1 of 1

(iv)

x

2

+

2y

3

and

x

y

3

- 3

= 0

Soln.

Elimination method

x

2

+

2y

3

=

-1

Multiplying throughout by 6 in eq (i), we get

3x

+

4y

=

-6

... (i)

x

y

3

=

3

Multiplying throughout by 3 in eq (ii), we get

3x

y

=

9

Subtracting eqn (iv) from eqn (iii)

... (ii)

3x

+

4y

=

-6

... (iii)

3x

y

=

9

... (iv)

+

5y

=

-15

y

=

-15

5

y

=

-3

Substituting y = – 3 in eqn (iii), we get

3x

+

4(-3)

=

-6

3x

12

=

-6

3x

=

-6

+

12

3x

=

6

x

=

2

Solution is x = 2, y = – 3

+1 = 0

To remove ‘2’ & ‘3’ from denominator multiply by LCM of 2 & 3

LCM of 2 & 3 is 6

To remove ‘3’ from denominator multiply by LCM of 3

LCM of 3 is 3

... (iii)

... (iv)

Are the Coefficient of X is same

Yes !!

Are the Coefficient of Y is same

No !!

1

3x

+

4y

=

-6

... (iii)

Now which two

equations have

to be solved ??

equations (iii) & (iv)…

+

+

x

=

6

3

2

Constant → R.H.S.

− 12 → 12

In eliminition method, keep variable on L.H.S and constant on R.H.S

How to get the value of ‘x’ ?

Either equation (iii) or equation (iv)

We need to either remove x or remove y

Which variable can be removed ?

Whichever variable’s coefficient is same

Here coefficent of

x is same

To remove x, we need to subtract

As Signs are same

Number the equation as equation (iii)

Number the equation as equation (iv)