I am having difficulty solving a math equation, can anyone help? show solutions please!?

Question

If 3x = 8y and 12y=7z and kz = 72 x
SOlve for k

Answer 1

k=112

3x = 8y
(3/2)*3x = (3/2)*8y
9x/2 = 12y and 12y = 7z
so 9x/2 = 7z
x/z = 7*2/9 = 14/9

and

kz = 72x
k = 72*(x/z) = 72*(14/9) = 112

Answer 2

3x=8y <=> x=8/3y (1)
12y=7z <=> y=7/12z (2)
Because of (1) kz=72*3/8y <=> kz=192y
and because of (2) kz=192*7/12z <=> kz=112z <=> k=112

Answer 3

Okay this problem involves some deductive reasoning. First of isolate your k to get:

k = 72x/z

now the job is to see if one of the variables will be able to cancel out.
from the first equation we can get x = 8y/3
The second equation yields z = 12y/7
looks promising for the y's to cancel out. so now we plug this into the k = 72x/z equation

k = 72(8/3)(y)/(12y/7)--> 192y/(12y/7) -->192y(7/12y)

So, even with only three equations and 4 unknown variables we were still able to solve for the one in question.
the y's cancel out and the math gives k = 112

Answer 4

Now

x = (8/3).y and y = (7/12).z

so

x = (8/3).(7/12).z = (14/9).z

Hence

kz = 72x = 72.(4/9)z = 112.z

Hence k=112

Answer 5

12y = 4.5x = 7z, or 9x = 14z. Then 72x = 14*8z = 112z.
k = 112

Answer 6

k = 72 x / z
k = (72 x) (7) / (12 y)
k = (192 y) (7) / (12 y)
k = 16 x 7
k = 112

Answer 7

x=(8/3)y

(12/7)y = z

k = 72x/z

=(12/7)y / (8/3)y

=36/56 = 18/28 = 9/14

Answer 8

3x = 8y, therefore 9/2x = 12y

12y = 9/2x = 7z, therefore 72x = 196z

k = 196...which is wrong...bugger it.

Answer 9

81 > -9w [plus 9w] 81 + 9w > 0 [factorise the 9] 9 (9 + w) > 0 [divide by 9] 9 + w > 0 [subtract 9] w > - 9 so w is greater than (-9). that means w = {-8, -7, - 6......}

Answer 10

3x=8y
3x/8=y -------------------- (1)

12y=7z
12(3x/8)=7z
36x/8=7z
36x/8times1/7=z
36x/56=z ----------------------------- (2)

kz=72x
k(36x/56)=72x
36kx/56=72x
36kx=72xtimes56
36k(x)=4032(x)
36k=4032
k=112



i guess so ;p