Timothy has 5/8 as many coins as pearly. After pearly gave 1/4 of her coins to timothy, Timothy has 10 coins more than pearly. How many coins did timothy have at first.

t = 5/8 p

t + 1/4 p = 10 + 3/4 p

now just solve for t

Thank you Steve. I am able to solve the problem.

I would like to know why did you add 10 to right side. Timothy has 10 more coins so i am thinking it should be added to left side.

timothy's coins were 10 more than pearly's. So, if the new amounts were t' and p',

10 more than pearly = 10+p'

t' = 10+p'

To solve this problem, let's break it down step by step.

Let's assume that the number of coins Pearly originally had is equal to x.

According to the problem, Timothy has 5/8 as many coins as Pearly, which means Timothy has (5/8)x coins.

After Pearly gives 1/4 of her coins to Timothy, Pearly will have (1-1/4)x = (3/4)x coins left, and Timothy will have (5/8)x + (1/4)x = (7/8)x coins.

We also know that Timothy has 10 coins more than Pearly after the transaction, so we can set up the equation:

(7/8)x - (3/4)x = 10

To solve for x, we need to find a common denominator. In this case, the least common denominator is 8:

(7/8)x - (3/4)x = 10

Multiplying the numerator and denominator of the first fraction by 2 and simplifying:

(14/16)x - (3/4)x = 10

Now we can combine the fractions:

(14/16 - 12/16)x = 10

(2/16)x = 10

Now we can solve for x by multiplying both sides of the equation by the reciprocal of 2/16, which is 16/2:

x = 10 * (16/2)

x = 10 * 8

x = 80

Therefore, Pearly originally had 80 coins.

Since Timothy has 5/8 as many coins as Pearly, Timothy originally had (5/8) * 80 = 50 coins.