-x + y = 5

To determine if the point (1, 6) is a solution to the system of equations given by:

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

we'll substitute \(x = 1\) and \(y = 6\) into both equations.

For the first equation: \[ -x + y = 5 \] Substituting \(x = 1\) and \(y = 6\): \[ -1 + 6 = 5 \] \[ 5 = 5 \quad \text{(True)} \]

For the second equation: \[ y = (x - 3)^2 + 2 \] Substituting \(x = 1\): \[ y = (1 - 3)^2 + 2 \] \[ y = (-2)^2 + 2 \] \[ y = 4 + 2 \] \[ y = 6 \quad \text{(True)} \]

Since (1, 6) satisfies both equations, we conclude that (1, 6) is indeed a solution to the system of equations.

Solve the system of equations using substitution.

To solve the system of equations using substitution, we have:

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

We can start by rearranging Equation 1 to express \(y\) in terms of \(x\):

From Equation 1: \[ y = x + 5 \quad \text{(Equation 3)} \]

Next, we can substitute Equation 3 into Equation 2:

Substituting \(y\) in Equation 2: \[ x + 5 = (x - 3)^2 + 2 \]

Next, we simplify and solve for \(x\):

1. Expand the right side: \[ x + 5 = (x - 3)(x - 3) + 2 \] \[ x + 5 = x^2 - 6x + 9 + 2 \] \[ x + 5 = x^2 - 6x + 11 \]

2. Move all terms to one side of the equation: \[ 0 = x^2 - 6x - x + 11 - 5 \] \[ 0 = x^2 - 7x + 6 \]

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

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

Now, we have two possible solutions for \(x\): \(x = 6\) and \(x = 1\).

Next, we will find the corresponding \(y\) values by substituting each \(x\) back into Equation 3:

For \(x = 1\): \[ y = 1 + 5 = 6 \] So, one solution is \((1, 6)\).

For \(x = 6\): \[ y = 6 + 5 = 11 \] So, another solution is \((6, 11)\).

Thus, the system of equations has two solutions:

• \((1, 6)\)
• \((6, 11)\)

Graph the system of equations on a piece of graph paper. On the graph, label the
solution(s) to the system of equations.

While I can't create graphs directly, I can guide you through the steps to graph the system of equations and label the solutions.

Given Equations

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

Step 1: Graph Equation 1

To graph the first equation, rearrange it to slope-intercept form \(y = mx + b\):

\[ y = x + 5 \]

1. Find intercepts:

   • y-intercept: Set \(x = 0\): \[ y = 0 + 5 \Rightarrow (0, 5) \]
   • x-intercept: Set \(y = 0\): \[ 0 = -x + 5 \Rightarrow x = 5 \Rightarrow (5, 0) \]

2. Plot the points (0, 5) and (5, 0) on the graph paper and draw a line through these points.

Step 2: Graph Equation 2

To graph the second equation \(y = (x - 3)^2 + 2\), recognize that this is a quadratic equation (a parabola) that opens upwards.

1. Find the vertex: The vertex form is \(y = a(x - h)^2 + k\), where \( (h, k) \) is the vertex. Here, the vertex is at \( (3, 2) \).

2. Find additional points:

   • When \(x = 2\): \[ y = (2 - 3)^2 + 2 = 1 + 2 = 3 \Rightarrow (2, 3) \]
   • When \(x = 4\): \[ y = (4 - 3)^2 + 2 = 1 + 2 = 3 \Rightarrow (4, 3) \]
   • When \(x = 1\): \[ y = (1 - 3)^2 + 2 = 4 + 2 = 6 \Rightarrow (1, 6) \]
   • When \(x = 5\): \[ y = (5 - 3)^2 + 2 = 4 + 2 = 6 \Rightarrow (5, 6) \]
   • When \(x = 6\): \[ y = (6 - 3)^2 + 2 = 9 + 2 = 11 \Rightarrow (6, 11) \]

3. Plot the points \( (3, 2) \), \( (2, 3) \), \( (4, 3) \), \( (1, 6) \), \( (5, 6) \), and \( (6, 11) \) and draw a smooth curve through these points, forming a parabola.

Step 3: Label the Solutions

After plotting both lines and the parabola, look for the points where they intersect, which should be the solutions:

• One point of intersection is \( (1, 6) \).
• Another point of intersection is \( (6, 11) \).

