10/x+y+2/x-y=4; 15/x+y-5/x-y=-2 Solve the simultaneous equation
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SOLUTION IS IN THE ATTACHMENT..
Sometimes equation are not linear but they can be reduced to a pair of linear equation by making some suitable substitutions.
•if the given equation involves 1/(x+y), 1/(x-y) , then put 1/(x+y)= p and 1/(x -y) = q to convert them into linear form. After solving, put the values of p and q in above substitutions , to get linear equations in x and y. Now solve these equations to get the value of x and y.
HOPE THIS WILL HELP YOU….
Sometimes equation are not linear but they can be reduced to a pair of linear equation by making some suitable substitutions.
•if the given equation involves 1/(x+y), 1/(x-y) , then put 1/(x+y)= p and 1/(x -y) = q to convert them into linear form. After solving, put the values of p and q in above substitutions , to get linear equations in x and y. Now solve these equations to get the value of x and y.
HOPE THIS WILL HELP YOU….
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Given equations are ,
10/( x + y ) + 2/( x - y ) = 4 ---( 1 )
15/( x + y ) - 5/( x - y ) = -2 --( 2 )
Let ,
1/(x+y) = a , 1/(x-y) = b
10a + 2b = 4
Divide each term with 2 , we get
5a + b = 2
=> b = 2 - 5a ----( 3 )
15a - 5b = -2 -----( 4 )
Substitute b = 2 - 5a in equation
( 4 ) , we get
15a - 5( 2 - 5a ) = -2
=> 15a - 10 + 25a = -2
=> 40a = -2 + 10
=> 40a = 8
=> a = 8/40
=> a = 1/5
Put a = 1/5 in equation ( 3 ) , we
get
b = 2 - 5 × 1/5
=> b = 2 - 1
b = 1
Therefore ,
1/( x + y ) = 1/5 => x + y = 5 --( 5 )
1/(x-y) = 1/1 => x - y = 1 ---( 6 )
Add equations ( 5 ) and ( 6 ) ,
We get
2x = 6
=> x = 6/2 = 3 ,
Put x = 3 in equation ( 5 ) , we get
3 + y = 5
=> y = 5 - 3 = 2
Therefore ,
x = 3 , y = 2
••••
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