# In a certain math class, each studen has a textbook, every two students share a books of tables, every 3 students share a problem book, and every 4 students share a mathematics dictionary. If the total number of books is 75, how many students are in the class? How many students share tables? Hor many students share a problem book? How mnay share a mathematics dicitonary?

I will start if for you
let TB be number texbooks, BT number of books of tables, PB, and MD are the others

TB + BT + PB + MD=75

BT =TB/2

PB =TB/3
MD = TB/4

You should be able to take it from here.

## To solve this problem, we can use algebra to represent the relationships given in the question. Let's denote the number of students in the class as "S", the number of textbooks as "TB", the number of books of tables as "BT", the number of problem books as "PB", and the number of mathematics dictionaries as "MD".

We are given that the total number of books is 75, so we can write the equation:

TB + BT + PB + MD = 75 ---(1)

We are also given that every two students share a book of tables, which means the number of books of tables is half the number of students, so we can write the equation:

BT = S/2 ---(2)

Similarly, every three students share a problem book, so we can write:

PB = S/3 ---(3)

Finally, every four students share a mathematics dictionary, so we have:

MD = S/4 ---(4)

Now we have a system of equations with four variables (TB, BT, PB, MD) and four equations (equations 1, 2, 3, and 4). We can solve this system of equations to find the values of TB, BT, PB, and MD, which will give us the number of students in the class (S) and the number of students sharing each type of book.

To do this, we'll use a technique called substitution. We'll solve equation (2) for BT, equation (3) for PB, and equation (4) for MD in terms of S. Then, we'll substitute these expressions into equation (1) and solve for S.

From equation (2), we have:

BT = S/2

From equation (3), we have:

PB = S/3

From equation (4), we have:

MD = S/4

Substituting these expressions into equation (1), we get:

TB + (S/2) + (S/3) + (S/4) = 75

To simplify this equation, we'll get rid of the fractions by multiplying everything by 12, the least common multiple of 2, 3, and 4:

12TB + 6S + 4S + 3S = 900

Combining like terms, we have:

12TB + 13S = 900

At this point, we have one equation (equation 5) with the variable TB and another equation (equation 6) with the variable S. We can solve these simultaneous equations to find the values of TB (number of textbooks) and S (number of students) in the class.

Now, solve equation 6 for TB:

TB = (900 - 13S) / 12 ---(7)

We can start by substituting equation 7 into equation 5:

(900 - 13S) / 12 + 13S = 75

To solve for S, we can multiply through by 12 to eliminate the fraction:

900 - 13S + 156S = 900

Combine like terms:

143S = 900

Divide both sides by 143:

S = 900 / 143

Simplifying the right side gives:

S ≈ 6.294

Since the number of students cannot be a decimal, we can round the answer to the nearest whole number:

S ≈ 6

Therefore, there are 6 students in the class.

To find the number of students sharing each type of book, we can substitute this value back into equations 2, 3, and 4:

BT = S/2 = 6/2 = 3

PB = S/3 = 6/3 = 2

MD = S/4 = 6/4 = 1.5

However, since we are dealing with whole numbers of students, we need to adjust:

BT ≈ 3

PB ≈ 2

MD ≈ 1

Therefore, in the class, 3 students share a book of tables, 2 students share a problem book, and 1 student shares a mathematics dictionary.