Consider the following system of equations:

Question

-x+y=5
y=(x-3)^2+2
A. Is (1, 6) a solution to the system of equations? Explain.

Answer 1

To determine if the point \((1, 6)\) is a solution to the given system of equations, we need to substitute \(x = 1\) and \(y = 6\) into each equation and see if both equations are satisfied.

The system of equations is:

1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)

Let's check the first equation:

1. Substitute \(x = 1\) and \(y = 6\) into \(-x + y = 5\): \[ -1 + 6 = 5 \] \[ 5 = 5 \quad \text{(True)} \]

Now, let's check the second equation:

2. Substitute \(x = 1\) into \(y = (x - 3)^2 + 2\): \[ y = (1 - 3)^2 + 2 \] \[ y = (-2)^2 + 2 \] \[ y = 4 + 2 \] \[ y = 6 \quad \text{(True)} \]

Since the point \((1, 6)\) satisfies both equations, we conclude that:

Yes, (1, 6) is a solution to the system of equations.

Answer 2

Solve the system of equations using substitution.

Answer 3

To solve the system of equations using substitution, we first rewrite the given equations:

1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)

We'll start by solving the first equation for \(y\):

\[ y = x + 5 \]

Now we can substitute this expression for \(y\) into the second equation:

\[ x + 5 = (x - 3)^2 + 2 \]

Next, we simplify the equation:

1. Expand the right side: \[ (x - 3)^2 = x^2 - 6x + 9 \] So, \[ x + 5 = x^2 - 6x + 9 + 2 \] Simplifying further, we have: \[ x + 5 = x^2 - 6x + 11 \]

2. Now, move all terms to one side: \[ 0 = x^2 - 6x + 11 - x - 5 \] \[ 0 = x^2 - 7x + 6 \]

3. We factor the quadratic equation: \[ 0 = (x - 6)(x - 1) \]

4. Setting each factor to zero gives: \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

Now we have two values for \(x\): \(x = 6\) and \(x = 1\). We will find the corresponding \(y\) values for both \(x\) values using \(y = x + 5\).

For \(x = 6\): \[ y = 6 + 5 = 11 \quad \Rightarrow \quad (6, 11) \]

For \(x = 1\): \[ y = 1 + 5 = 6 \quad \Rightarrow \quad (1, 6) \]

Thus, the solutions to the system of equations are: \[ (1, 6) \quad \text{and} \quad (6, 11) \]

Answer 4

Could a system of two equations that includes a linear equation and a quadratic equation
ever have an infinite number of solutions? If so, how could you tell from the graph? If not,
explain why.

Answer 5

A system of two equations that consists of one linear equation and one quadratic equation cannot have an infinite number of solutions. Here’s why:

1. Types of Equations:

   • A linear equation typically represents a straight line in a two-dimensional space.
   • A quadratic equation usually represents a parabola (which opens upwards or downwards).

2. Graph Intersection:

   • The intersection points between a line and a parabola represent the solutions to the system of equations.
   • Normally, the potential intersections can occur at zero, one, or two distinct points. However, they cannot form a scenario where there are infinitely many intersection points.

3. Why Infinite Solutions are Impossible:

   • For a system of equations to have infinitely many solutions, the two equations must be essentially the same equation—meaning one can be transformed into the other through algebraic manipulation. This occurs during systems of linear equations (where two lines are the same).
   • A line and a parabola cannot be identical because a parabola is not a straight line. Even at the vertex of the parabola, the slope will differ from that of the linear equation.

Graphical Illustration:

• Graphing a Linear Equation: This yields a straight line.
• Graphing a Quadratic Equation: This yields a U-shaped curve.

Scenarios:

• No Intersection: The line is above or below the parabola; the system has no solutions.
• Tangent Intersection: The line touches the parabola at exactly one point; the system has one solution.
• Crossing Intersection: The line intersects the parabola at two points; the system has two distinct solutions.

Conclusion:

Therefore, while a system of two equations can have no solutions, one solution, or two solutions, it cannot have an infinite number of solutions due to the fundamentally different nature of linear and quadratic equations. This can be confirmed graphically by observing the shapes of the graphs and their intersection points.