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anonymous

  • 5 years ago

someone please help me solve this step by step 3x+y=15 4x+5y=9

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  1. anonymous
    • 5 years ago
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    By graphing? or algebraically?

  2. anonymous
    • 5 years ago
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    by graphing i tried x=0 solve for y but its not right

  3. anonymous
    • 5 years ago
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    You have to graph each line separately

  4. anonymous
    • 5 years ago
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    the first equation i came up with (4,3) and did the x=0 solve y for the second

  5. anonymous
    • 5 years ago
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    For the first equation: \[x =0 \implies 3(0) + y = 15 \implies y = 15\] \[y =0 \implies 3x + 0 = 15 \implies 3x = 15 \implies x = 5\] This gives the pionts (0,15) and (5,0) So now you can graph the line that corrisponds to the equation 3x+y=15 by drawing the line between those two points.

  6. anonymous
    • 5 years ago
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    Now do the same for the second equation. and see where the two lines cross.

  7. anonymous
    • 5 years ago
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    So what do you get for y when you plug in 0 for x in the second equation?

  8. anonymous
    • 5 years ago
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    1.8

  9. anonymous
    • 5 years ago
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    Just leave it as a fraction for now. It's easier to see =)

  10. anonymous
    • 5 years ago
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    9/5 is correct though. Now plug in 0 for y and solve for x.

  11. anonymous
    • 5 years ago
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    2.3

  12. anonymous
    • 5 years ago
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    Not quite. As I said, leave it as a fraction. \[y =0 \implies 4x + 5(0) = 9 \implies 4x = 9 \implies x = 9/4 = 2.25\]

  13. anonymous
    • 5 years ago
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    9/4

  14. anonymous
    • 5 years ago
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    Ok, so what are the two points you have found on this line?

  15. anonymous
    • 5 years ago
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    idk im confused, i hate this stuff

  16. anonymous
    • 5 years ago
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    (0,1.8) (2.3,0)

  17. anonymous
    • 5 years ago
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    Yes, exactly right!

  18. anonymous
    • 5 years ago
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    except you keep rounding 2.25 to 2.3 for some reason.

  19. anonymous
    • 5 years ago
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    thought i was suppose to

  20. anonymous
    • 5 years ago
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    So if you draw a line between those two points it'll give you the line that corresponds to the second equation.

  21. anonymous
    • 5 years ago
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    That's fine

  22. anonymous
    • 5 years ago
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    I think 2 and 1/4 is easier to graph than 2 and 3/10'ths but it depends on your graph paper I suppose.

  23. anonymous
    • 5 years ago
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    Now you're wanting to find where these two lines intersect.

  24. anonymous
    • 5 years ago
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    according to my graph they dont

  25. anonymous
    • 5 years ago
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    They must. They don't have the same slope, so they are not parallel. Any two distinct non-parallel lines must intersect somewhere.

  26. anonymous
    • 5 years ago
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    You will need to continue the line beyond the points that you graphed. It will go straight through those 2 points, but continue on infinitely.

  27. anonymous
    • 5 years ago
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    equation 1 i got (0,15) (5,0) drew the line 2nd i got (0,1.8) (2.25,0)

  28. radar
    • 5 years ago
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    Look real close around the point (6,-3)

  29. anonymous
    • 5 years ago
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    i only drew the lines to the points not thru to the end of the graph

  30. radar
    • 5 years ago
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    You must extend the lines to the point where they intersect. That point is the solution.

  31. anonymous
    • 5 years ago
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    ok so then it is considered consistent and dependent

  32. anonymous
    • 5 years ago
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    cant figure out how to use my new graphing calc

  33. radar
    • 5 years ago
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    The each equation by itself has many solutions all points on the line are solutions. But when you consider the two equations together, the point they intersect is the only point that is a solution to both equations. I am not familiar with term consistent, maybe you can tell me what that term means??

  34. anonymous
    • 5 years ago
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    Consistent means that the equations are either the same line (overlapping) or non-parallel. If they are parallel and do not overlap, they are not consistent because there will be no solution. If they are the same line (each one overlaps the other) then they are dependent.

  35. anonymous
    • 5 years ago
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    This system is consistent (they aren't parallel) and independent ( they aren't the same line)

  36. radar
    • 5 years ago
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    Ah so, so you are right tina=ann

  37. anonymous
    • 5 years ago
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    yeah thats it, and you can just call me tina

  38. radar
    • 5 years ago
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    ok

  39. radar
    • 5 years ago
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    I believe tina uses her calculator and that is why she comes up with decimal answers rather than the exact fraction

  40. radar
    • 5 years ago
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    tina, I have a graphing calculator, bought it new, still in blister pack at a garage sale for $10.00. I haven't learnt how to use it! lol

  41. anonymous
    • 5 years ago
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    yes i use calc, cant figure out how to use my new graphing calc, just spent 150 cant figure out how to use it

  42. anonymous
    • 5 years ago
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    your profile says you graduated from Strayer, that is where i am going

  43. radar
    • 5 years ago
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    I guess you use the book and try and follow their examples, eventually you will learn it.

