Each course at college X is worth either 4 or 5 credits. The members of the men's swim team are taking a total of 52 courses that are worth a total of 22 credits. How many 4-credit courses and how many 5-credit courses are being taken?
If there are x at 4 pts and y at 5 pts, we have
x+y=52
4x+5y=22
Sorry. 52 courses must add up to a lot more than 22 credits.
Fix the numbers and solve the equations.
number of 4-credit courses --- x
number of 5-credit courses --- y
x+y = 52
4x + 5y = 22
This has no positive x and y solutions.
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How can a total of 52 courses yield only 22 credits, unless they failed most of the courses ????
Even if courses were failed, no combination of whole number values of x and y will satisfy the equation
4x + 5y = 22
Let's assume the number of 4-credit courses taken as x and the number of 5-credit courses taken as y.
According to the given information, we can form the following equations:
x + y = 52 (equation 1) -- representing the total number of courses taken
4x + 5y = 22 (equation 2) -- representing the total credits earned
To solve these equations, we can use the method of substitution or elimination.
Let's solve using the method of substitution:
From equation 1, we have x = 52 - y.
Substituting this value of x in equation 2, we get:
4(52 - y) + 5y = 22
208 - 4y + 5y = 22
208 + y = 22
y = 22 - 208
y = -186
Since the number of courses and credits cannot be negative, there is no solution to this problem.
To solve this problem, let's assign variables to represent the number of 4-credit courses and 5-credit courses being taken.
Let's say the number of 4-credit courses is represented by 'x', and the number of 5-credit courses is represented by 'y'.
According to the problem, the total number of courses being taken is 52. Therefore, we can write the equation:
x + y = 52 ---(Equation 1)
Also, the total number of credits for these courses is 22. Each 4-credit course contributes 4x credits, and each 5-credit course contributes 5y credits. So we can write the equation:
4x + 5y = 22 ---(Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2), and we can solve for the values of 'x' and 'y'.
One method to solve this system of equations is by substitution.
First, solve Equation 1 for 'x':
x = 52 - y
Substitute this value of 'x' in Equation 2:
4(52 - y) + 5y = 22
208 - 4y + 5y = 22
208 + y = 22
y = 22 - 208
y = -186
Now substitute the value of 'y' back into Equation 1 to solve for 'x':
x + (-186) = 52
x = 52 + 186
x = 238
Therefore, there are 238 4-credit courses and -186 5-credit courses being taken.
However, it doesn't make sense to have a negative number of courses. Hence, it appears that there is no valid solution to this problem.