Ten of Phil's stamps are of the $0.37 denomination.
This is a simple algebraic word problem that requires a little set up before you can work toward its solution. You'll need to put together two equations with x and y values, and you'll need to solve them independently before you can solve the question entirely.
To begin, you'll want to isolate each detail.
Phil has 15 stamps
These stamps come in two denominations: $0.37 and $0.23
The value of these stamps is $4.85
Next, figure out what you need to solve. It's easy in this case. You need to figure out how many stamps Phil has of each denomination, and you need to figure it out based on the $4.85 total value. So your variables are x (the $0.37 stamps) and y (the $0.23 stamps). As such, your two equations will look as follows:
X + y = 15
0.37x + 0.23y = 4.85
Now it's time to use the substitution method. In this method, you take the simplest equation (x + y = 15) and you solve it so that you can substitute the new value of x (or y, but you're solving x) into the second, more complicated equation.
X + y = 15 becomes x = 15 - y
You'll notice that the y became a negative number. This is what happens when you transfer a number or variable from one side of the equation to the other.
Your next step is to substitute the new equation into the existing one, so:
0.37x + 0.23y = 4.85 becomes 0.37(15-y) + 0.23y = 4.85
Solving it further,
5.55 - 0.37y + 0.23y = 4.85
-0.37y + 0.23y = 4.85 - 5.55
-0.14y = 0.7
y = 5
When x + 5 = 15, then x = 15 - 5, or 10. Therefore, Phil has ten $0.37 stamps and five $0.23 stamps.
This is a simple algebraic word problem that requires a little set up before you can work toward its solution. You'll need to put together two equations with x and y values, and you'll need to solve them independently before you can solve the question entirely.
To begin, you'll want to isolate each detail.
Phil has 15 stamps
These stamps come in two denominations: $0.37 and $0.23
The value of these stamps is $4.85
Next, figure out what you need to solve. It's easy in this case. You need to figure out how many stamps Phil has of each denomination, and you need to figure it out based on the $4.85 total value. So your variables are x (the $0.37 stamps) and y (the $0.23 stamps). As such, your two equations will look as follows:
X + y = 15
0.37x + 0.23y = 4.85
Now it's time to use the substitution method. In this method, you take the simplest equation (x + y = 15) and you solve it so that you can substitute the new value of x (or y, but you're solving x) into the second, more complicated equation.
X + y = 15 becomes x = 15 - y
You'll notice that the y became a negative number. This is what happens when you transfer a number or variable from one side of the equation to the other.
Your next step is to substitute the new equation into the existing one, so:
0.37x + 0.23y = 4.85 becomes 0.37(15-y) + 0.23y = 4.85
Solving it further,
5.55 - 0.37y + 0.23y = 4.85
-0.37y + 0.23y = 4.85 - 5.55
-0.14y = 0.7
y = 5
When x + 5 = 15, then x = 15 - 5, or 10. Therefore, Phil has ten $0.37 stamps and five $0.23 stamps.
