Question

Solve the following equations:
$xy+ab=2ax,x^2{y}^2+a^2{b}^2=2b^2{y}^2$.

• A. $(b,a)$
• B. $(a,b)$
• C. $(-2a,b)$
• D. $(a,-b)$

Answer 1

The correct option is A $(b,a)$

$x^2{y}^2+a^2{b}^2=2b^2{y}^2$ ......(i)

$xy+ab=2ax$ ......(ii)

Squaring both sides, we get

$(xy+ab{)}^2=(2ax{)}^2$

$\Rightarrow x^2{y}^2+a^2{b}^2+2abxy=4a^2{x}^2$ .....(iii)

Substituting value of (i) in equation (iii),

$2b^2{y}^2+2abxy-4a^2{x}^2=0\Rightarrow b^2{y}^2+abxy-2a^2{x}^2=0\Rightarrow b^2{y}^2-abxy+2abxy-2a^2{x}^2=0\Rightarrow by(by-ax)+2ax(by-ax)=0\Rightarrow (by+2ax)(by-ax)=0\Rightarrow by=-2ax,ax\Rightarrow y=-\frac{2ax}{b},\frac{ax}{b}$

Substituting $y$ in (ii),

(a) $y=-\frac{2ax}{b}$

$x(-\frac{2ax}{b})+ab=2ax2a{x}^2+2abx-a{b}^2=0x=\frac{-2ab\pm \sqrt{4a^2{b}^2-4\left(2a\right)(-a{b}^2)}}{2\left(2a\right)}x=\frac{-2ab\pm \sqrt{12}ab}{4a}x=\frac{-b\pm \sqrt{3}b}{2}y=-\frac{2ax}{b}\Rightarrow y=a\mp \sqrt{3}a$

(b) $y=\frac{ax}{b}$

$x\left(\frac{ax}{b}\right)+ab=2axa{x}^2-2abx+a{b}^2=0a{x}^2-abx-abx+a{b}^2=0ax(x-b)-ab(x-b)=0(ax-ab)(x-b)=0x=b,by=\frac{a\left(b\right)}{b}=a$

So, the values of $x$ are $\frac{-b\pm \sqrt{3}b}{2},b$ and the corresponding values of $y$ are $a\mp \sqrt{3}a,a$.