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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