantee you get the exact answers (the point may have fractional coordinates)

20. anonymous

I am most comfortable with substitution =) So I solve both for a variable (let's say y, so y=something) and then just substitute (if y=x and y=2 then 2=x)?

21. jim_thompson5910

you are correct

22. jim_thompson5910

let me know what you get

23. anonymous

Awesome thanks! Alright I'll do that real quick...

24. anonymous

I got y=2x-13 and y=(-5/3)

25. jim_thompson5910

x,y, & z are all whole numbers for #25

26. anonymous

True. And the answer for this is x=0, y=1, z=7... let's see what I did wrong.

27. anonymous

Ah. I combined x and y. I'll redo it

28. jim_thompson5910

how did you get y=2x-13? which equation did you solve for y?

29. jim_thompson5910

alright

30. anonymous

I solved the first one to get that.

31. anonymous

|dw:1442627727801:dw| Here's what I did to solve the first eq.

32. jim_thompson5910

You made a mistake in distributing x+3y-(3x - 2y + 9) = -4 turns into x+3y-3x + 2y - 9 = -4

33. jim_thompson5910

notice the + 2y and  - 9

34. anonymous

OH! I was wondering about that. So the negative sign gets distributed!

35. jim_thompson5910

yeah think of -(3x - 2y + 9) as -1*(3x - 2y + 9)

36. jim_thompson5910

and that -1 multiplies with each term inside

37. anonymous

All right =) Here, I'll try that again

38. anonymous

I got y=(2/5x)+1

39. anonymous

and y=(8/5x)+1 for the other which when set equal to eachother make x=0 which is correct

40. jim_thompson5910

I'm getting the same equations when I solve for y

41. anonymous

Yay! When I plugged x=0 into the other eq.'s, it worked! Thank you so much!

42. jim_thompson5910

no problem

43. anonymous

I think I should be able to do the rest now. Thanks again =)

44. jim_thompson5910

sure thing