its a simultaneous equation
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? PLEASE HELP ME.
let X = number of college ruled notebooks
let Y = number of graph paper notebooks
we know we sold 50 in total so that means
X + Y = 50
we know the price of each 2.38 for the college and 0.88 for the graph paper
2.38*X + 0.88*Y = 59.00
now we can use either elimination method or matrix math .. its simpler with these types for elimination
X = 50 - Y
then substitute this into the second equation and solve for Y
2.38*X + 0.88*Y = 59.00
2.38*(50 - Y) + 0.88*Y = 59.00
119 - 2.38Y + 0.88Y = 59.00
Y(-2.38 + 0.88) = 59.00 - 119
Y = -60/-1.5
Y = 40
we can use either of the equations to find x now but the first is easier
X = 50 - Y
X = 50 - 40
X = 10
so you have 40 graph paper books and 10 ruled books
to check see if the amounts come to 59 in the second equation
2.38*10 = 23.8
40*.88 = 35.2
23.8 + 35.2 = 59.00
so its confirmed the answer is correct