Math, asked by Friendlyboyprem143, 1 month ago

5/x-1+1/y-2=2. 6/x-1-3/y-2=1 give me ans pl zz​

Answers

Answered by mathdude500
3

\large\underline{\sf{Solution-}}

Given pair of linear equation is

\rm :\longmapsto\:\dfrac{5}{x - 1}  + \dfrac{1}{y - 2}  = 2 -  -  - (1)

and

\rm :\longmapsto\:\dfrac{6}{x - 1}   -  \dfrac{3}{y - 2}  = 1 -  -  - (2)

Let assume that

\red{\rm :\longmapsto\:\dfrac{1}{x - 1}  = u -  -  - (3)}

and

\red{\rm :\longmapsto\:\dfrac{1}{y - 2}  = v -  -  - (4)}

So, equation (1) and (2) can be rewritten as

\rm :\longmapsto\:5u + v = 2 -  -  - (5)

and

\rm :\longmapsto\:6u - 3v = 1

can be further rewritten as

\rm :\longmapsto\:3(2u - v) = 1

\rm :\longmapsto\:2u - v = \dfrac{1}{3}  -  -  - (6)

On adding equation (5) and (6), we get

\rm :\longmapsto\:7u = 2 + \dfrac{1}{3}

\rm :\longmapsto\:7u = \dfrac{6 + 1}{3}

\rm :\longmapsto\:7u = \dfrac{7}{3}

\bf\implies \: \boxed{ \bf{ \: u \:  =  \: \dfrac{1}{3}}} -  -  - (7)

On substituting the value of u in equation (5), we get

\rm :\longmapsto\:\dfrac{5}{3} + v = 2

\rm :\longmapsto\:2 - \dfrac{5}{3} =   v

\rm :\longmapsto\:\dfrac{6 - 5}{3} =   v

\bf\implies \: \boxed{ \bf{ \: v \:  =  \: \dfrac{1}{3}}} -  -  - (8)

So, equation (3) and (4), can be rewritten as

\rm :\longmapsto\:\dfrac{1}{x - 1}  = \dfrac{1}{3}  \:  \:  \: and \:  \:  \: \dfrac{1}{y - 2} =  \dfrac{1}{3}

\rm :\longmapsto\:x - 1 = 3 \:  \:  \: and \:  \:  \: y - 2 = 3

\rm :\longmapsto\:x = 3 + 1 \:  \:  \: and \:  \:  \: y = 3 + 2

\bf\implies \: \boxed{ \bf{ \: x = 4 \:  \:  \: and \:  \:  \: y = 5}}

Verification :-

Consider,

\rm :\longmapsto\:\dfrac{5}{x - 1}  + \dfrac{1}{y - 2}  = 2 -  -  - (1)

On substituting x = 4 and y = 5, we get

\rm :\longmapsto\:\dfrac{5}{4 - 1}  + \dfrac{1}{5 - 2}  = 2

\rm :\longmapsto\:\dfrac{5}{3}  + \dfrac{1}{3}  = 2

\rm :\longmapsto\:\dfrac{5 + 1}{3} = 2

\rm :\longmapsto\:\dfrac{6}{3} = 2

\bf\implies \:2 = 2

Hence, Verified

Consider,

\rm :\longmapsto\:\dfrac{6}{x - 1}   -  \dfrac{3}{y - 2}  = 1 -  -  - (2)

On substituting the values of x and y, we get

\rm :\longmapsto\:\dfrac{6}{4 - 1}   -  \dfrac{3}{5 - 2}  = 1

\rm :\longmapsto\:\dfrac{6}{3}   -  \dfrac{3}{3}  = 1

\rm :\longmapsto\:2 - 1 = 1

\rm :\longmapsto \: 1 = 1

Hence, Verified

Answered by sohamc060
3

Answer:

Solution

Given pair of linear equation is

1

5

=2 2 --- (1)

X

2

and

6

3

=1--- (2)

- 1 y-2

Let assume that

1

-(3)

=u-

1

and

1

-- (4)

y - 2

So, equation (1) and (2) can be rewritten as

5u+v = 2--- (5)

and

6u- 3v ==1

can be further rewritten as

3(2u - v) = 1

1

(6)

2u- v =

3

On adding equation (5) and (6), we get

1

7u=2+

3

6+1

Tu=

3

7u =

3

1

u= - (7)

3

On substituting the value of u in equation (5),

we get

5

+v= 2

3

5

2

3

V

6 -5

6-5

3

-- (8)

3

V =

So, equation (3) and (4), can be rewritten as

1

1

and

1

- 1

y-2

3

X

: x-1 =3 and y - 2 = 3

x= 3+1 and y = +2

X= 4 and

y 5

Verification

Consider,

5

1

- (1)

-1 Ty- 2.

X

On substituting x = 4 and y = 5, we get

5

1

= 2

5 2

4 1

5

1

3 2

5+ 2

3

2

3

2= 2

Hence,Verified

Consider,

6

3

X-1y-21-- (2)

On substituting.the values of x andy, we get

6

3

1

41 5-2

6

3

= 1

3

3

21=1

1=1

Step-by-step explanation:

hope it's helpful

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