  44. radar
    • 5 years ago
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    Mine is the TI-86

  45. radar
    • 5 years ago
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    Oh really, in VA or DC

  46. anonymous
    • 5 years ago
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    ok new problem solve by substitution 6x-2y=-56 5x+48=y

  47. anonymous
    • 5 years ago
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    chesapeke Va

  48. anonymous
    • 5 years ago
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    mine is TI-Nspire

  49. radar
    • 5 years ago
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    I went to the Manassas,VA campus, graduated at the Kennedy Center in DC..Ok lets look at the substitution problem. I think we should write the last equation like; y=5x+48 do you agree?

  50. anonymous
    • 5 years ago
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    i think so i got (-40,-192) for the solution

  51. radar
    • 5 years ago
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    Humor me as I wade through this, you are ahead of me! lol I want to subsitute for y in the first equation so it will look like this: 6x -2(5x+48)=-56 6x-10x-96=-56 -4x=40 x=-10 Did I do that right?

  52. anonymous
    • 5 years ago
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    yes you did i must of had a brain fart because i forgot to divide

  53. radar
    • 5 years ago
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    I don't think I violated any laws of algebra! Now going back to the second equation where we had y=5x+48 substitute are newly found value of x in that we get: y=5(-10)+48=-2 y=-2 The solution according to the above : x=-10, y=-2

  54. anonymous
    • 5 years ago
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    and then you add 60+48 to get 108 right?

  55. anonymous
    • 5 years ago
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    oh yeah im way off

  56. anonymous
    • 5 years ago
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    is that 10 or -10

  57. radar
    • 5 years ago
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    It is a -50 added to 48. Please go over what I did and question me on any thing I did.

  58. anonymous
    • 5 years ago
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    im just relearning this and i wasnt ever good at it to begin with

  59. radar
    • 5 years ago
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    You question the solution of -10 for x?

  60. radar
    • 5 years ago
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    I will put it here so you can ask away. I want to subsitute for y in the first equation so it will look like this: 6x -2(5x+48)=-56 6x-10x-96=-56 -4x=40 x=-10

  61. radar
    • 5 years ago
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    It didn't paste so well lol should be a new line after the -56 Note that the (5x+48) was substituted for y inthe first equation. Do you see that?

  62. radar
    • 5 years ago
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    However, in the first equation the coefficient of y was a -2 So you have to multiply (5x+48) by -2 It then becomes -10x-96

  63. anonymous
    • 5 years ago
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    ok i got it i new a new one for you two angles are supplementary one angle is 3 degrees less than twice the other. find the measures of the angles

  64. radar
    • 5 years ago
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    So now the first equation with the substitution of y looks like this: 6x-10x-96=-56 Combining similar terms it becomes: -4x-96=-56 and so on.

  65. radar
    • 5 years ago
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    Two angles are supplementary if their sum is what? Is it 90 or 180 degrees?

  66. anonymous
    • 5 years ago
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    180 complementary is 90

  67. radar
    • 5 years ago
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    Let x = one of the angles this then means that 180-x is the other angle. Does seem sensible?

  68. anonymous
    • 5 years ago
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    i was taught x+y=180 x=2y-3 (2y-3)+y=180 3y-3=180 3y=180 y=61 x=2(61)-3 x=119 is this right?

  69. radar
    • 5 years ago
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    I am looking at your work

  70. radar
    • 5 years ago
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    You did good until you got to 3y-3=180 good so far. Now look at that step after that and find your error

  71. radar
    • 5 years ago
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    Did you find your error in that next step

  72. anonymous
    • 5 years ago
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    i added 3 instead of subtracting?

  73. radar
    • 5 years ago
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    Well you should of added 3 to both sides and then you would of had 3y=183

  74. anonymous
    • 5 years ago
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    i did and 3y=183 3(61)=183

  75. anonymous
    • 5 years ago
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    so y=61

  76. radar
    • 5 years ago
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    I just saw where you had 3y=180 some how you got the right answer lol

  77. radar
    • 5 years ago
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    The other angle is supplementary so subtract the 61 from 180 and it is solved!

  78. anonymous
    • 5 years ago
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    i wrote it down right on my paper i got lots of math problems. i just need someone to help me solve them or keep it on track so i get the right answers

  79. radar
    • 5 years ago
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    Did you say that you have graduated from Strayer or that you are presently going.

  80. anonymous
    • 5 years ago
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    presently going

  81. radar
    • 5 years ago
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    I graduated around 1992. I no longer live in VA or work in DC, I am retired and live in the Ozarks of MO. Way out in the woods! lol

  82. radar
    • 5 years ago
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    Good luck tina

  83. anonymous
    • 5 years ago
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    thank you, im not good at this kind of math

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