# 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)

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### 2 Answers

thankyou very much

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)