TS ICET 24 May 2018 Shift 1 Paper

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Question : 126 of 200

Marks: +1, -0

If

4/{\x-3} +6/{\y-4}

= 5 and

5/{\x-3} +3/{\y-4}

= 1 yhen (x,y) =

(3,4)
(4,5)
(5,6)
(6,7)

Solution:

Let

\;{1}/{x-3}=a ; \;{1}/{y-4}=b

Given equations become

4a + 6b = 5;

→ (1)

5a- 3b = 1

→ (2)
Solving (1) and (2)

a=\;{1}/{2} ; b=\;{1}/{2}

∴ x-3=2 ; y-4=2

x=5 y=6

(x, y)=(5,6)

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