If $$y \ne 0$$ then the number of values of the pair $$(x, y)$$ such that $$x + y + \frac{x}{y} = \frac{1}{2}$$ and $$\left( {x + y} \right)\frac{x}{y} = - \frac{1}{2},$$ is
A.
1
B.
2
C.
0
D.
none of these
Answer :
2
Solution :
$$xy + {y^2} + x = \frac{y}{2},{x^2} + xy = - \frac{y}{2}.$$
Adding, $${\left( {x + y} \right)^2} + x = 0,$$ and subtracting $${y^2} - {x^2} = y - x.$$
Solving these equations, $$x = - 1, y = 2$$ and $$x = - \frac{1}{4} = y.$$
Releted MCQ Question on Algebra >> Quadratic Equation
Releted Question 1
If $$\ell ,m,n$$ are real, $$\ell \ne m,$$ then the roots by the equation: $$\left( {\ell - m} \right){x^2} - 5\left( {\ell + m} \right)x - 2\left( {\ell - m} \right) = 0$$ are