Write your own system of equations.1.) One equation should be linear and one should be quadratic.

Question

2.) describe how you can determine whether a given point is a solution to your system of equations by substitute the point into both equations and show your steps.
3.) graph the solution set for the system of equations and label the solution

Answer 1

Sure! Let's create a system of equations consisting of one linear equation and one quadratic equation.

System of Equations:

1. Linear Equation: \( y = 2x + 3 \)
2. Quadratic Equation: \( y = x^2 - 4 \)

Step 1: Determine a solution

To determine whether a given point is a solution to the system of equations, we can substitute the \( x \) and \( y \) coordinates of that point into both equations.

Let’s say we want to check if the point \( (1, 5) \) is a solution.

Substitute into the linear equation:

1. Substitute \( x = 1 \) into the linear equation: \[ y = 2(1) + 3 \] \[ y = 2 + 3 = 5 \] The calculated \( y \) value is \( 5 \), which matches with the \( y \) value of the point \( (1, 5) \).

Substitute into the quadratic equation:

2. Substitute \( x = 1 \) into the quadratic equation: \[ y = (1)^2 - 4 \] \[ y = 1 - 4 = -3 \] The calculated \( y \) value is \( -3 \), which does not match with the \( y \) value of the point \( (1, 5) \).

Conclusion:

Since the point \( (1, 5) \) satisfies the linear equation but not the quadratic equation, it is not a solution for the system of equations.

Step 2: Finding a solution point

Let's find a solution point by equating the two equations: \[ 2x + 3 = x^2 - 4 \] Rearranging gives: \[ x^2 - 2x - 7 = 0 \] We can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1, b = -2, c = -7 \): \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-7)}}{2(1)} \] \[ x = \frac{2 \pm \sqrt{4 + 28}}{2} \] \[ x = \frac{2 \pm \sqrt{32}}{2} \] \[ x = \frac{2 \pm 4\sqrt{2}}{2} \] \[ x = 1 \pm 2\sqrt{2} \]

This gives us two possible \( x \) values:

1. \( x_1 = 1 + 2\sqrt{2} \)
2. \( x_2 = 1 - 2\sqrt{2} \)

Next, let's find the corresponding \( y \) values. Using the linear equation:

1. For \( x_1 = 1 + 2\sqrt{2} \): \[ y_1 = 2(1 + 2\sqrt{2}) + 3 = 2 + 4\sqrt{2} + 3 = 5 + 4\sqrt{2} \]
2. For \( x_2 = 1 - 2\sqrt{2} \): \[ y_2 = 2(1 - 2\sqrt{2}) + 3 = 2 - 4\sqrt{2} + 3 = 5 - 4\sqrt{2} \]

Final Solutions:

The system has two solutions:

1. \( \left( 1 + 2\sqrt{2}, 5 + 4\sqrt{2} \right) \)
2. \( \left( 1 - 2\sqrt{2}, 5 - 4\sqrt{2} \right) \)

Step 3: Graphing the Solution Set

To graph the system of equations:

1. Linear Equation:

   • The equation \( y = 2x + 3 \) is a straight line.

2. Quadratic Equation:

   • The equation \( y = x^2 - 4 \) is a parabola opening upwards.

Intersection Points

Plotting these equations on a coordinate plane reveals their intersection points, which correspond to the solutions we derived.

• Label the points of intersection as \( A \left( 1 + 2\sqrt{2}, 5 + 4\sqrt{2} \right) \) and \( B \left( 1 - 2\sqrt{2}, 5 - 4\sqrt{2} \right) \).

Graphing Instructions:

1. Draw the linear line according to the slope-intercept form \( y = 2x + 3 \).
2. Draw the parabola for \( y = x^2 - 4 \) (it intersects the y-axis at -4).
3. Locate the intersection points computed earlier and label them \( A \) and \( B \).

Example Graph Illustration:

It's generally not possible to provide an actual graphical representation here, but you can sketch the graph on paper or use graphing software to visualize this.

Conclusion

This gives us a complete view of how to create a system of equations with one linear and one quadratic equation, how to check if points are solutions, and how to identify and graph solutions.