David and Marie had $300 altogether. After David spent 2/5 of his money and Marie spent $20, they each had the same amount of money. How much money did Marie have at first?
david --- x
Marie --- 300-x
David spent 2/5 of his money, so his has (3/5)x left
marie spent $20, so she has 300-x - 20 left
(3/5)x = 300-x-20
times 5
3x = 1400 - 5x
8x = 1400
x = 175, so
David had $175 and
Marie had 300-175 or $125
check:
David has 175 , he spends 2/5 of that or 70 , so he has 105 dollars left
Marie has 300-175 or 125 dollars, she spends 20, leaving her also with 105
Yeahhh!
To find out how much money Marie had at first, we need to break down the information given in the problem and solve step by step.
Let's assume that David had x dollars initially, and Marie had y dollars initially.
According to the problem, they had 300 dollars altogether:
x + y = 300 (Equation 1)
After David spent 2/5 (or 2/5 * x) of his money, he would have 3/5 (or 3/5 * x) of his money remaining.
Similarly, after Marie spent $20, she would have y - 20 dollars remaining.
The problem states that after spending, David and Marie had the same amount of money left:
3/5 * x = y - 20 (Equation 2)
Now, we have two equations with two variables. We can solve this system of equations to find the values of x and y.
First, we can solve Equation 2 for x:
3/5 * x = y - 20
Multiply both sides by 5/3:
x = (5/3) * (y - 20)
Next, substitute this value of x into Equation 1:
(5/3) * (y - 20) + y = 300
Multiply through by 3 to eliminate the fraction:
5 * (y - 20) + 3y = 900
Distribute the 5:
5y - 100 + 3y = 900
Combine like terms:
8y - 100 = 900
Add 100 to both sides:
8y = 1000
Divide by 8:
y = 125
So, Marie initially had $125.
Therefore, Marie had $125 initially.