Answer :
To find the pair of points representing the solution set of the system of equations:
1. First, we have the equations:
[tex]\[ y = x^2 - 2x - 19 \][/tex]
[tex]\[ y + 4x = 5 \][/tex]
2. We substitute the expression for [tex]\(y\)[/tex] from the second equation into the first equation. From the second equation, solve for [tex]\(y\)[/tex]:
[tex]\[ y = 5 - 4x \][/tex]
3. Substitute [tex]\( y = 5 - 4x \)[/tex] into the first equation:
[tex]\[ 5 - 4x = x^2 - 2x - 19 \][/tex]
4. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 19 - 5 + 4x = 0 \][/tex]
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
5. Factorize or use the quadratic formula to solve for [tex]\(x\)[/tex]. The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -24 \)[/tex].
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 96}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{8}{2} = 4 \][/tex]
[tex]\[ x = \frac{-12}{2} = -6 \][/tex]
6. Substitute these [tex]\(x\)[/tex]-values back into the second equation to find the corresponding [tex]\(y\)[/tex]-values:
For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 5 - 4(4) = 5 - 16 = -11 \][/tex]
For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = 5 - 4(-6) = 5 + 24 = 29 \][/tex]
The pair of points representing the solutions to the system of equations are [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].
Therefore, the missing pair of points is:
[tex]\[ (4, -11) \][/tex]
1. First, we have the equations:
[tex]\[ y = x^2 - 2x - 19 \][/tex]
[tex]\[ y + 4x = 5 \][/tex]
2. We substitute the expression for [tex]\(y\)[/tex] from the second equation into the first equation. From the second equation, solve for [tex]\(y\)[/tex]:
[tex]\[ y = 5 - 4x \][/tex]
3. Substitute [tex]\( y = 5 - 4x \)[/tex] into the first equation:
[tex]\[ 5 - 4x = x^2 - 2x - 19 \][/tex]
4. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 19 - 5 + 4x = 0 \][/tex]
[tex]\[ x^2 + 2x - 24 = 0 \][/tex]
5. Factorize or use the quadratic formula to solve for [tex]\(x\)[/tex]. The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -24 \)[/tex].
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 96}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ x = \frac{-2 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{8}{2} = 4 \][/tex]
[tex]\[ x = \frac{-12}{2} = -6 \][/tex]
6. Substitute these [tex]\(x\)[/tex]-values back into the second equation to find the corresponding [tex]\(y\)[/tex]-values:
For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 5 - 4(4) = 5 - 16 = -11 \][/tex]
For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = 5 - 4(-6) = 5 + 24 = 29 \][/tex]
The pair of points representing the solutions to the system of equations are [tex]\((-6, 29)\)[/tex] and [tex]\((4, -11)\)[/tex].
Therefore, the missing pair of points is:
[tex]\[ (4, -11) \][/tex]
