Respuesta :
Let be "x" and "y" the two positive numbers mentioned in the exercise.
You need to remember the following definitions:
- A Difference is the result of a Subtraction.
- A Product is the result of a Multiplication.
In this case, you know that their Difference is 12 and their Product is 693. So you can set up this System of Equations:
[tex]\begin{cases}x-y=12 \\ xy=693\end{cases}[/tex]You can solve the System of Equation using the Substitution Method:
1. You can solve for the variable "x" from the first equation:
[tex]x=12+y[/tex]2. Substitute this new equation into the second original equation and simplify:
[tex]\begin{gathered} xy=693 \\ (12+y)y=693 \\ y^2+12y=693 \end{gathered}[/tex]3. Now you have to solve for "y", in order to find its value:
- Rewrite the equation in this form:
[tex]ax^2+bx+c=0[/tex]Since it is the variable "y":
[tex]ay^2+by+c=0[/tex]Then:
[tex]y^2+12y-693=0[/tex]4. Use the Quadratic Formula to find the values of "y". The formula would be:
[tex]y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]In this case:
[tex]\begin{gathered} a=1 \\ b=12 \\ c=-693 \end{gathered}[/tex]Then, substituting values and evaluating, you get:
[tex]\begin{gathered} y=\frac{-12\pm\sqrt[]{(12)^2-(4)(1)(-693)}}{(2)(1)} \\ \\ y_1=-33 \\ y_2=21 \end{gathered}[/tex]5. Since the numbers mentioned in the exercise are both positive, you must choose only the positive values. Therefore:
[tex]y=21[/tex]6. To find the value of "x", you have to substitute the value of "y" into the equation
[tex]x=12+y[/tex]Then:
[tex]x=12+(21)[/tex]7. Finally, you have to solve the Addition:
[tex]x=33[/tex]The answer is:
[tex]33\text{ and }21[/tex]