Final Graph

1. Mark the point \( (1, 6) \) with a label: "Solution 1".
2. Mark the point \( (6, 11) \) with a label: "Solution 2".

This procedure will help you visualize the system of equations and clearly indicate the solutions on the graph. If you use graph paper and follow these steps, you'll end up with a clear graph!

Simplify the answer

Certainly! Here’s a simplified guide to graphing the system of equations and identifying the solutions.

Given Equations

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

Steps to Graph

Step 1: Graph the first equation

Convert to slope-intercept form: \[ y = x + 5 \]

Find key points:

• y-intercept: \((0, 5)\)
• x-intercept: \((5, 0)\)

Plot the points and draw the line through (0, 5) and (5, 0).

Step 2: Graph the second equation

This is a quadratic function.

Find the vertex: \[ (3, 2) \quad \text{(This is the lowest point since the parabola opens upwards.)} \]

Choose additional points:

• When \(x = 2\): \((2, 3)\)
• When \(x = 4\): \((4, 3)\)
• When \(x = 1\): \((1, 6)\)
• When \(x = 5\): \((5, 6)\)
• When \(x = 6\): \((6, 11)\)

Plot the vertex and additional points and draw a smooth curve (the parabola) through them.

Step 3: Identify and label the solutions

Solutions:

1. \( (1, 6) \)
2. \( (6, 11) \)

Label these points on your graph as "Solution 1" at \( (1, 6) \) and "Solution 2" at \( (6, 11) \).

Summary

• Equation 1: Linear function \(y = x + 5\).
• Equation 2: Quadratic function \(y = (x - 3)^2 + 2\).
• Solutions are: \( (1, 6) \) and \( (6, 11) \).

This simplified process will help you graph the equations and find the solutions. You can now proceed to draw your graph using graph paper!

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.

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

1. Nature of the Equations

• A linear equation (such as \(y = mx + b\)) represents a straight line in the Cartesian plane. It can intersect other lines or curves at a finite number of points, but it is defined by a first-degree polynomial.

• A quadratic equation (such as \(y = ax^2 + bx + c\)) represents a parabola, which can open upwards or downwards and may intersect lines at most at two points. These intersections correspond to the solutions of the system.

2. Possible Intersections

When you graph a linear equation and a quadratic equation:

• The linear equation can intersect the quadratic equation in 0, 1, or 2 points.
• This means the system could have:
  • No solutions (the line is above, below, or tangent to the parabola without intersecting),
  • One solution (the line is tangent to the parabola),
  • Two solutions (the line intersects the parabola at two distinct points).

3. Infinite Solutions

To have infinite solutions, two equations must represent the same line or curve. This occurs when:

• The two equations are actually the same equation, which would mean their forms are not just similar but equivalent through algebraic manipulation (e.g., multiplying a linear equation by a non-zero constant).

However, since one equation is linear and the other is quadratic, the only way they may look similar is if a particular linear equation forms part of the quadratic equation (as in the linear case being the tangent line of the parabola at an infinite number of points), but practically they can only be tangent at one point, not infinite.

Conclusion

In summary, a system comprising a linear equation and a quadratic equation can only have a finite number of intersections (solutions), which can be zero, one, or two. It can never yield an infinite number of solutions, as that would require both equations to represent the same geometric shape, which is not possible with one being linear and the other being quadratic. If you graph them, you would observe at most two intersections, and if they coincide in forms, they would appear as a single straight line, indicating equivalent equations, but this does not happen with the forms you provided.

Simplify what you just said

Sure! Here's a simplified explanation:

Can a Linear and Quadratic Equation Have Infinite Solutions?

• Linear Equation: Represents a straight line.
• Quadratic Equation: Represents a parabola (a U-shaped curve).

Possible Intersections:

• A linear equation can intersect a quadratic equation:
  • 0 points: The line is above or below the parabola.
  • 1 point: The line is tangent to the parabola (just touches it).
  • 2 points: The line intersects the parabola at two points.

Infinite Solutions:

A system of one linear and one quadratic equation cannot have infinite solutions because:

• Infinite solutions would mean they are the same line or curve.
• Since one is a line and the other is a parabola, they can't be identical.

Conclusion:

In summary, a system with one linear and one quadratic equation can have at most two solutions, but it cannot have infinite solutions. If you graph them, you’ll see they can intersect in at most two places.