## iiamentertainment one year ago What is the solution of the system of equations? { 2x + 2y + 3z = -6 3x + 5y + 4z = 3 2x + 3y + 4z = - 10} Three equations

1. iiamentertainment

So 2x + 2y + 3z = -6 2y + 3z = -2x - 6 2y = -2x -3z - 6 y = -2x -3z - 6/2 y = -x -3/2z - 3 ?

2. iiamentertainment

@CalebBeavers

3. iiamentertainment

4. abb0t

What class is this for?

5. iiamentertainment

Algebra 2

6. abb0t

Well, you can use 1 of 2 methods. Substitution, or Elimination. Are you familiar with either of them?

7. iiamentertainment

im familar with substitution

8. abb0t

Ah, I almost want to suggest using gauss-elimination, but that's beyond algebra 2. Well, I think it might be easier to use elimination actually to eliminate one equation first and then a second to get 2-variable equations. Does that make sense?

9. abb0t

Hint, start by eliminating the x's since equations (1) and (3) both begin with 2. Multiply the first (or second) by -1.

10. iiamentertainment

Um im not that good with elimination , butf you walk me through and let me solve it i think ill be ok

11. abb0t

Yes. I will walk you through it, mate. Start by eliminating one variable in the system of equations. As I said before, It might be easier to eliminate the x from equations (1) and (3) since they both begin with the number 2 [NOTE: Do NOT do anything to eq (2), yet]. You can do this by multiplying either (1) or (3) by -1. Try it and let me know which one you choose and what ur answer is.

12. iiamentertainment

-1(2x + 2y + 3z)= -6 -2x - 2y - 3z = -6 ?

13. abb0t

You're 90% correct, when you multiply (-1) to the eq, you also have to multiply it to the -6 and you know (-1)(-6) = 6. making eq(1) -2x-2y-3z = 6 I'm going to use the term: eq(#) just so I type less :P But notice that eq(1) and eq(2) can be added. And the x' term cancels out nicely. So that you get a new equation with only y's and z's. When you add everything! Including the 6 and -10. Try it.

14. iiamentertainment

2x + 3y + 4z = - 10 -2x - 2y - 3z = 6 y + 1 = -4 y = -5 ?

15. abb0t

i have to step out for a moment, but I will tell you the next few steps. Next, do the same thing to eliminate x! from eq(2)! To get a second equation with only y's and z's! Notice that now you have TWO system of equations which you can use elimination (again) or substitution (which you said you like) to solve for either y or z! Once you have your y or z value, plug it back into the original 2 system equation to solve for either y or z (which ever variable you chose to solve for). Once you solved for the 2nd variable you should have TWO variables: y AND z! Now plug in everything into ANY of the equations of the 3-systems and solve for x! You can check by plugging everything in and you should get either -6, 3, or -10! If you are correct, you should get those answers!

16. iiamentertainment

@blondie16 can you help , im confused

17. LogicalApple

You are certainly on the right track. You seem to have forgotten the 'z' variable when adding "4z" and "-3z". This gives you a z. So your equation becomes: y + z = -4

18. iiamentertainment

Ok thats what i did wrong ok cool, then what

19. LogicalApple

What @abb0t suggested next was to use elimination once more. This time with eq(1) and eq(2). The focus here is to, again, eliminate the 'x' variable. equation 1: 2x + 2y + 3z = -6 equation 2: 3x + 5y + 4z = 3 If you multiply eq(1) by -3/2 you will be able to then add it to eq(2) to cancel out the 'x' again. What do you obtain when you multiply eq(1) through by -3/2 ?

20. iiamentertainment

-3/2(2x + 2y + 3z) = -6 -3x - 3y - 4.5z = 9 3x + 5y + 4z = 3 2y -.5z = 6 ?

21. iiamentertainment

@LogicalApple

22. LogicalApple

The 9 + 3 should be a 12 so 2y - 0.5z = 12

23. iiamentertainment

24. LogicalApple

-3x + 3x = 0 -3y + 5y = 2y -4.5z + 4z = -0.5z 9 + 3 = 12 You added everything else correctly but the last numbers

25. iiamentertainment

Oh Ok , Next Step?

26. LogicalApple

In any event you are now down to a system of 2 equations with 2 variables y + z = -4 2y - 0.5z = 12 You can use substitution at this point.

27. LogicalApple

Notice how the elimination method eliminates one of the variables, leaving you with only two to deal with. From here you could set y = -4 - z in the first equation and substitute this value of y into the second equation and solve for z

28. iiamentertainment

2x + 2(-4 -z) + 3z = -6 like that

29. LogicalApple

No, I meant the second equation listed here: y + z = -4 2y - 0.5z = 12 Not any of the original equations

30. iiamentertainment

2(-4 -z) - 0.5z = 12 -8 - 2z - .5z = 12 -8 + 3z = 12 3z = 20 z = 6.67

31. LogicalApple

-2z - 0.5z = -2.5z So you should have -8 - 2.5z = 12

32. iiamentertainment

-8 - 2.5z = 12 -2.5z = 20 z = -8

33. LogicalApple

Yes, that's it!

34. LogicalApple

And we know: y + z = -4

35. iiamentertainment

y + (-8) = -4 y = 4

36. iiamentertainment

@LogicalApple

37. LogicalApple

Yes, that is also correct. Now you know what y and z are. Use these values in any one of the original three equations and you will obtain the value for x.

38. iiamentertainment

2x + 2(4) + 3(-8) = -6 2x + 8 - 24 = -6 2x - 16 = -6 2x = 10 x = 5 y = 4 z = -8 @LogicalApple

39. LogicalApple

Fantastic. (5, 4, -8) is the correct solution

40. abb0t

Correct :)

41. biasophia

solve the system of equations by elimination 3x+y+2z=3 2x-3y-z=3 x+2y+z=4