Question

solve the following pair of equations by reducing them to a pair of linear equations
1) 1/3x+y+1/3x-y =3/4
1/2(3x+y)-1/2(3x-y)=-1/8
please answer this ​

Answer 1

Given:

(3x+y)

1

+

(3x−y)

1

=

4

3

,

2(3x+y)

1

+

2(3x−y)

1

=

8

−1

let

3x+y

1

=aand

3x−y

1

=b

Thena+b=

4

3

⇒4a+4b=3........eq1

2

a

−

2

b

=

8

−1

2

a−b

=

8

−1

a−b=

8

−2

⇒a−b=

4

−1

4a−4b=−1..............eq2

Adding eq1 and eq2 we get

8a=2

a=

8

2

⇒

4

1

Substituting the value of a in eq1 we get

4×

4

1

+4b=3

4b=2

b=

4

2

⇒

2

1

Then

3x+y

1

=

4

1

3x+y=4................eq3

3x−y

1

=

2

1

3x−y=2...............eq4

Adding eq3 and eq4 we get

6x=6

x=1

Substituting x=1 in eq3 we get

3×1+y=4

y=1

Thus x=1 and y=1