Question

Solve the following pair of linear equations by the substitution method.

`sqrt2x + sqrt3y = 0`

`sqrt3x - sqrt8y = 0`

Answer 1

`sqrt2x + sqrt3y = 0`     ...(i)

`sqrt3x - sqrt8y = 0`      ...(ii)

From equation (i)

`sqrt2x + sqrt3y = 0`

or `sqrt2x = -sqrt3y`

or x = `-sqrt3/2`

Now on putting the value of x in equation (ii)

`sqrt3x - sqrt8y = 0`

or `sqrt3(-(sqrt3y)/sqrt2) - sqrt8y = 0`

or `-3y - sqrt16y = 0`

or -3y – 4y = 0

or -7y = 0

or y = 0

Now putting y = 0 in equation (i)

x = `-(sqrt3y)/sqrt2`

or x = `(sqrt3(0))/sqrt2 = 0`

Therefore, the solution of the given pair of linear equations is

x = 0 and y = 0