What is the value of $2x+3y$?

Question

What is the value of $2x+3y$ if

$x+y=6$ & $x^2+3xy+2y=60$ ?

My trial: from given conditions: substitute $y=6-x$ in $x^2+3xy+2y=60$ $$x^2+3x(6-x)+2(6-x)=60$$ $$x^2-8x+24=0$$ $$x=\frac{8\pm\sqrt{8^2-4(1)(24)}}{2(1)}=4\pm2i\sqrt2$$ this gives us $y=2\mp2i\sqrt2$ we now have $x=4+2i\sqrt2, y=2-2i\sqrt2$ or $x=4-2i\sqrt2, y=2+2i\sqrt2$

substituting these values i got $2x+3y=14-2i\sqrt2$ or $$2x+3y=14+2i\sqrt2$$

But my book suggests that $2x+3y$ should be a real value that I couldn't get. Can somebody please help me solve this problem? Is there any mistake in the question.?

Thank you.

Answer 1

we get $$x^2+3x(6-x)+2(6-x)=60$$ simplifying we obtain $$-2x^2+16x-48=0$$ or $$x^2-8x+24=0$$ can you solve this? you will get $$x_1=4+2\sqrt{2}i$$ or $$x_2=4-2\sqrt{2}i$$ and and $$y_1=2-2\sqrt{2}i$$ $$y_2=2+2\sqrt{2}i$$ so $x+y=6$

Answer 2

It seems correct to me

$$x^2+3x(6-x)+2(6-x)=60\iff x^2+18x-3x^2+12-2x-60=0$$ $$\iff-2x^2+16x-48=0\iff x^2-8x+24=0\implies x=4\pm2i\sqrt2$$

Answer 3

Assuming a typo ($y$ instead of $y^2$), we restore

$$\begin{cases}x+y=6,\\x^2+3xy+2y^2=(x+y)(x+2y)=60,\end{cases}$$ then

$$2x+3y=\frac{60}6+6.$$