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Question

Solve the following system of equation for x and y.
$$\displaystyle \frac{5}{x-1}+\frac{1}{y-2}=2,\frac{6}{x-1}-\frac{3}{y-2}=1$$


Solution

Let $$\dfrac{1}{x-1}=a$$ and $$\dfrac{1}{y-2}=b$$
$$5a+b=2 ..... (1) .... \times 3$$
$$6a-3b=1 ..... (2) ...... \times 1$$
$$15a+3b=6$$
$$6a-3b$$
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$$
$$21a=7$$
$$\therefore \boxed{a=\dfrac{1}{3}}$$
Putting $$a=\dfrac{1}{3}$$ in $$(1)$$
$$\dfrac{5}{3}+b=2$$
or, $$b=2-\dfrac{5}{3}=\dfrac{1}{3}$$
Substituting value of $$a$$ and $$b$$
$$\dfrac{1}{x-1}=\dfrac{1}{3}\Rightarrow x-1=3 ..... (3)$$
$$\dfrac{1}{y-2}=\dfrac{1}{3}\Rightarrow y-2=3 ...... (4)$$
From $$(3), x=4$$
From $$(4), y=5$$

Mathematics

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