A student has some $1 and $5 bills in his wallet. He has a total of 15 bills that are worth $47. How many of each type of bill does he have?
1.) Write a system of equations for the problem above.
2.) Solve algebraically (substitution or elimination methods)
Let x be the number of $1 bills and y be the number of $5 bills.
We can set up the following equations based on the given information:
1) x + y = 15 (Equation 1, representing the total number of bills)
2) 1x + 5y = 47 (Equation 2, representing the total value of the bills)
To solve algebraically using the substitution method, we can solve Equation 1 for x in terms of y:
x = 15 - y
Substitute this into Equation 2:
1(15 - y) + 5y = 47
15 - y + 5y = 47
15 + 4y = 47
Combine like terms:
4y = 32
y = 8
Substitute y = 8 back into Equation 1 to solve for x:
x + 8 = 15
x = 7
Therefore, the student has 7 $1 bills and 8 $5 bills.
1.) Let's assume the student has x $1 bills and y $5 bills.
The first equation can be derived from the total number of bills:
x + y = 15
The second equation can be derived from the total value of the bills:
1x + 5y = 47
2.) We can use the substitution method to solve the system of equations.
From the first equation, we can solve for x:
x = 15 - y
Substitute this value of x into the second equation:
1(15 - y) + 5y = 47
15 - y + 5y = 47
15 + 4y = 47
Rearranging the equation:
4y = 47 - 15
4y = 32
Solving for y:
y = 32 / 4
y = 8
Now substitute this value of y back into the first equation to solve for x:
x + 8 = 15
x = 15 - 8
x = 7
Therefore, the student has 7 $1 bills and 8 $5 bills.
To solve this problem, we can set up a system of equations based on the given information.
Let's denote the number of $1 bills as 'x' and the number of $5 bills as 'y'.
1) Write a system of equations:
The first equation represents the total number of bills:
x + y = 15 (since the student has a total of 15 bills)
The second equation represents the total value of the bills:
1x + 5y = 47 (since the total value of the bills is $47)
Now, we have a system of equations:
x + y = 15
1x + 5y = 47
2) Solve algebraically using substitution or elimination methods:
Let's use the elimination method in this case.
Multiply the first equation by 5 to make the coefficients of 'y' the same in both equations:
5(x + y) = 5(15)
5x + 5y = 75
Now, subtract the second equation (1x + 5y = 47) from the new equation (5x + 5y = 75):
(5x + 5y) - (1x + 5y) = 75 - 47
5x + 5y - x - 5y = 28
4x = 28
Divide both sides of the equation by 4:
4x/4 = 28/4
x = 7
Now, substitute the value of 'x' back into the first equation to solve for 'y':
7 + y = 15
y = 15 - 7
y = 8
So, the student has 7 $1 bills and 8 $5 bills.