Question

its the pic for the same question I asked before. please tell​

Answer 1

Step-by-step explanation:

$\frac{5}{x - 1} + \frac{1}{y - 2} = 2 \: ; \: \frac{6}{x - 1} - \frac{3}{y - 2} = 1 \\ 5( \frac{1}{x - 1} ) + 1( \frac{1}{y - 2} ) = 2 \: ; \: 6(\frac{1}{x - 1} ) - 3( \frac{1}{y - 2} ) = 1 \\Let \: \frac{1}{x - 1} = p \: and \: \frac{1}{y - 2} = q \\ 5p + q = 2 - eq1\: ; \: 6p - 3q = 1 - eq2 \\Multiply \: eq1 \: by \: 6 \: and \: eq2 \: by \: 5 \\ 6 \times (5p + q = 2)-> 30p + 6q = 12 - eq3 \\ 5 \times (6p - 3q = 1) - > 30p - 15q = 5 - eq4 \\ Add \: eq3 \: and \: eq4 , \: we \: get \\ 21q = 7 \\ q = \frac{7}{21} = \frac{1}{3} \\ Substitute \: q = \frac{1}{3} \: in \: eq2 \\6p - 3q = 1 \\ 6p - 3 \times \frac{1}{3} = 1 \\ 6p - 1 = 1 \\ 6p = 1 + 1 \\ 6p = 2 \\ p = \frac{2}{6} = \frac{1}{3} \\ \frac{1}{x - 1} = p \: - > \frac{1}{x - 1} = \frac{1}{3} \: - > x - 1 = 3 ->x=3+1 - >x=4\\ \frac{1}{y - 2} = q \: - > \frac{1}{y - 2} = \frac{1}{3} \: - > y - 2 = 3->y=3+2->y=5$