Question

Solve the following systems of equations.
5(x + y) + 2xy = -19, 3xy + x + y = -35

Answer 1

$5(x+y)+2xy=-\mathrm{19...}\left(i\right)$

$3xy+x+y=-\mathrm{35...}\left(ii\right)$

Simplifying equation (i)

$5(x+y)+2xy=-19$

or, $5(x+y)=-19-2xy$

or, $x+y=\frac{-19-2xy}{5}...\left(iii\right)$

Putting this value of $x+y$ in equation (ii)

$3xy+(x+y)=-35$

or, $3xy+\left(\frac{-19-2xy}{5}\right)=-35$

or, $\frac{15xy-19-2xy}{5}=-35$

or, $\frac{13xy-19}{5}=-35$

or, $13xy-19=-35\times 5$

or, $13xy=-175+19$

or, $xy=\frac{-156}{13}$

or, $xy=-\mathrm{12...}\left(iv\right)$

Hence, every pair of $x$ of $y$ will be solution of

this equation which satisfy equation (iv).i.e, there

will be infinite solutions.