Question

Solve the following systems of equations.
$x+y=7,$ $y=\frac{6}{x}$ and $x>y$.

Answer 1

Given value is $x+y=7$ and $xy=6$

now we have ,

$(x-y{)}^2=(x+y{)}^2-4xy$

$(x-y{)}^2=(7{)}^2-4\times 6$

$(x-y{)}^2=49-24$

$(x-y{)}^2=25$

$(x-y)=\pm 5$

But $x>y$

$\Rightarrow (x-y)=5$

Now we have $(x+y)=7and(x-y)=5$ add both the equation

so, $2x=12$

$x=6$

$x+y=7$

$y=7-6=1$

so $x=6andy=1$ hence this the required solution