Answer :
x, y - the numbers
The product is 18.
[tex]xy=18[/tex]
The quotient is 2.
[tex]\frac{x}{y}=2 \ \ \ |\times y \\ x=2y[/tex]
Substitute 2y for x in the 1st equation:
[tex]2y \times y=18 \\ 2y^2=18 \ \ \ |\div 2 \\ y^2=9 \\ y=\pm \sqrt{9} \\ y=3 \ \lor \ y=-3 \\ \\ x=2y \\ x=2 \times 3 \ \lor \ x=2 \times (-3) \\ x=6 \ \lor \ x=-6 \\ \\ (x,y)=(6,3) \hbox{ or } (x,y)=(-6,-3)[/tex]
The numbers are 6 and 3 or -6 and -3.
The product is 18.
[tex]xy=18[/tex]
The quotient is 2.
[tex]\frac{x}{y}=2 \ \ \ |\times y \\ x=2y[/tex]
Substitute 2y for x in the 1st equation:
[tex]2y \times y=18 \\ 2y^2=18 \ \ \ |\div 2 \\ y^2=9 \\ y=\pm \sqrt{9} \\ y=3 \ \lor \ y=-3 \\ \\ x=2y \\ x=2 \times 3 \ \lor \ x=2 \times (-3) \\ x=6 \ \lor \ x=-6 \\ \\ (x,y)=(6,3) \hbox{ or } (x,y)=(-6,-3)[/tex]
The numbers are 6 and 3 or -6 and -3.
[tex]\begin{cases} x*y=18 \\ \frac{x}{y}=2 \end{cases}\\\\\begin{cases} 2y*y=18 \\ x=2y \end{cases}\\\\ \begin{cases} 2y^2=18\ \ /:2 \\ x=2y \end{cases}\\\\\begin{cases} y^2=9 \\ x=2y \end{cases}\\\\ y^2-9=0\\\\(y-3)(y+3)=0\\y-3=0\ \ \vee \ \ y+3=0\\y=3 \ \ \vee \ \ y=-3[/tex]
[tex]\begin{cases} y=3 \\ x=2y \end{cases}\ \ \vee \ \ \ \ \begin{cases} y=-3 \\ x=2y \end{cases}\\\\ \begin{cases} y=3 \\ x=2*3 \end{cases}\ \ \vee \ \ \ \ \begin{cases} y=-3 \\ x=2*(-3) \end{cases}\\\\ \begin{cases} y=3 \\ x=6 \end{cases}\ \ \vee \ \ \ \ \begin{cases} y=-3 \\ x=-6\end{cases}\\\\[/tex]
[tex]\begin{cases} y=3 \\ x=2y \end{cases}\ \ \vee \ \ \ \ \begin{cases} y=-3 \\ x=2y \end{cases}\\\\ \begin{cases} y=3 \\ x=2*3 \end{cases}\ \ \vee \ \ \ \ \begin{cases} y=-3 \\ x=2*(-3) \end{cases}\\\\ \begin{cases} y=3 \\ x=6 \end{cases}\ \ \vee \ \ \ \ \begin{cases} y=-3 \\ x=-6\end{cases}\\\\[/tex]
