The Everton college store paid $1668 for an order of 44 calculators. The store paid $8 for each scientific calculator. The others, all graphing calculators, cost the store $55 each. How many of each type of calculator was ordered?
Sorry! Actually, it was $8 the store paid, intead of $18. All else is correct. Thank you.
One posting at a time is enough, see below.
No problem! Let's solve it with the corrected information.
Let's assume that x represents the number of scientific calculators and y represents the number of graphing calculators.
We know that the college store paid $8 for each scientific calculator, so the total cost for the scientific calculators is 8x dollars.
Similarly, the college store paid $55 for each graphing calculator, so the total cost for the graphing calculators is 55y dollars.
The total cost for the order is $1668, so we can write the equation:
8x + 55y = 1668
We also know that there were a total of 44 calculators ordered, so we can write another equation:
x + y = 44
Now we have a system of two equations:
8x + 55y = 1668
x + y = 44
To solve this system, we can use the method of substitution or elimination.
Let's use the elimination method to solve this system.
Multiply the second equation by 8 to make the coefficients of x in both equations the same:
8x + 8y = 352
Now subtract this equation from the first equation:
8x + 55y - 8x - 8y = 1668 - 352
47y = 1316
Divide both sides by 47 to solve for y:
y = 1316/47
y ≈ 28
Now substitute the value of y back into one of the original equations to solve for x:
x + 28 = 44
x = 44 - 28
x = 16
Therefore, there were 16 scientific calculators and 28 graphing calculators ordered.