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help_people

  • one year ago

Create a system of equations that includes one linear equation and one quadratic equation. Part 1. Show all work to solving your system of equations algebraically. Part 2. Graph your system of equations, and show the solution graphically to verify your solution.

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  1. help_people
    • one year ago
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    @midhun.madhu1987

  2. welshfella
    • one year ago
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    try using y = 2x + 1 y = x^2 - x - 6

  3. midhun.madhu1987
    • one year ago
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    General form of a linear equation is: ax + by + c = 0 General form of a Quadratic equation is: a(x^2) + bx + c = 0

  4. help_people
    • one year ago
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    ok basically like @welshfella said

  5. help_people
    • one year ago
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    how would i solve algebraically though?

  6. midhun.madhu1987
    • one year ago
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    Equate both equtions: 2x + 1 = x^2 - x - 6 Now arrange it...by combining like terms...

  7. help_people
    • one year ago
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    I'm not sure/// comine 2x-x?

  8. help_people
    • one year ago
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    how can you combine you put the stuff on the other side of the equal sign

  9. midhun.madhu1987
    • one year ago
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    x^2 - x- 2x - 6-1 = 0 x^2 - 3x - 7 = 0

  10. midhun.madhu1987
    • one year ago
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    Now you can solve this quadratic equation using QUadratic Formula...

  11. help_people
    • one year ago
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    wait can you tell me what we have doen so far and wha i should put ?

  12. midhun.madhu1987
    • one year ago
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    Solve this system of equations algebraically: y = x2 - x - 6 (quadratic equation in one variable of form y = ax2 + bx + c ) y = 2x - 2 (linear equation of form y = mx + b)

  13. midhun.madhu1987
    • one year ago
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    Substitute from the linear equation into the quadratic equation and solve. y = x2 - x - 6 2x - 2 = x2 - x - 6 2x = x2 - x - 4 0 = x2 - 3x - 4 0 =(x - 4)(x + 1) x - 4 = 0 OR x + 1 =0 x = 4 x = -1

  14. midhun.madhu1987
    • one year ago
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    Find the y-values by substituting each value of x into the linear equation. y = 2(4) - 2 = 6 POINT (4,6) y = 2(-1) - 2 = -4 POINT (-1,-4)

  15. help_people
    • one year ago
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    oh so that's part 1?

  16. midhun.madhu1987
    • one year ago
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    Yes...

  17. help_people
    • one year ago
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    that makes sense srry i was a little confused

  18. help_people
    • one year ago
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    what would i do after>

  19. midhun.madhu1987
    • one year ago
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    Now you can plot the points graphically and check...

  20. help_people
    • one year ago
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    ok :) ty!

  21. help_people
    • one year ago
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    wait @midhun.madhu1987 what do they mean how the solution graphically ?

  22. welshfella
    • one year ago
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    https://www.desmos.com/screenshot/rljjhtzr7b to graph you plot points for a range of x and y the above link shows the graphs where they intersect are the solutions

  23. help_people
    • one year ago
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    oh so thats what i should use ?

  24. welshfella
    • one year ago
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    as you see they intersect at (4,6) and (-1,-4) which agrees with the values you got earlier. I'm not sure Desmos Graphing is a very good piece of software but i dont know if your teacher will allow it.

  25. welshfella
    • one year ago
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    He/she might only accept a drawing on graph paper.

  26. help_people
    • one year ago
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    yes he will i always use it :)

  27. welshfella
    • one year ago
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    oh ok

  28. help_people
    • one year ago
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    yep could you help with another one ?

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