A university bookstore recently sold a wirebound graph-paper notebook for $0.88, and a college-ruled notebook for $2.38. At the start of spring semester, a combination of 50 of these notebooks were sold for a total of $59.00. How many of each type were sold?
number of wire-bound --- x
number of ruled ---- 50-x
solve for x :
.88x + 2.38(50-x) = 59
I would multiply by 100
88x + 238(50-x) = 5900
etc.
To solve this problem, let's assign variables to the unknowns.
Let's call the number of wirebound graph-paper notebooks sold 'x' and the number of college-ruled notebooks sold 'y'.
We know that the university bookstore sold a wirebound graph-paper notebook for $0.88 and a college-ruled notebook for $2.38.
So, we can translate the given information into equations:
Equation 1: price of x wirebound graph-paper notebooks + price of y college-ruled notebooks = total cost of notebooks sold
0.88x + 2.38y = 59.00
Equation 2: x + y = 50 (since a combination of 50 notebooks were sold)
Now we have a system of two equations with two variables. We can solve this system to find the values of x and y.
There are multiple methods to solve this system, such as substitution or elimination. We will use the substitution method in this explanation.
From Equation 2, we can rewrite it as:
y = 50 - x
Now we substitute this value of y into Equation 1:
0.88x + 2.38(50 - x) = 59.00
Simplifying the equation:
0.88x + 119 - 2.38x = 59.00
Combine like terms:
119 - 59.00 = 2.38x - 0.88x
Simplify further:
60 = 1.5x
Divide both sides by 1.5 to solve for x:
x = 40
Now that we have the value of x, we can substitute it back into Equation 2 to find y:
y = 50 - x
y = 50 - 40
y = 10
Therefore, the university bookstore sold 40 wirebound graph-paper notebooks and 10 college-ruled notebooks.