IDNLearn.com: Where your questions meet expert advice and community support. Find the answers you need quickly and accurately with help from our knowledgeable and experienced experts.
Sagot :
To find the solutions to the system of equations:
1. [tex]\( y = (x + 4)^2 - 1 \)[/tex]
2. [tex]\( y = -3x - 13 \)[/tex]
we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations.
### Step-by-Step Solution:
1. Set the equations equal to each other:
Since both expressions equal [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ (x + 4)^2 - 1 = -3x - 13 \][/tex]
2. Simplify the equation:
Expand and simplify the equation:
[tex]\[ (x + 4)^2 - 1 = -3x - 13 \][/tex]
[tex]\[ x^2 + 8x + 16 - 1 = -3x - 13 \][/tex]
[tex]\[ x^2 + 8x + 15 = -3x - 13 \][/tex]
Move all terms to one side:
[tex]\[ x^2 + 11x + 28 = 0 \][/tex]
3. Solve the quadratic equation:
We solve the quadratic equation [tex]\( x^2 + 11x + 28 = 0 \)[/tex].
4. Roots of the quadratic equation:
The roots of the quadratic equation [tex]\(x^2 + 11x + 28 = 0\)[/tex] are [tex]\(x = -4\)[/tex] and [tex]\(x = -7\)[/tex].
5. Calculate the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x = -4 \)[/tex]:
[tex]\[ y = (x + 4)^2 - 1 \][/tex]
[tex]\[ y = (-4 + 4)^2 - 1 \][/tex]
[tex]\[ y = 0^2 - 1 \][/tex]
[tex]\[ y = -1 \][/tex]
For [tex]\( x = -7 \)[/tex]:
[tex]\[ y = (x + 4)^2 - 1 \][/tex]
[tex]\[ y = (-7 + 4)^2 - 1 \][/tex]
[tex]\[ y = (-3)^2 - 1 \][/tex]
[tex]\[ y = 9 - 1 \][/tex]
[tex]\[ y = 8 \][/tex]
Therefore, the points that satisfy both equations are [tex]\((-4, -1)\)[/tex] and [tex]\((-7, 8)\)[/tex].
6. Check the solutions against the given options:
- [tex]\((-7, -4)\)[/tex] and [tex]\((8, -1)\)[/tex] - No, these do not match our solutions.
- [tex]\((-7, 8)\)[/tex] and [tex]\((-4, -1)\)[/tex] - Yes, these match our solutions.
- [tex]\((-4, -7)\)[/tex] and [tex]\((-1, 8)\)[/tex] - No, these do not match our solutions.
- [tex]\((8, -7)\)[/tex] and [tex]\((-1, -4)\)[/tex] - No, these do not match our solutions.
Thus, the correct pair of points that solve the system of equations is:
[tex]\[ \boxed{(-7, 8) \text{ and } (-4, -1)} \][/tex]
1. [tex]\( y = (x + 4)^2 - 1 \)[/tex]
2. [tex]\( y = -3x - 13 \)[/tex]
we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations.
### Step-by-Step Solution:
1. Set the equations equal to each other:
Since both expressions equal [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ (x + 4)^2 - 1 = -3x - 13 \][/tex]
2. Simplify the equation:
Expand and simplify the equation:
[tex]\[ (x + 4)^2 - 1 = -3x - 13 \][/tex]
[tex]\[ x^2 + 8x + 16 - 1 = -3x - 13 \][/tex]
[tex]\[ x^2 + 8x + 15 = -3x - 13 \][/tex]
Move all terms to one side:
[tex]\[ x^2 + 11x + 28 = 0 \][/tex]
3. Solve the quadratic equation:
We solve the quadratic equation [tex]\( x^2 + 11x + 28 = 0 \)[/tex].
4. Roots of the quadratic equation:
The roots of the quadratic equation [tex]\(x^2 + 11x + 28 = 0\)[/tex] are [tex]\(x = -4\)[/tex] and [tex]\(x = -7\)[/tex].
5. Calculate the corresponding [tex]\( y \)[/tex] values:
For [tex]\( x = -4 \)[/tex]:
[tex]\[ y = (x + 4)^2 - 1 \][/tex]
[tex]\[ y = (-4 + 4)^2 - 1 \][/tex]
[tex]\[ y = 0^2 - 1 \][/tex]
[tex]\[ y = -1 \][/tex]
For [tex]\( x = -7 \)[/tex]:
[tex]\[ y = (x + 4)^2 - 1 \][/tex]
[tex]\[ y = (-7 + 4)^2 - 1 \][/tex]
[tex]\[ y = (-3)^2 - 1 \][/tex]
[tex]\[ y = 9 - 1 \][/tex]
[tex]\[ y = 8 \][/tex]
Therefore, the points that satisfy both equations are [tex]\((-4, -1)\)[/tex] and [tex]\((-7, 8)\)[/tex].
6. Check the solutions against the given options:
- [tex]\((-7, -4)\)[/tex] and [tex]\((8, -1)\)[/tex] - No, these do not match our solutions.
- [tex]\((-7, 8)\)[/tex] and [tex]\((-4, -1)\)[/tex] - Yes, these match our solutions.
- [tex]\((-4, -7)\)[/tex] and [tex]\((-1, 8)\)[/tex] - No, these do not match our solutions.
- [tex]\((8, -7)\)[/tex] and [tex]\((-1, -4)\)[/tex] - No, these do not match our solutions.
Thus, the correct pair of points that solve the system of equations is:
[tex]\[ \boxed{(-7, 8) \text{ and } (-4, -1)} \][/tex]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.