Question

2nd question answer ​

Answer 1

Answer:

x= 1 and y =2

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\color{red}\[ \left.\begin{array}{l} \tt \dfrac{5}{x-1}+\dfrac{1}{y-2}=2 \quad \ldots(1) \\ \\ \tt \dfrac{6}{x-1}-\dfrac{3}{y-2}=1 \quad \ldots(2) \end{array}\right\} \text { Let } \tt \dfrac{1}{x-1}= u \: \: and \: \: \: \dfrac{1}{y - 2} = v\]$

So, our equations become

$\tt\[ 5 u+v=2 \quad \ldots(3) \\ \tt \quad 6 u-3 v=1 \quad \ldots(4) \]$

Thus, our equations are

$\color{blue} \text{\( \tt 5 u+v=2 \) ...(3) } \\ \color{blue} \text{\( \tt 6 u-3 v=1 \) ...(4)}$

$\color{navy} \[ \begin{array}{l} \text{ From (3) }\\ \tt 5 u+v=2 \\ \tt v=2-5 u \end{array} \]$

Putting value of v in (4)

$\pmb{ \color{olive}\[ \begin{array}{l} \tt 6 u-3 v=1 \\ \\ \tt 6 u-3(2-5 u)=1 \\ \\ \tt 6 u-6+15 u=1 \\ \\ \tt 6 u+15 u=1+6 \\ \\ \tt 21 u=7 \end{array} \]}$

$\pmb{ \color{olive}\[ \begin{array}{l} \tt u=\dfrac{7}{21} \\ \\ \tt u=\dfrac{1}{3} \end{array} \]}$

$\text{Putting \( \tt u=\dfrac{1}{3} \) in equation (3).}$

$\pmb{ \color{magenta} \[ \begin{array}{l} \tt \[ 5 u+v=2 \] \\ \\ \tt 5\left(\dfrac{1}{3}\right)+v=2 \\ \\ \tt\dfrac{5}{3}+v=2 \\ \\ \tt v=2-\dfrac{5}{3} \\ \\ \tt v=\dfrac{2(3)-5}{3} \\ \\ \tt v=\dfrac{6-5}{3} \\ \\ \tt v=\dfrac{1}{3} \\ \\ \text{ Hence, \( \boxed{\tt u=\frac{1}{3} } \: \: \& \: \: \: \boxed{ \tt v=\frac{1}{3} }\)}\end{array} \]}$

But we need to find x & y.

$\pmb{\color{darkcyan}\[ \begin{array}{l|| l} \tt u =\dfrac{1}{x-1} & \tt v =\dfrac{1}{y-2} \\ \\ \tt \dfrac{1}{3}=\dfrac{1}{x-1} & \tt \dfrac{1}{3}=\dfrac{1}{y-2} \\ \\ \tt x-1=3 & \tt y-2=3 \\ \\ \tt x=3+1 & \tt y=3+2 \\ \\ \tt x=4 & \tt y=5 \end{array} \]}$

So, x=4, y=5 is the solution of our equations