Question

5. Solve the given simultaneous linear equation using elimination method.

+ 2 = −4 3 − 5 = −1

Answer 1

Answer:

The given equations are

x+y

5

−

x−y

2

=−1 ...........(i)

x+y

15

+

x−y

7

=10 .....(ii)

Let

x+y

1

=a,

x−y

1

=b

⇒5a−2b=−1 .........(iii)

15a+7b=10 ...........(iv)

Multiplying (iii) by 7 and (iv) by 2

35a−14b=−7

30a+14b=20

and then adding, we get

65a=13

⇒a=

65

13

=

5

1

Putting a=

5

1

in (iii), we get

5×

5

1

−2b=−1

⇒2=2b⇒b=1

Now

x+y

1

=

5

1

⇒x+y=5 .........(v)

x−y

1

=1

⇒x−y=1 .......(vi)

Adding (v) and (vi), we get

2x=6

⇒x=3 and 3x+y=5

⇒y=2

Hence the required solution is x=3 and y=2.