The Everton college store paid 1532 dollars for an order of42 calculators.The store paid 18 dollars for each scientific calculator,and tthe other all graphing calculators, cost the store 54 dollars each, How many of each calsulator was ordered?
x + y = 42
18x + 54y = 1532
Solve that pair of equations for x (the number of scientific calculators) and y (the number of graphing calculators)
To solve this problem, we can use a system of equations.
Let's assume the number of scientific calculators ordered is 'x' and the number of graphing calculators ordered is 'y'.
From the problem, we know that the store ordered a total of 42 calculators, so we can write the first equation as:
x + y = 42 ---(equation 1)
Next, we know that the scientific calculators cost $18 each and the graphing calculators cost $54 each. The total cost of the order is $1532. Therefore, the second equation is:
18x + 54y = 1532 ---(equation 2)
Now we have a system of two equations that we can solve using substitution or elimination methods.
Let's use the elimination method to solve this system.
Multiply equation 1 by 18, so that the coefficients of 'x' in both equations will be the same:
18x + 18y = 756 ---(equation 3)
Now, subtract equation 3 from equation 2:
(18x + 54y) - (18x + 18y) = 1532 - 756
Simplifying the equation, we get:
36y = 776
Divide both sides of the equation by 36:
y = 776 / 36
Simplify further:
y ≈ 21.56
Since 'y' represents the number of calculators, it cannot be a fractional value. We need to find a whole number solution.
Let's substitute the value of 'y' back into equation 1:
x + 21.56 = 42
x ≈ 42 - 21.56
x ≈ 20.44
Again, 'x' represents the number of calculators, so it cannot be a fractional value. Therefore, there is no solution where both 'x' and 'y' are whole numbers.
It appears that there might be an error or inconsistency in the given information, as it does not lead to a whole number solution. Please verify the data and try again.