A librarian is expanding some sections of the city library. He buys books at a special price from a dealer who charges one price for any hardback book and another price for any paperback book. For the children's section, Mr. Shah purchased 31 new hardcover books and 35 new paperback books, which cost a total of $613. He also purchased 38 new hardcover books and 85 new paperback books for the adult fiction section, spending a total of $1,004. What is the special price for each type of book?
To solve this problem, we can set up a system of equations using the given information.
Let's assume the special price for each hardcover book is "x" dollars, and the special price for each paperback book is "y" dollars.
For the children's section, Mr. Shah purchased 31 hardcover books and 35 paperback books, which cost a total of $613.
So, we can write the equation:
31x + 35y = 613 --------(1)
For the adult fiction section, Mr. Shah purchased 38 hardcover books and 85 paperback books, which cost a total of $1,004.
So, we can write the equation:
38x + 85y = 1004 ----------(2)
Now, we can solve these equations to find the values of "x" and "y".
To do that, let's multiply equation (1) by 38, and equation (2) by 31, so that the coefficients of "x" in both equations will be the same:
(31x + 35y) * 38 = 613 * 38
(38x + 85y) * 31 = 1004 * 31
Expanding these equations, we get:
1178x + 1330y = 23294 ---------(3)
1178x + 2635y = 31124 ---------(4)
Now, let's subtract equation (3) from equation (4):
(1178x + 2635y) - (1178x + 1330y) = 31124 - 23294
1305y = 7830
Dividing both sides by 1305, we get:
y = 6
Now, substitute the value of y into equation (1) or (2) to find the value of x. Let's use equation (1):
31x + 35 * 6 = 613
31x + 210 = 613
31x = 613 - 210
31x = 403
x = 403 / 31
x ≈ 13
Therefore, the special price for each hardcover book is approximately $13, and the special price for each paperback book is $6.
To find the special price for each type of book, let's assign variables to represent the cost of a hardback book and a paperback book.
Let:
x = cost of a hardback book
y = cost of a paperback book
Now let's form two equations based on the given information:
Equation 1: 31x + 35y = 613 (for the children's section)
Equation 2: 38x + 85y = 1004 (for the adult fiction section)
To solve these equations, we can use the method of substitution or elimination. Let's use substitution here.
From Equation 1, solve for x:
31x = 613 - 35y
x = (613 - 35y) / 31
Substitute this value of x in Equation 2:
38[(613 - 35y) / 31] + 85y = 1004
Now, let's solve this equation to find the value of y.
Multiply through by 31 to eliminate the fraction:
38(613 - 35y) + 85*31y = 1004*31
Simplify and solve for y:
23294 - 1330y + 2635y = 31124
2635y - 1330y = 31124 - 23294
1305y = 7830
y = 7830 / 1305
y = 6
Therefore, the cost of a paperback book is $6.
Now substitute this value of y back into Equation 1 to find the value of x:
31x + 35*6 = 613
31x + 210 = 613
31x = 613 - 210
31x = 403
x = 403 / 31
x = 13
So, the cost of a hardback book is $13.
Therefore, the special price for each type of book is:
Hardback book: $13
Paperback book: $6
Yes , thank you so much!
The grade 10 curriculum has this as a two equations in 2 unknowns system of equations.
Let x be the hardback books and y the paperbacks.
31x + 35y = 613
38x + 85y = (1004 - 613)
I would suggest using "elimination method" to solve for y.
Mult equation 1 by -38, and multiply equation 2 by 31 and then add the two equations together.
is this enough of a hint??