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ineedbiohelpandquick

solving by elimination 1) 2x-y=3 4x+3y=21 when i solve it i get x=1.2 and y= -.6 but when i check it back in to the equation it is wrong will someone help me!

  • 2 years ago
  • 2 years ago

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  1. jim_thompson5910
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    2x-y=3 4x+3y=21 3*(2x-y)=3*3 4x+3y=21 6x-3y=9 4x+3y=21 ----------- 10x = 30 10x = 30 x = 30/10 x = 3 Now that we know that x = 3, we can use it to find y 2x-y = 3 2(3)-y = 3 6 - y = 3 -y = 3-6 -y = -3 y = 3 So x = 3 and y = 3

    • 2 years ago
  2. ineedbiohelpandquick
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    thank you so much! would you be willing to explain this one? x-y=2 -2x+2y=5

    • 2 years ago
  3. jim_thompson5910
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    x-y=2 -2x+2y=5 2(x-y)=2*2 -2x+2y=5 2x-2y=4 -2x+2y=5 ----------- 0x+0y = 9 0 = 9 ... which is FALSE So there are no solutions. This means that the system is inconsistent.

    • 2 years ago
  4. ineedbiohelpandquick
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    okay the zero threw me off when i had done it! do you know how to solve by using back substitution by chance?

    • 2 years ago
  5. jim_thompson5910
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    Solving by substitution: x-y=2 x = 2 + y ------------------------------------------ -2x+2y = 5 -2(2+y) + 2y = 5 -4 - 2y + 2y = 5 -4 = 5 ... which is FALSE So again, the system is inconsistent and there are no solutions.

    • 2 years ago
  6. ineedbiohelpandquick
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    here is one... 2x-y+5z=24 y+2z=4 z=6

    • 2 years ago
  7. ineedbiohelpandquick
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    for using back-substitution

    • 2 years ago
  8. jim_thompson5910
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    y+2z=4 y+2(6)=4 y + 12 = 4 y = 4-12 y = -8 ------------------------------------------------------- 2x-y+5z = 24 2x - (-8) + 5(6) = 24 2x + 8 + 30 = 24 2x + 38 = 24 2x = 24-38 2x = -14 x = -14/2 x = -7 So the solutions are x = -7, y = -8 and z = 6

    • 2 years ago
  9. ineedbiohelpandquick
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    you are incredible thank you!

    • 2 years ago
  10. jim_thompson5910
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    You're welcome

    • 2 years ago
  11. ineedbiohelpandquick
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    hey are you still on here? can you help me with solve using row-operations... x+y+z=6 2x-y+z=3 3x -z=0

    • 2 years ago
  12. jim_thompson5910
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    x+y+z=6 2x-y+z=3 3x -z=0 1 1 1 6 2 -1 1 3 3 0 -1 0 1 1 1 6 2 -1 1 3 0 -3 -4 -18 R3 - 3R1 1 1 1 6 0 -3 -1 -9 R2 - 2R1 0 -3 -4 -18 1 1 1 6 0 1 1/3 3 (-1/3)*R2 0 -3 -4 -18 1 1 1 6 0 1 1/3 3 0 0 -3 -9 R3 + 3R2 1 1 1 6 0 1 1/3 3 0 0 1 3 (-1/3)*R3 x+y+z = 6 y+(1/3)z = 3 z = 3 ============================================= ============================================= y+(1/3)z = 3 y+(1/3)(3) = 3 y + 1 = 3 y = 3-1 y = 2 ---------------- x+y+z = 6 x+2+3 = 6 x + 5 = 6 x = 1 So the solutions are x = 1, y = 2 and z = 3

    • 2 years ago
  13. ineedbiohelpandquick
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    May i ask you what the R stands for

    • 2 years ago
  14. jim_thompson5910
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    R stands for the row So R1 means "row 1", R2 means "row 2", and R3 means "row 3"

    • 2 years ago
  15. ineedbiohelpandquick
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    okay so how did you determine subtract row 3- 3 by row1

    • 2 years ago
  16. jim_thompson5910
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    When going from 1 1 1 6 2 -1 1 3 3 0 -1 0 to 1 1 1 6 2 -1 1 3 0 -3 -4 -18 R3 - 3R1 I did this to make that lower left corner value go from 3 to 0 Notice that 3 - 3*1 = 3-3 = 0 This basically "eliminates" the x variable in equation 3 which gets me one step closer to solving for x, y, and z

    • 2 years ago
  17. ineedbiohelpandquick
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    okay can you tell me what you get for this one then... 6y+4z=-18 3x+3y = 9 2x -z =12

    • 2 years ago
  18. jim_thompson5910
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    using the same method of row reduction or some other method?

