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    So I'm slightly confused in the order of steps you have to do them in. Can someone please explain why this doesn't work?

    Here's the question..

    v= u + at Make a the subject

    So why can't I divide first so that I get this answer = v/t -u = a?

    I mean in terms of BIDMAS this would be correct :confused:

    However the answers is v-u/t=a.
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    (Original post by Samistrawberry)
    So I'm slightly confused in the order of steps you have to do them in. Can someone please explain why this doesn't work?

    Here's the question..

    v= u + at Make a the subject

    So why can't I divide first so that I get this answer = v/t -u = a?

    I mean in terms of BIDMAS this would be correct :confused:

    However the answers is v-u/t=a.
    When you divide an equation by t you have to divide every terms
    on both side

    so:

    v=u+at

    dividing by t

    \displaystyle \frac{v}{t}=\frac{u}{t}+a
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    v-u=at

    a=(v-u)/t

    You go backwards in BIDMAS when rearranging formulae.
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    (Original post by Samistrawberry)
    So I'm slightly confused in the order of steps you have to do them in. Can someone please explain why this doesn't work?

    Here's the question..

    v= u + at Make a the subject

    So why can't I divide first so that I get this answer = v/t -u = a?

    I mean in terms of BIDMAS this would be correct :confused:

    However the answers is v-u/t=a.
    You can divide first but you would get \frac{v}{t}=\frac{u}{t}+a if you did.

    The answer is not, as you have it, v-u/t=a but rather a=\frac{v-u}{t}
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    (Original post by Samistrawberry)
    So I'm slightly confused in the order of steps you have to do them in. Can someone please explain why this doesn't work?

    Here's the question..

    v= u + at Make a the subject

    So why can't I divide first so that I get this answer = v/t -u = a?

    I mean in terms of BIDMAS this would be correct :confused:

    However the answers is v-u/t=a.
    Because you want to go backward of bidmas. To obtain the u+at, they multiplied t to a, then added u. But when you want to make a the subject, that means they want you to go backward to go back to 'original a'. So they added u at last, so you subtract u first. They multiplied by t first, so you divide by t second

    Edit: also your way must be v/t=u/t - a which goes to v-u/t like answer says

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    (Original post by C0balt)
    X

    (Original post by BabyMaths)
    x

    (Original post by ILovePancakes)
    X


    (Original post by ztibor)
    X
    Thanks guys

    Got it.. It's cause we have to go backwards for the BIDMAS.
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    (Original post by C0balt)
    Because you want to go backward of bidmas. To obtain the u+at, they multiplied t to a, then added u. But when you want to make a the subject, that means they want you to go backward to go back to 'original a'. So they added u at last, so you subtract u first. They multiplied by t first, so you divide by t second

    Edit: also your way must be v/t=u/t - a which goes to v-u/t like answer says

    Posted from TSR Mobile
    Some more help please..?

    So I was going through the solution of 4x-1=y(x-3) and the guy did the brackets first :confused:

    (It was a video I was watching)
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    (Original post by Samistrawberry)
    Some more help please..?

    So I was going through the solution of 4x-1=y(x-3) and the guy did the brackets first :confused:

    (It was a video I was watching)
    Uh what was the full question?
    Was it expand the thing and simplify or solve simultaneous equation (in which case you need another equation)
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    https://www.youtube.com/watch?v=hMQ9x9ieEFE

    Here's the link to the vid.. But that was the full qs.
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    (Original post by C0balt)
    Uh what was the full question?
    Was it expand the thing and simplify or solve simultaneous equation (in which case you need another equation)
    Posted from TSR Mobile
    Sorry forgot to mention he was trying to make x the subject.
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    (Original post by Samistrawberry)
    Sorry forgot to mention he was trying to make x the subject.
    Oh because you want to put everything that doesn't contain x to the other side.
    When its y(x-3), it's counted as one term but as you can probably see, you will get 3y out of it which does not contain x.

    So you expand, put everything that has x on left, and everything that does not have x on right (make sure you change signs when moving sides because what you are doing is adding or subtracting the value from one side to get rid of them, but you must do the same thing on the other side because if you didn't, the equation is no longer equation - value each side will be different)

    After you do this, you'll get 4x-xy = -3y-1
    But you still have y and 4 on left. Which you want to be at right. To do this, you will have to see that x is common factor in the two terms so you can factorise it by x. Which gives x(4-y). Now you can divide both sides by (4-y) which will be your answer.
    Keep in mind you have to perform the operation on both sides and every term. That is why you factorise in this case.

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