Question

please find x and y ​

Answer 1

Answer:

X1 and x2 please make brainly answer

Answer 2

Answer

$\sf \dashrightarrow \dfrac{5}{x - 1} + \dfrac{1}{y - 2} = 2 \: \: --- (i)$

$\sf \dashrightarrow \dfrac{6}{x - 1} - \dfrac{3}{y - 2} = 1 \: \: --- (ii)$

Let $\sf \dfrac{1}{x - 1}$ be u.

Let $\sf \dfrac{1}{y - 2}$ be v.

So, the equations become

$\sf \dashrightarrow 5u + 1v = 2$

$\sf \dashrightarrow 6u - 3v = 1$

First, find the values of u and v by substitution method.

$\sf \dashrightarrow 5u + 1v = 2$

$\sf \dashrightarrow 5u = 2 - 1v$

$\sf \dashrightarrow u = \dfrac{2 - 1v}{5}$

Now, let's find the value of v by second equation.

$\sf \dashrightarrow 6u - 3v = 1$

$\sf \dashrightarrow 6 \bigg( \dfrac{2 - 1v}{5} \bigg) - 3v = 1$

$\sf \dashrightarrow \dfrac{12 - 6v}{5} - 3v = 1$

$\sf \dashrightarrow \dfrac{12 - 6v - 15v}{5} = 1$

$\sf \dashrightarrow \dfrac{12 - 21v}{5} = 1$

$\sf \dashrightarrow 12 - 21v = 1 \times 5$

$\sf \dashrightarrow 12 - 21v = 5$

$\sf \dashrightarrow -21v = 5 - 12$

$\sf \dashrightarrow -21v = -7$

$\sf \dashrightarrow v = \dfrac{-7}{-21}$

$\sf \dashrightarrow v = \dfrac{1}{3}$

Now, let's find the value of u by first equation.

$\sf \dashrightarrow 5u + 1v = 2$

$\sf \dashrightarrow 5u + 1 \bigg( \dfrac{1}{3} \bigg) = 2$

$\sf \dashrightarrow 5u + \dfrac{1}{3} = 2$

$\sf \dashrightarrow 5u = 2 - \dfrac{1}{3}$

$\sf \dashrightarrow 5u = \dfrac{6 - 1}{3}$

$\sf \dashrightarrow 5u = \dfrac{5}{3}$

$\sf \dashrightarrow u = \dfrac{\dfrac{5}{3}}{5}$

$\sf \dashrightarrow u = \dfrac{5}{3} \times \dfrac{1}{5}$

$\sf \dashrightarrow u = \dfrac{5}{15}$

$\sf \dashrightarrow u = \dfrac{1}{3}$

We know that,

$\sf \dashrightarrow \dfrac{1}{x - 1} = u$

$\sf \dashrightarrow \dfrac{1}{x - 1} = \dfrac{1}{3}$

$\sf \dashrightarrow 1 \times 3 = 1 (x - 1)$

$\sf \dashrightarrow 3 = 1x - 1$

$\sf \dashrightarrow 1x = 3 + 1$

$\sf \dashrightarrow 1x = 4$

$\sf \dashrightarrow x = 4$

Now, let's find the value of y.

$\sf \dashrightarrow \dfrac{1}{y - 2} = v$

$\sf \dashrightarrow \dfrac{1}{y - 2} = \dfrac{1}{3}$

$\sf \dashrightarrow 1 \times 3 = 1 (y - 2)$

$\sf \dashrightarrow 3 = 1y - 2$

$\sf \dashrightarrow 1y = 3 + 2$

$\sf \dashrightarrow 1y = 5$

$\sf \dashrightarrow y = 5$

Hence, the values of x and y are 4 and 5 respectively.