1. Which of the following is an example of a system of equations consisting of a linear and a quadratic equation?(1 point)
A. {y=2x^2=0
y=x^2-6x+3
B. {y-10=3x
y-18=5x^-6x
C. {y=2x+1
5x+y+2x=6
D. 3x+5x=2
6 squerd x+y=1
2. Given the graph of a system consisting of a linear and a quadratic equation, if the line is tangent to the original quadratic, how many solutions does the system have?
A. 2 solutions
B. 1 solution
C. 0 solutions
D. infinitely many solutions
3. How does the number of solutions in a linear system compare to a system with a linear and a quadratic equation?(1 point)
A. Both can have 0, 1, or infinite solutions.
B. Both can have 0, 1, or 2 solutions.
C. A linear system can have 0, 1, or infinite solutions. A linear-quadratic system can have 0, 1, or 2 solutions.
D. A linear system can have 0, 1, or 2 solutions. A linear-quadratic system can have 0, 1, or infinite solutions .
4. How does the number of solutions in a linear system compare to a system with a linear and a quadratic equation?(1 point)
A. (1,2)
B. There are no solutions
C. (0,1)
D. (-1,-2)
E. (0,2)
5. Determine the solution(s) to the linear-quadratic system of equations below.
{3x−y=3y
y=x^2+6x−7
A. (10,153)
B. (1,0)
C. (-4,-15)
D. (-1,-12)
E. (4,33)
F. (-10,33)
1. The correct answer is A. {y=2x^2, y=x^2-6x+3}
2. The correct answer is B. 1 solution
3. The correct answer is C. A linear system can have 0, 1, or infinite solutions. A linear-quadratic system can have 0, 1, or 2 solutions.
4. The question appears to be repeated. The answer choices do not relate to the question.
5. The correct answer is C. (-4, -15)
1. The correct answer is A. {y=2x^2=0, y=x^2-6x+3}
2. If the line is tangent to the original quadratic, the system will have 1 solution. Therefore, the correct answer is B. 1 solution.
3. The correct answer is C. A linear system can have 0, 1, or infinite solutions, whereas a linear-quadratic system can have 0, 1, or 2 solutions.
4. This question appears to be incomplete as only answer choices are given without the corresponding question. Please provide the complete question.
5. To determine the solution(s) to the linear-quadratic system of equations, we can substitute the value of y from the linear equation into the quadratic equation and solve for x.
From the linear equation, we have:
3x - y = 3y
3x - 4y = 0
x = (4/3)y
Substituting this value of x into the quadratic equation, we have:
y = (4/3)y^2 + 6(4/3)y - 7
y = (4/3)y^2 + 8y - 7
Now we can solve this quadratic equation. Rearranging and simplifying the equation:
(4/3)y^2 + 5y - 7 = 0
Using the quadratic formula, we have:
y = (-5 ± √(5^2 - 4*(4/3)*(-7))) / (2*(4/3))
y = (-5 ± √(25 + 112/3)) / (8/3)
y = (-5 ± √(225/3 + 112/3)) / (8/3)
y = (-5 ± √(337/3)) / (8/3)
Simplifying further:
y = (-5 ± √(337)/√(3)) / (8/3)
y = (-5 ± √(337)*√(3)) / (8/3)
y = (-15 ± √(337)*√(3)) / 8
y = (-15 ± √(1011)) / 8
Now we can substitute the values of y back into the linear equation to find the corresponding values of x:
For y = (-15 + √(1011)) / 8:
x = (4/3)((-15 + √(1011)) / 8)
x = (-20 + 4√(1011)) / 6
x = (-10 + 2√(1011)) / 3
For y = (-15 - √(1011)) / 8:
x = (4/3)((-15 - √(1011)) / 8)
x = (-20 - 4√(1011)) / 6
x = (-10 - 2√(1011)) / 3
Therefore, the solutions to the linear-quadratic system of equations are:
E. (4, 33)
F. (-10, 33)
1. A. {y=2x^2=0, y=x^2-6x+3}
To determine if a system of equations consists of a linear and a quadratic equation, we need to check the forms of the equations. A linear equation has the form y = mx + b, where m is the slope and b is the y-intercept. A quadratic equation has the form y = ax^2 + bx + c, where a, b, and c are constants.
In option A, we have two equations:
- y = 2x^2 = 0 (This equation has an error, as there are two equal signs.)
- y = x^2 - 6x + 3 (This is a quadratic equation.)
Therefore, option A contains a quadratic equation, but not a linear equation.
2. B. 1 solution
If the line is tangent to the original quadratic equation in a system of equations, it means that the line intersects the quadratic equation at exactly one point. This results in only one solution for the system.
3. C. A linear system can have 0, 1, or infinite solutions. A linear-quadratic system can have 0, 1, or 2 solutions.
A linear system of equations can have varying numbers of solutions, including 0 solutions (parallel lines), 1 solution (intersecting lines), or infinite solutions (coinciding lines).
A linear-quadratic system of equations can have 0, 1, or 2 solutions. The quadratic equation can intersect the linear equation at zero, one, or two points.
4. B. There are no solutions
This question does not provide any specific equations to analyze. Without the equations, we cannot determine the number of solutions.
5. C. (-4, -15) and D. (-1, -12)
To solve the linear-quadratic system of equations, we need to find the values of x and y that satisfy both equations simultaneously.
The equations given are:
- 3x - y = 3
- y = x^2 + 6x - 7
We can use substitution or elimination to solve this system. Let's use substitution for this example:
From equation 1, we can rewrite it as: y = 3x - 3.
Now, substitute this into equation 2:
3x - 3 = x^2 + 6x - 7
Rearrange the equation:
x^2 + 3x - 4 = 0
Factor the quadratic equation:
(x + 4)(x - 1) = 0
Set each factor equal to zero:
x + 4 = 0 --> x = -4
x - 1 = 0 --> x = 1
Now, substitute these x-values back into equation 2 to find the corresponding y-values:
For x = -4:
y = (-4)^2 + 6(-4) - 7
y = 16 - 24 - 7
y = -15
For x = 1:
y = 1^2 + 6(1) - 7
y = 1 + 6 - 7
y = 0
Therefore, the solutions to the linear-quadratic system of equations are (-4, -15) and (-1, -12).