Math, asked by divyangana56, 9 months ago

solve 1/2x-1/y=-1 , 1/x+1/2x=8 by reducing it to a pair of linear equation ​

Answers

Answered by sahilsinghhaz12394
0

Step-by-step explanation:

1

Secondary School 

 

Math 

 

5 points

Solve by reducing them to a pair of linear equations. 1/2x + 1/3y =2 and 1/3x + 1/2y = 13/6

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 bySwetasingh57 20.08.2019

Answers

rihan99 

 

Virtuoso

Answer:

the value of x=2 and y=3

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THE BRAINLIEST ANSWER!

ashu3998825 

 

Expert

Answer:

let 1/x = a and 1/y= b

1/2x + 1/3y =2

a/2 + b/3 = 2

3a + 2b = 12 .....…......(1)

1/3x +1/2y =13/6

a/3 + b/ 2 = 13/6

2a +3b = 13 ........…......(2)

from equation (1 ) we have,

a = (12-2b)/3 ...….......(3)

substituting the value of a in equation (2)

2(12-2b)/3 + 3b =13

24 - 4b +9b = 39

5b = 39-24 =15

b = 15/5 = 3

b=3

from equation (3),

a = (12- 2× 3 )/3

a = 6/3 = 2

a=2

1/x = a => 1/a = x

x = 1/2

1/y = b => 1/b = y

y = 1/3

Answered by gyanendramahnar
0

Answer:

The value of x and y is \frac{1}{6}\ {and}\ \frac{1}{4}

6

1

and

4

1

.

To find:

Solve the Equation: \frac{1}{2 x}-\frac{1}{y}=-1 \text { and } \frac{1}{x}+\frac{1}{2 y}=8=?

2x

1

y

1

=−1 and

x

1

+

2y

1

=8=?

Solution:

Given: \frac{1}{2 x}-\frac{1}{y}=-1 \text { and } \frac{1}{x}+\frac{1}{2 y}=8

2x

1

y

1

=−1 and

x

1

+

2y

1

=8

\frac{1}{2 x}-\frac{1}{y}=-1

2x

1

y

1

=−1

\frac{y-2 x}{2 x y}=-1

2xy

y−2x

=−1

y-2 x=-2 x yy−2x=−2xy

2 \mathrm{x}-\mathrm{y}=2 \mathrm{xy} \ldots \ldots \ldots (1)2x−y=2xy………(1)

\frac{1}{x}+\frac{1}{2 y}=8

x

1

+

2y

1

=8

\frac{2 y+x}{2 x y}=8

2xy

2y+x

=8

2 y+x=16 x y2y+x=16xy

x+2 y=16 x y \dots \ldots \ldots \ldots(2)x+2y=16xy…………(2)

Multiplying equation (1) by 2 and then adding it to equation (2), we get

5 \mathrm{x}=20 \mathrm{xy}5x=20xy

y=\frac{1}{4}y=

4

1

Replacing the value of y=\frac{1}{4}y=

4

1

in equation (1),

2 x-y=2 x y2x−y=2xy

2 x-\frac{1}{4}=2 x y2x−

4

1

=2xy

8 x-1=\frac{8 x y}{4}8x−1=

4

8xy

6 \mathrm{x}=16x=1

x=\frac{1}{6}x=

6

1

please mark me as brainiest

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