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?
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