Question

Solve the following systems of equations.
$\frac{1}{x}+\frac{1}{y}=\frac{1}{3},x^2+y^2=160$

Answer 1

$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}...\left(i\right)$

$x^2+y^2=\mathrm{160...}\left(ii\right)$

Solving equation (i)

$\frac{1}{x}+\frac{1}{y}=\frac{1}{3}$

or, $\frac{y+x}{xy}=\frac{1}{3}$

or, $x+y=\frac{xy}{3}...\left(iii\right)$

Simplifying equation (ii)

$x^2+y^2=160$

or, $x^2+y^2=160$

or, $x^2+y^2+2xy-2xy=160$

$(x+y{)}^2-2xy=160$ $[a^2+b^2+2ab=(a+b{)}^2]$

Putting the value of $(x+y)$ from equation (iii)

${\left(\frac{xy}{3}\right)}^2-2xy=160$

or, ${\left(\frac{xy}{3}\right)}^2-2xy=160$

or, ${\left(\frac{xy}{3}\right)}^2-2\times \frac{xy}{3}\times 3+3^2-3^2-160=0$

or ${(\frac{xy}{3}-3)}^2-9-160=0$

or, ${(\frac{xy}{3}-3)}^2=169$

or $\frac{xy}{3}-3=\pm \sqrt{169}$

or, $\frac{xy}{3}-3=\pm 13$

when $\frac{xy}{3}-3=+13$

or, $\frac{xy}{3}=13+3$

or, $xy=16\times 3$

or, $xy=48....\left(iv\right)$

when $\frac{xy}{3}-3=-13$

or $\frac{xy}{3}=-13+3$

or, $xy=-10\times 3$

or, $xy=-30...\left(v\right)$

Hence, the solution of the given pair of equations

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

of the equation (iv) or (v)

So, there will be infinite solutions.