Question

Solve the following systems of equations.
x + y + xy = 5, $x^2+y^2+xy=7$

Answer 1

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

$x^2+y^2+xy=7...\left(ii\right)$

we know that $a^2+b^2+2ab=(a+b{)}^2$

Simplifying equation (ii)

$x^2+y^2+xy=7$

or, $x^2+y^2+2xy-xy=7$

or, $(x+y{)}^2-xy=\mathrm{7...}(iii)$

Simplifying equation (i)

$x+y+xy=5$

or, $x+y=5-xy...\left(iv\right)$

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

$(5-xy{)}^2-xy=7$

or, $25+x^2{y}^2-10xy-xy=7$

or, $x^2{y}^2-11xy+25-7=0$

or $x^2{y}^2-11xy+18=0$

or, $x^2{y}^2-2\times xy\times \frac{11}{2}+(\frac{11}{2}{)}^2-(\frac{11}{2}{)}^2+18=0$

or, $(xy-\frac{11}{2}{)}^2-\frac{121}{4}+18=0$

or, $(xy-\frac{11}{2}{)}^2-\frac{49}{4}$

or, $(xy-\frac{11}{2})=\pm \sqrt{\frac{49}{4}}$

or, $xy-\frac{11}{2}=\pm \frac{7}{2}$

When $xy-\frac{11}{2}=+\frac{7}{2}$

$xy=\frac{7}{2}+\frac{11}{2}$

$xy=\frac{18}{2}$

$xy=9...\left(v\right)$

When $xy-\frac{11}{2}=\frac{-7}{2}$

$xy=\frac{11}{2}-\frac{7}{2}$

$xy=\frac{y}{2}$

$xy=2....\left(vi\right)$

Hence, the solution of the given pair of equations

will be every pair of $x$ & $y$ which satisfies any of

the equations (v) or (vi)

So, there will be infinite solutions.