    • 2 years ago
  19. ineedbiohelpandquick
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    row reduction :)

    • 2 years ago
  20. jim_thompson5910
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    alright thanks, one sec

    • 2 years ago
  21. jim_thompson5910
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    6y+4z=-18 3x+3y = 9 2x -z =12 0 6 4 -18 3 3 0 9 2 0 -1 12 0 6 4 -18 1 1 0 3 (1/3)*R2 2 0 -1 12 1 1 0 3 R1 <--> R2 0 6 4 -18 2 0 -1 12 1 1 0 3 0 6 4 -18 0 -2 -1 6 R3 - 2*R1 1 1 0 3 0 -2 -1 6 R2 <--> R3 0 6 4 -18 1 1 0 3 0 1 1/2 -3 (-1/2)*R2 0 6 4 -18 1 1 0 3 0 1 1/2 -3 0 0 1 0 R3 - 6*R2 x+y = 3 y + (1/2)z = -3 z = 0 ------------------------------------------------------- y + (1/2)z = -3 y + (1/2)(0) = -3 y = -3 -------------------- x+y = 3 x + (-3) = 3 x - 3 = 3 x = 3+3 x = 6 So the solutions are x = 6, y = -3, and z = 0

    • 2 years ago
  22. ineedbiohelpandquick
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    THANK YOU! You don't know how much you have helped me! there is only one more I can't seem to get right do you have time for one more? :)

    • 2 years ago
  23. jim_thompson5910
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    sure whats the problem

    • 2 years ago
  24. ineedbiohelpandquick
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    same thing so solving by row operations :)... its x-2y+5z=2 4x -z=0

    • 2 years ago
  25. jim_thompson5910
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    x-2y+5z=2 4x -z=0 1 -2 5 2 4 0 -1 0 1 -2 5 2 0 8 -21 -8 R2 - 4*R1 x-2y+5z = 2 8y-21z=-8 ------------------------------------------------------- 8y-21z=-8 8y = -8 + 21z y = (21z - 8)/8 ----------------------- x-2y+5z = 2 x-2( (21z - 8)/8 )+5z = 2 x - 4(21z - 8) + 5z = 2 x - 84z + 32 + 5z = 2 x - 79z + 32 = 2 x - 79z = 2-32 x - 79z = -30 x = 79z - 30 Now if you let z = s, where s is any number, then the solutions are x = 79s - 30, y = (21s - 8)/8, z = s This means that there are an infinite number of solutions. So the system is consistent and dependent.

    • 2 years ago
  26. jim_thompson5910
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    A very quick way to determine if you'll have a dependent system is noticing that there are more variables than equations.

    • 2 years ago
  27. ineedbiohelpandquick
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    so the answer is consistent and dependent or x = 79z - 30 sorry I am a little confused haha

    • 2 years ago
  28. jim_thompson5910
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    Well both aspects make up the total answer. The idea here is that there are an infinite number of points where the two equations x-2y+5z=2 and 4x -z=0 intersect So this makes the system dependent (since one equation technically "depends" on the other -- it changes as the other one does) Because at least one solution exists, this makes the system consistent. The answers of x = 79s-30, etc... are the more explicit forms of the solutions and are more specific than saying "there are an infinite number of solutions" This is because you can't just say that something random like (1,2,3) is a solution even though I did say "there are an infinite number of solutions". All of the solutions fit a very specific algebraic form.

    • 2 years ago
  29. ineedbiohelpandquick
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    That makes sense! Thank you again! I appreciate all your help!

    • 2 years ago
  30. jim_thompson5910
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    You're welcome, glad it's all clicking.

    • 2 years ago
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