A coin bank for donations for the protection of the reef contained Php 1,570. There were twice as many Php 5-coins as Php 10-coins and 14 less Php 1-coins as Php 5-coins. How many of each kind were there?

There are 144 Php 5-coins, 72 Php 10-coins and 130 Php-1 coins

Step-by-step explanation:

The amount of the coins in the coin bank = Php 1,570

The number of Php 5-coins = 2 × The number of Php 10-coins

The number of Php 5 coins = The number of Php 1-coins + 14

Let 'x' represent the number of Php-5 coins, let 'y' represent the number of Php 10-coins, and let 'z' represent the number of Php 1-coins, we get;

x = 2 × y...(1)

x = z + 14...(2)

5·x + 10·y + z = 1,570...(3)

From equation (1) equation (2), and equation three, we have;

y = x/2

z = x - 14

5·x + 10·x/2 + x - 14 = 1,570

11·x - 14 = 1,570

x = (1,570 + 14)/11 = 144

The number of Php 5-coins, x = 144

y = x/2

∴ y = 144/2 = 72

The number of Php 10-coins, y = 72

z = x - 14

∴ z = 144 - 14 = 130

The number of Php 1-coins, z = 130

the solution is h=10 because you add 9 to each side of the equation.