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27 July, 09:42

What is the product of 4n^+3=7n?

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  1. 27 July, 10:29
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    So if you solve for n

    4n^2+3=7n

    make into trinomial

    subtract 7n from both sides

    4n^2-7n+3=0

    if we can factor this, then we can asume that the factors are equal to zero becase if

    xy=0 then assume x and/or y=0 so

    to factor an equation in ax^2+bx+c where a is greater than 1 then

    b=t+z

    a times c=t times z

    to factor 4n^2-7n+3 you do

    4 times 3=12

    what 2 numbers multiply to get 12 and add to get - 7

    the numbers are - 3 and - 4 so

    split up the - 7

    4n^2-4n-3n+3

    group

    (4n^2-4n) + (-3n+3)

    undistribute

    (4n) (n-1) + (-3) (n-1)

    reverse distributive property

    ab+ac=a (b+c)

    (4n-3) (n-1) = 0

    set each to zero

    4n-3=0

    add 3 to both sides

    4n=3

    divide both sides by 4

    n=3/4

    n-1=0

    add 1 to both sides

    n=1

    n=3/4 or 1
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