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At first, the amount of money Bryant had was 60% the amount of money Michael had. They went for lunch together and Bryant paid for 25% of the bill and Michael paid for the rest. After paying for the lunch, Bryant had $56 left and Michael had 25% of his money left. How much did they pay for lunch?

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  1. Let the money Michael had first=x
    Bryant=60/100x= 3/5x

    let the bill paid for lunch=y
    Bryant=25/100y= 1/4y
    Michael=75/100y= 3/4y

    3/5x-1/4y=56
    12x-5y=1120 ....eqn(i)

    x-3/4y=25/100x
    x-3/4y=1/4x
    4x-3y=x
    3x-3y=0....eqn(ii)
    solve by elimination method by multiplying eqn(ii) by 4

    12x-5y=1120
    12x-12y=0

    7y=1120
    y=160
    the amount paid for lunch= $160

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