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    sooolost's Avatar
    sooolost Posts: 1, Reputation: 1
    New Member
     
    #1

    Feb 26, 2007, 10:59 PM
    breakdown of solving these Algebra equations
    maybe this isn't the right site to go to, but I am in such desperate need of help solving some Algebra equations:

    S=L + 2B; B

    also:

    finding for y

    2y + 4x = 8
    Thanks
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #2

    Feb 27, 2007, 07:35 AM
    are you confused about anything specific? Show us what you have so far and we can help you along. I'll work a similar example problem and hopefully that will help you along.

    solve the following equation for y:
    14 - 6y = -2x
    first we need to get only terms that contain y on one side so lets remove the 14
    to remove the 14 we need to cancel it out in some way
    to do that we will subtract 14 since (14 - 14) = 0
    but we always need to do the same operation to the left and right hand sides of the equation so that they will remain equal so we need to subtract 14 from the right also
    14 - 6y - 14 = -2x -14
    so the 14 and -14 on the left will cancel and equal 0 so we can remove them
    -6y = -2x - 14
    now we have to get the y by itself. Currently we have -6 * y. to cancel the -6 we can divide by -6 but we must remember to do the same to the left and right sides
    -6y/-6=(-2x-14)/-6
    the -6/-6 = 1 so the left becomes just y
    y=(-2x-14)/-6
    now we can simplify the right hand side. We can factor a -2 out of the numerator
    y=-2(x+7)/-6
    now we can simplify -2/-6=1/3
    y=(x+7)/3

    Hopefully you were able to follow that example. Try your problems and let us know how you're coming along or if you have specific questions.

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