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  1. #1

    Default School

    So umm just throwing this out there but... Anyone good at algebra 1 willing to help me with a prob?

  2. #2

  3. #3


    Sweet thanks!!

    Explain in words how to graph the line 3x - 5y = 10 to someone who has never taken an Algebra class. Use correct vocabulary as you explain the process and what the solution (the graph) tells you about the equation of the line.

  4. #4


    Ok, here goes nothing!

    Since you're solving for y, move the 3x to the right side. The 5y is negative (3x - 5y) so it becomes

    -5y = -3x + 10

    Divide by -5 to get y by itself.

    y = -3x/-5 + 10/-5

    Work it out. -3 and -5 cancel to 3/5. Leave it as a fraction, since they're most precise. 10/-5 becomes -2.

    y = (3/5)x - 2

    To find the y-intercept, or where the line meets y, take the number that's added or subtracted from the x - in this case, it's -2.

    The slope of the line is whatever the x is multiplied by. In this case it's 3/5, or up three units, right five units.
    Last edited by Point; 12-Feb-2008 at 01:53. Reason: Thanks Lukie

  5. #5

  6. #6


    Hopefully you don't just take that and write it down - study hard. I'm bad at math but if I study enough I do just fine.

  7. #7


    Yup... the way Point has done it is generally the most effective way to solve these sorts of problems.

    However I got y = (3/5)x - 2 ... minus 2!

    It always helps to use a method where you deal with as little negatives as possible!

    Because you got 3x - 5y = 10
    3x = 10 + 5y (bring the 5y over, get rid of all the negatives)
    3x - 10 = 5y (move the 10 to the other side - negative is unavoidable - to isolate the 5y)
    (3/5)x - 2 = y (divide through by 5 to get the function of y.)

    I think you just put in the wrong sign Point.

    Simply make the equation as a function of y (That is, have it as y = ...)

    As a side note, it always helps to know general formula's too.

    You (should :p) know the general equation of a straight line is y = mx + c

    Where: m = gradient of the line
    x = your variable
    c = a constant, and also where the line crosses the y-axis.

  8. #8

  9. #9


    I like your cookie... That's all I have to say

  10. #10


    Quote Originally Posted by Point Blanch View Post
    Oh, crap! I guess I still need practice too!
    I always tell my students when they turned negative numbers into positive ones: "If you ever open a bank, please let me know immediately - I want to be your first customer!" :-D


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