Math, asked by surevanshika, 1 month ago

solve the following pair of equation by elimination method
4/√x + 6/√y = 4
4/√x -9√y = -1​

Answers

Answered by abhiharwansh5
0

Answer: Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)  

2x

1

+  

3y

1

=2;  

3x

1

+  

2y

1

=  

6

13

 

(ii)  

x

 

2

+  

y

 

3

=2;  

x

 

4

−  

y

 

9

=−1

(iii)  

x

4

+3y=14;  

x

3

−4y=23

(iv)  

x−1

5

+  

y−2

1

=2;  

x−1

6

−  

y−2

3

=1

(v)  

xy

7x−2y

=5;  

xy

8x+7y

=15

(vi) 6x+3y=6xy;2x+4y=5xy

(vii)  

x+y

10

+  

x−y

2

=4;  

x+y

15

−  

x−y

5

=−2

(viii)  

3x+y

1

+  

3x−y

1

=  

4

3

;  

2(3x+y)

1

−  

2(3x−y)

1

=  

8

−1

 

Hard

Answer

(i)

2x

1

+  

3y

1

=2

3x

1

+  

2y

1

=  

6

13

 

Lets assume  

x

1

=p and  

y

1

=q, both equations will become

2

p

+  

3

q

=2

⇒3p+2q=12    ... (A)

3

p

+  

2

q

=  

6

13

 

⇒2p+3q=13   ...(B)

Multiply (A) by 3 and (B) by 2, we get

9p+6q=36

4p+6q=26

Subtracting these both, we get

5p=10

⇒p=2

⇒x=  

p

1

=  

2

1

 

∴q=  

6

26−8

=3

⇒y=  

q

1

=  

3

1

 

(ii)  

x

 

2

+  

y

 

3

=2

x

 

4

−  

y

 

9

=−1

Lets assume  

x

 

1

=p and  

y

 

1

=q, both equations will become

2p+3q=2   ...(A)

4p−9q=−1    ...(B)

Multiply (A) by 3, we get

6p+9q=6

4p−9q=−1

Adding these both, we get

10p=5

⇒p=  

2

1

 

⇒  x=  

p  

2

 

1

=4

∴q=  

3

1

 

⇒y=  

q  

2

 

1

=9

(iii)

x

4

+3y=14

   

x

3

−4y=23

Lets assume  

x

1

=p, both equations become

4p+3y=14   ...(A)

3p−4y=23   ...(B)

Multiply  (A) by 4 and (B) by 3, we get

16p+12y=56

9p−12y=69

Adding these both, we get

25p=125

⇒p=5

⇒x=  

p

1

=  

5

1

 

⇒y=−2

(iv)  

x−1

5

+  

y−2

1

=2

x−1

6

−  

y−2

3

=1

Lets assume  

x−1

1

=p and  

y−2

1

=q, both equations become

5p+q=2

6p−3q=1

Multiply (A) by 3, we get

15p+3q=6

6p−3q=1

Adding both of these, we get

21p=7

⇒p=  

3

1

 

⇒   x=4

∴q=  

3

1

 

⇒   y=5

(v)

xy

7x−2y

=5

⇒  

y

7

−  

x

2

=5

xy

8x+7y

=15

⇒    

y

8

+  

x

7

=15

Let t=  

x

1

 and r=  

y

1

 equation will be

7r−2t=5;8r+7t=15

On solving we get

t=1r=1

∴x=1andy=1

(vi)

Dividing both side by xy then we have

y

6

+  

x

3

=6;  

y

2

+  

x

4

=5

Let t=  

x

1

andr=  

y

1

 equation will be

6r+3t=6;2r+4t=5 on solving we get

t=1r=  

2

1

 

∴x=1andy=2

(vii)

Let t=  

x+y

1

andr=  

x−y

1

 then equation will be

10t+2r=4;15t−5r=−2 on solving we have

t=  

5

1

andr=1

∴x+y=5andx−y=1

Hence, x=4,y=1

(viii)

Let t=  

3x+y

1

andr=  

3x−y

1

 then equation will be

t+r=  

4

3

;  

2

t

−  

2

r

=  

8

−1

 

On solving we have

t=  

4

1

andr=  

2

1

 

∴3x+y=4and3x−y=2

Hence, x=1,y=1

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