Question

Solve the following systems of equations:

`x + 2y = 3/2`

`2x + y = 3/2`

The given system of equation is

`x + 2y = 3/2` ..`.(i)

`2x + y = 3/2`  ...(ii)

Let us eliminate y from the given equations. The Coefficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.

Multiplying (i) by 1 and (ii) by 2, we get

`x + 2y = 3/2` ...(iii)

4x + 2y = 3 ....(iv)

Subtracting (iii) from (iv), we get

`4x  - x + 2y - 2y = 3 - 3/2`

`=> 3x = (6 - 3)/2`

`=> 3x = 3/2`

`=> x = 3/(2 xx3)`

`=> x = 1/2`

Putting  x = 1/2 in equation (iv), we get

`4 xx 1/2 + 2y = 3`

`=> 2 + 2y = 3`

`=> 2 + 2y = 3`

`=> 2y = 3- 2`

`=> y  =1/2`

Hence, solution of the given system of equation is `x = 1/2` and `y = 1/2`