Find answers to your most challenging questions with the help of IDNLearn.com's experts. Our community is here to provide detailed and trustworthy answers to any questions you may have.

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

[tex]\[
\begin{array}{l}
y = x^2 - 2x - 19 \\
y + 4x = 5
\end{array}
\][/tex]

The pair of points representing the solution set of this system of equations is [tex]$(-6, 29)$[/tex] and [tex]$\square$[/tex].


Sagot :

To solve the system of equations:

[tex]\[ \begin{array}{l} y = x^2 - 2x - 19 \\ y + 4x = 5 \end{array} \][/tex]

we follow these steps:

1. Express [tex]\( y \)[/tex] from the second equation:
[tex]\[ y = 5 - 4x \][/tex]

2. Substitute [tex]\( y \)[/tex] from the second equation into the first equation:
[tex]\[ 5 - 4x = x^2 - 2x - 19 \][/tex]

3. Rearrange into standard quadratic form:
[tex]\[ x^2 - 2x - 19 - 4x + 5 = 0 \\ x^2 - 6x - 14 = 0 \][/tex]

4. Solve the quadratic equation ([tex]\(x^2 - 6x - 14 = 0\)[/tex]):
The quadratic formula is [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. For the equation [tex]\( x^2 - 6x - 14 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -14 \)[/tex].

[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \\ x = \frac{6 \pm \sqrt{36 + 56}}{2} \\ x = \frac{6 \pm \sqrt{92}}{2} \\ x = \frac{6 \pm 2\sqrt{23}}{2} \\ x = 3 \pm \sqrt{23} \][/tex]

The numerical solutions can be calculated as:
- [tex]\( x_1 = -6 \)[/tex] (part of the known solution)
- [tex]\( x_2 = 4 \)[/tex] (the other unknown solution)

5. Find the corresponding [tex]\( y \)[/tex] value for [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 5 - 4x \\ y = 5 - 4(4) \\ y = 5 - 16 \\ y = -11 \][/tex]

Therefore, the solution set includes the points [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].

So, the complete pair of points representing the solution set of the system of equations is [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].

The correct answer is [tex]\( \boxed{(4, -11)} \)[/tex].