Help with a math problem plz. Problem: ab=3, bc=4, ac=5, find all possible (a,b, c). plz show work. thx?

ab/bc = 3/4


a/c = 3/4


==%26gt; a=(3c/4)





ac=5


ac= (3c/4).c = 5


c^2 = 5*4/3


c =+ or - 2 sqr(5/3)





a= + or - (3*(2)sqrt(5/3) )/4 = + or - (3/2)sqrt(5/3)





ab=3


b=3/a = + or - 2sqrt(3/5)








==%26gt; at a= (3/2)sqrt(5/3)


b= 2sqrt(3/5)


c=2sqr(5/3)





and





at a= -(3/2)sqrt(5/3)


b= -2sqrt(3/5)


c=-2sqr(5/3)

Help with a math problem plz. Problem: ab=3, bc=4, ac=5, find all possible (a,b, c). plz show work. thx?
You are trying to solve a system of equations:





(1) ab=3


(2) ac=5


(3) bc=4





So a = 3/b so 3c/b = 5


But bc = 4 so c = 4/b





Putting this into the 3* ( 4/b) /b= 5 gives 12/b^2 = 5 or b^2 = 12/5 so b = +/- 2*root(3)/root(5) = +/- 2*root(15)/5





This means that a = +/- root(15)/2 and has the same sign as b





Then (root(15)/2) * c = 5 implies that (root(3)/2) * c = root(5)





so c= +/- 2root(5)/root(3) = +/- 2root(15)/3 and has the same sign as a.





There are 2 ordered pairs:





{root(15)/2, 2*root(15)/5, 2root(15)/3) }





and





{-root(15)/2, -2*root(15)/5, -2root(15)/3) }





Note rote refers to squareroot