Prove that if ab>0, then the equation a/x-1 + b/x-3 = 0 has at least 1 solution in the interval (1,3)

To prove that  a/(x-1)+b/(x-3) = 0  has a solution in the
interval (1 ,3).

Multiply both sides of the give equation
 by (x-1)(x-3):

a(x-3) + b(x-1) =
0.

ax-3a+bx-b = 0.

Add 3a+b to
both sides:

ax+bx =
3a+b.

x(a+b) = 3a+b.

x =
(3a+b)/(a+b).

x =
{3(a/b)+1}/(a/b+1)........(1)

Since ab> 0,  a and b
are of the same sign. Therefore  0 < a/b < 1 Or a/b
>1.

If a/b> 1, then   2
<{3a/b+1}/(a/b  +1) <
3....................(2)

If a/b = 1, then (3a/b +1)/(a/b
+1) = (3+1)/(1+1) = 2.....(3)

If 0 < a/b <1 ,
then 1 < {3(a/b) +1}/((a/b)+1) <
2..........(4)

From (1) and (2)  we see that whenab
> 0 and  a>b  there is solution in (2,3)
.

From (1) and (3) when  ab> 0 a = b , there is a
solution x =2.

From (1) and (4) , when ab> 0, a
< b rhere is a solution fox in (1 , 2).

Combining
the 3 cases, when ab> 0 , x has a solution in (1,3)