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प्रश्न
Solve the following pair of linear equations by the substitution method.
`sqrt2x + sqrt3y = 0`
`sqrt3x - sqrt8y = 0`
उत्तर
`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
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