Consider the following system of equations:

To solve the system of equations

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

we can substitute the expression for \(y\) from the second equation into the first equation.

Starting with equation 1:

\[ y = -x + 5 \]

Now we substitute this into equation 2:

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

Next, we will expand and simplify the right side:

\[ (x - 3)^2 = x^2 - 6x + 9 \]

So, the second equation becomes:

\[ -x + 5 = x^2 - 6x + 9 + 2 \] \[ -x + 5 = x^2 - 6x + 11 \]

Now, we can set up the equation by bringing all terms to one side:

\[ 0 = x^2 - 6x + 11 + x - 5 \]

Combining like terms:

\[ 0 = x^2 - 5x + 6 \]

Now we factor the quadratic equation:

\[ 0 = (x - 2)(x - 3) \]

Setting each factor to zero gives us the possible values for \(x\):

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

Next, we find the corresponding \(y\) values for each \(x\) value using \(y = -x + 5\):

1. For \(x = 2\): \[ y = -2 + 5 = 3 \]

2. For \(x = 3\): \[ y = -3 + 5 = 2 \]

Thus, the solutions to the system of equations are:

\[ (2, 3) \quad \text{and} \quad (3, 2) \]

In conclusion, the solutions are:

\[ \boxed{(2, 3) \text{ and } (3, 2)} \]

Is (1,6)
a solution to the system of equations?

To determine if the point \((1, 6)\) is a solution to the system of equations, we need to check if it satisfies both equations:

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

Let's substitute \(x = 1\) and \(y = 6\) into both equations.

Checking the first equation:

\[ -x + y = 5 \] Substituting in the values: \[ -1 + 6 = 5 \quad \Rightarrow \quad 5 = 5 \quad \text{(True)} \]

Checking the second equation:

\[ y = (x - 3)^2 + 2 \] Substituting in the values: \[ 6 = (1 - 3)^2 + 2 \] Calculating the right side: \[ (1 - 3)^2 = (-2)^2 = 4 \quad \Rightarrow \quad 6 = 4 + 2 \] \[ 6 = 6 \quad \text{(True)} \]

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

\[ (1, 6) \text{ is a solution to the system of equations.} \]

Solve the system of equations using substitution.

To solve the system of equations using substitution, let's first rewrite the two equations we are working with:

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

Step 1: Solve one equation for one variable.

We can solve the first equation for \(y\):

\[ y = x + 5 \]

Step 2: Substitute this expression into the second equation.

Now we substitute \(y\) in the second equation:

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

Step 3: Expand and simplify the equation.

Now, expand the right-hand side:

\[ x + 5 = (x^2 - 6x + 9) + 2 \] \[ x + 5 = x^2 - 6x + 11 \]

Step 4: Set the equation to zero.

Rearranging gives us:

\[ 0 = x^2 - 6x - x + 11 - 5 \] \[ 0 = x^2 - 7x + 6 \]

Step 5: Factor the quadratic equation.

Next, we factor the quadratic:

\[ 0 = (x - 1)(x - 6) \]

Step 6: Solve for \(x\).

Setting each factor equal to zero gives:

1. \(x - 1 = 0 \quad \Rightarrow \quad x = 1\)
2. \(x - 6 = 0 \quad \Rightarrow \quad x = 6\)

Step 7: Substitute back to find \(y\).

Now, we will substitute back to find the corresponding \(y\) values.

For \(x = 1\):

\[ y = 1 + 5 = 6 \]

The first solution is \((1, 6)\).

For \(x = 6\):

\[ y = 6 + 5 = 11 \]

The second solution is \((6, 11)\).

Final Solutions:

The solutions to the system of equations are:

\[ \boxed{(1, 6) \text{ and } (6, 11)} \]

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 not.

A system of two equations, consisting of one linear equation and one quadratic equation, generally cannot have an infinite number of solutions. Here's why, along with a graphical explanation:

Why Not Infinite Solutions?

1. Nature of the Equations:

   • A linear equation represents a straight line in a two-dimensional space (e.g., \(y = mx + b\)).
   • A quadratic equation represents a parabola (e.g., \(y = ax^2 + bx + c\)).

2. Intersection Points:

   • The possible intersections between a straight line and a parabola can be:
     • No Intersection: The line does not touch the parabola at all.
     • One Intersection: The line is tangent to the parabola at exactly one point.
     • Two Intersections: The line intersects the parabola at two distinct points.

Since the straight line is a linear function and the parabola is a quadratic function, they can share at most two intersection points in a 2D Cartesian coordinate system but cannot share an infinite number of points.

Graphical Representation:

• If you graph a linear equation and a quadratic equation, it will look like this:

  • Linear Equation: A straight line.

  • Quadratic Equation: A U-shaped curve (upward or downward facing).

• Possible Graph Scenarios:

  • No Solutions: The line might be completely above or below the parabola.

  • One Solution: The line could touch the parabola at one point (tangent).

  • Two Solutions: The line might cross the parabola at two distinct points.

These graphical scenarios illustrate that while a linear function and a quadratic function can intersect at various points, it's impossible for them to intersect an infinite number of times—they are fundamentally different in their shapes and equations.

Conclusion:

In summary, a system of one linear equation and one quadratic equation cannot have an infinite number of solutions, and this can be viewed clearly in a graph where the behavior of lines versus parabolas indicates a maximum of two points of intersection or none at all.

shorter

A system of two equations, consisting of one linear equation and one quadratic equation, cannot have an infinite number of solutions.

Explanation:

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

Intersection Scenarios:

• No Solutions: The line does not intersect the parabola.
• One Solution: The line is tangent to the parabola (touches at one point).
• Two Solutions: The line crosses the parabola at two points.

Since a line can intersect a parabola at most twice, it cannot produce an infinite number of intersection points.

Graphically:

When graphed:

• The linear equation is a straight line.
• The quadratic equation is a curve.

This visual representation shows that they can intersect at a maximum of two points but never infinitely.

Conclusion:

In summary, a linear equation and a quadratic equation can intersect at most twice, so they cannot have an infinite number of solutions.