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Let $x$ and $y$ be the desired real numbers. Using the given information, we can write two equations, one for the sum of the numbers and one for the product of the numbers.
$Sum:Product: x+y=-13xy=42 $
We can solve this system of equations using the Substitution Method. First, let's isolate $y$ in the first equation and substitute it into the second equation.
The related quadratic equation can be written as $x_{2}+13x+42=0.$ Our next step is to solve the quadratic equation.
We can do this by graphing. Let's begin by finding the axis of symmetry using the formula $x=-2ab .$ For our equation, $a=1$ and $b=13.$
Next, we will make a table of values using $x$ values around the axis of symmetry which is $x=-6.5.$

${x+y=-13xy=42 (I)(II) $

${y=-13−xxy=42 $

${y=-13−xx(-13−x)=42 $

${y=-13−x-13x−x_{2}=42 $

${y=-13−x-13x−x_{2}−42=0 $

CommutativePropAdd$(II):$ Commutative Property of Addition

${y=-13−x-x_{2}−13x−42=0 $

ChangeSigns$(II):$ Change signs

${y=-13−xx_{2}+13x+42=0 $

$x=-2ab $

$x=-2(1)13 $

$x=-6.5$

$x$ | $x_{2}+13x+42$ | $y$ |
---|---|---|

$-11$ | $(-11)_{2}+13(-11)+42$ | $20$ |

$-10$ | $(-10)_{2}+13(-10)+42$ | $12$ |

$-6.5$ | $(-6.5)_{2}+13(-6.5)+42$ | $-0.25$ |

$-3$ | $(-3)_{2}+13(-3)+42$ | $12$ |

$-2$ | $(-2)_{2}+13(-2)+42$ | $20$ |

Now, we will plot the points and graph the equation that passes through the points.

It seems like the zeros of the equation are $-7$ and $-6.$ Let's investigate if they satisfy the original problem. $Sum:Product: -7+(-6)=-13-7⋅(-6)=42 ✓✓ $ Since the numbers satisfy both requirements stated in the exercise we have found the numbers we are looking for, and they are $-7$ and $-6.$