Shirley had 52 more coins than Jim after Jim gave 1/5 of his coins to her. If they had 260 coins in all, how many coins did Shirley have at first?
at the beginning:
coins Jim had ---- x
coins Shirley had -- y
x+y = 260 ---->#1
after giveaway:
Jim ---- (4/5)x
Shirley -- y + (1/5)x = (5y + x)/5
(5y+x)/5 - (4/5)x = 52
5y + x - 4x = 260
-3x + 5y = 260 --->#2
solve the two equations
To find out how many coins Shirley had at first, we need to set up a system of equations based on the given information.
Let's start by assigning variables to represent the number of coins Jim and Shirley had initially.
Let J = Jim's initial number of coins
Let S = Shirley's initial number of coins
Based on the given information, we can establish the following equations:
Equation 1: Shirley had 52 more coins than Jim after Jim gave 1/5 of his coins to her:
S = J + 52 (Shirley had 52 more coins)
Equation 2: They had a total of 260 coins in all:
J + S = 260 (Their total number of coins is 260)
Now, we have a system of two equations with two variables:
S = J + 52 (Equation 1)
J + S = 260 (Equation 2)
To solve this system of equations, we can use substitution or elimination method.
Let's solve it using the substitution method:
Substitute the value of S from Equation 1 into Equation 2:
J + (J + 52) = 260
2J + 52 = 260
2J = 260 - 52
2J = 208
Divide both sides by 2:
J = 104
Now, substitute the value of J back into Equation 1 to find S:
S = J + 52
S = 104 + 52
S = 156
Therefore, Shirley had 156 coins at first.