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help_people

  • one year ago

Hi this question I am not sure about, I am debating so may you please help me confirm the correct answer (Algebra)

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  1. help_people
    • one year ago
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    The table below shows four systems of equations: System 1 System 2 System 3 System 4 4x − 5y = 2 3x − y = 8 4x − 5y = 2 10x − 7y = 18 4x − 5y = 2 3x − 8y = 4 4x − 5y = 2 10x + 3y = 15 Which pair of systems will have the same solution? System 1 and system 2, because the second equation in system 2 is obtained by adding the first equation in system 1 to two times the second equation in system 1 System 2 and system 3, because the second equation in system 3 is obtained by adding the first equation in system 2 to two times the second equation in system 2 System 1 and system 2, because the second equation in system 2 is obtained by adding the first equation in system 1 to three times the second equation in system 1 System 2 and system 3, because the second equation in system 3 is obtained by adding the first equation in system 2 to three times the second equation in system 2

  2. help_people
    • one year ago
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    |dw:1433801425368:dw|

  3. help_people
    • one year ago
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    |dw:1433801513354:dw|

  4. help_people
    • one year ago
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    @Nnesha

  5. anonymous
    • one year ago
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    15

  6. help_people
    • one year ago
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    ? @0404diego

  7. anonymous
    • one year ago
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    hi

  8. help_people
    • one year ago
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    @mathstudent55

  9. help_people
    • one year ago
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    hi 04

  10. help_people
    • one year ago
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    i am not sure may you walk me through the process?

  11. help_people
    • one year ago
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    @perl

  12. mathstudent55
    • one year ago
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    Every system has the same equation: 4x - 5y = 2

  13. mathstudent55
    • one year ago
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    We need to look at the second equation of the 4 systems and compare them.

  14. mathstudent55
    • one year ago
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    4x − 5y = 2 3x − y = 8 4x − 5y = 2 10x − 7y = 18 4x − 5y = 2 3x − 8y = 4 4x − 5y = 2 10x + 3y = 15

  15. help_people
    • one year ago
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    well that have 10 x and 3x?

  16. help_people
    • one year ago
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    but all different ys?

  17. mathstudent55
    • one year ago
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    Look at choice A. Choice A tells you that the second equation of the second system is 2 times the second equation of the first system added to the first equation of the first system. If this is true, the equations are the same equation and the solutions will be the same. All you need to do is see if this is true. Multiply the second eq. of system 1 by 2. Then add it to the first eq. of system 1. What do you get?

  18. help_people
    • one year ago
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    who do i do that

  19. help_people
    • one year ago
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    like 12-6y=16

  20. mathstudent55
    • one year ago
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    Here is eq. 2 of system 1: 3x − y = 8 Multiply both sides of the equation by 2. What do you get?

  21. mathstudent55
    • one year ago
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    Multiply both sides by 2: 2(3x − y) = 2(8) What do you get?

  22. help_people
    • one year ago
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    6x-2y=16

  23. mathstudent55
    • one year ago
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    Great. Now add it to the first eq. of system 1. 4x - 5y = 2 (+) 6x - 2y =16 ----------------- What do you get?

  24. help_people
    • one year ago
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    10x+7y=18 @mathstudent55

  25. mathstudent55
    • one year ago
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    Check the +7y. Notice you are adding -5y and -2y. What is -5y + (-2y) = ?

  26. help_people
    • one year ago
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    -7y

  27. mathstudent55
    • one year ago
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    -5y + (-2y) = -5y - 2y = -7y Correct.

  28. mathstudent55
    • one year ago
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    The equation is: 10x - 7y = 18 Now compare this new equation with the second eq. of system, 2.

  29. help_people
    • one year ago
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    ok so what would i do? can you walk me through with it? @mathstudent55

  30. mathstudent55
    • one year ago
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    The equation you got by doing what choice A tells you is exactly the second eq. of system 2. That means the answer is choice A. The first and second systems have the same answer.

  31. help_people
    • one year ago
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    thank you so much may you help me w. more (its about squre roots)

  32. mathstudent55
    • one year ago
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    Sorry, but I have to go.

  33. mathstudent55
    • one year ago
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    You're welcome.

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