math question help?

1)

Let x be the number of students and thus x/4 be the number of non-students.

12x+15(x/4) = 567

Solve for x,

x = 36

2)

Let x be the length of the third piece. Thus, 3+2x is the length of the first piece and 3x is the length of the second piece. We have

(3+2x)+(3x)+x = 11

x = 8/6 = 4/3

3x = 4 inches, the length of the second piece.

Problem 1:

Let the number of non-students be x.

Then the number of students is 4x.

12 * 4x + 15*x=567

63x=567

x=9

Hence, the total number of students who can visit is 4x =36.

Problem 2:

Assume that third piece is x inches.

Then 2nd piece has the length 3x and

the 3rd piece has the length 2x + 3

Adding all this we should have 11 inches.

6x + 3 = 11

6x = 8

x = 8/6 =4/3

Hence the lenght of the 2nd piece is 4/3 *3 = 4 inches.

Number of students = x

Number of non-student chaperones = x/4

12x + 15x/4 = 567

48x+15x = 567*4

63x = 268

x = 36

36 students can visit the museum

Length of third piece = x

Length of first piece = 2x+3

Length of second piece = 3x

x + (2x+3) + 3x = 11

6x + 3 = 11

6x = 8

x = 4/3

Length of second piece = 3x = 3*4/3 = 4 inches

The second problem is:

1 piece=3+2x

2 piece=3x(3*1 1/3=3.99=4)

3 piece-x

3+2x+3x+x=11

6x=11-3

6x=8

x=8/6

x=1 1/3

The answer is 4 inches

well that is 4 students and one non student, that is 5 persons

at a cost of 63$

so 567/63 = 9 groupes of 4 studens =36

first piece = 3+2x

second piece =3x

third piece =x

11=3+2x+3x+x

11=3+6x

8=6x

x=1 1/3

3*x=3*1 1/3=4inches.

The steps are similar. For #1, set x to = # of students and y=# of non-students. Your equation will look like this:

12x+15y=567.

Since there are 4 students for every non-student, y=4x. Substitute in for y and solve for x.

#2, Similar Situation.

let x=piece #3. Equation is as follows:

(x+3)+3x+x=11.

Solve for x and plug back in to 3x to get your answer.