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.