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Math Help - A few equstions, i mean confusions...

  1. #1
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    A few equstions, i mean confusions...

    ex - 2y = 3z x=?

    ay + z = am - ny y=?


    a(y + 1) = b y=?

    (3/5)y + a = b y=?


    What I'm trying to figure out is well... How to do these equations, I am no less than stuck in doing any equations..

    So, would anyone please explain to me how to get started on these equations?
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Blakester View Post
    ex - 2y = 3z x=?

    ay + z = am - ny y=?


    a(y + 1) = b y=?

    (3/5)y + a = b y=?


    What I'm trying to figure out is well... How to do these equations, I am no less than stuck in doing any equations..

    So, would anyone please explain to me how to get started on these equations?
    These are called literal equations, since you are trying to solve for a particular variable when you have more than one variable in the equation.

    ex - 2y = 3z, solve for x

    First, add 2y to both sides:

    ex=3z+2y

    Finally, divide by e in order to isolate the variable x:

    x=\frac{3z+2y}{e}

    Now you try the next one, and let's see how you do.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Blakester View Post
    ay + z = am - ny Solve for           y

    So, would anyone please explain to me how to get started on these equations?
    Well, that second one is a little bit different, so I'll get you started and see if you can finish.

    First, get the y terms together. Add ny to both sides of the equation:

    ay+ny+z=am

    Now subtract  z from both sides.

    ay+ny=am-z

    Factor y from the left side and that leaves you with one more step.
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  4. #4
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    (a + n) = (am - z)

    To?

    y = (am - z)/(a + n)?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Blakester View Post


    (a + n){\color{red}y} = (am - z)

    To?

    y = (am - z)/(a + n)?
    You forgot the y...

    You should be able to do it now.

    --Chris
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by Blakester View Post
    a(y + 1) = b Solve for y


    So, would anyone please explain to me how to get started on these equations?
    Very good!

    Now, for the third one. This time you will have to expand the left side in order to free up the variable y.

    ay+a=b

    I think you can finish from here.

    And your last one:

    \frac{3y}{5} + a = b Solve for y

    First, multiply each term by 5 to eliminate the fraction.

    3y+5a=5b

    You can probably handle this one as well. Let us know.
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