Question

Solve the following equations:
$9x^2+33x-12=12xy-4y^2+22y,x^2-xy=18$.

• A. $-6,-3$
• B. $2,4$
• C. $\frac{1\pm \sqrt{-143}}{2},\frac{1\pm 3\sqrt{-143}}{4}$
• D. $-5,-9$

Answer 1

Answer: (A), (C)

$-6,-3$
$\frac{1\pm \sqrt{-143}}{2},\frac{1\pm 3\sqrt{-143}}{4}$

Given, $x^2-xy=18$ ........(i)

Also given $9x^2+33x-12=12xy-4y^2+22y$

$\Rightarrow 9x^2+4y^2-12xy+33x-22y-12=0\Rightarrow {(3x-2y)}^2+11(3x-2y)-12=0\Rightarrow {(3x-2y)}^2-(3x-2y)+12(3x-2y)-12=0\Rightarrow (3x-2y)\left(\right(3x-2y)-1)+12\left(\right(3x-2y)-1)=0\Rightarrow (3x-2y+12)(3x-2y-1)=0\Rightarrow 3x-2y-1=0\Rightarrow y=\frac{3x-1}{2}$

Substituting in (i), we get

$x^2-x\left(\frac{3x-1}{2}\right)=18\Rightarrow 2x^2-3x^2+x=36\Rightarrow x^2-x+36=0\Rightarrow x=\frac{1\pm \sqrt{1-4\left(1\right)\left(36\right)}}{2}\Rightarrow x=\frac{1\pm \sqrt{-143}}{2}\Rightarrow y=\frac{3x-1}{2}\Rightarrow y=\frac{3}{2}\left(\frac{1\pm \sqrt{-143}}{2}\right)-\frac{1}{2}\Rightarrow y=\frac{1\pm 3\sqrt{-143}}{4}$

Also $3x-2y+12=0$

$\Rightarrow y=\frac{3x+12}{2}$

Substituting in (i), we get

$x^2-x\left(\frac{3x+12}{2}\right)=18\Rightarrow 2x^2-3x^2-12x=36\Rightarrow x^2+12x+36=0\Rightarrow {(x+6)}^2=0\Rightarrow x=-6$

Then $y=\frac{3(-6)+12}{2}=-3$

So, the values of $x$ are $\frac{1\pm \sqrt{-143}}{2},-6$ and the corresponding values of $y$ are $\frac{1\pm 3\sqrt{-143}}{4},-3$.