To determine if (1, 6) is a solution to the given system of equations, we will substitute \( x = 1 \) and \( y = 6 \) into both equations.
The system of equations is:
- \( -x + y = 5 \)
- \( y = (x - 3)^2 + 2 \)
Checking (1, 6)
For the first equation \( -x + y = 5 \): \[ -x + y = -1 + 6 = 5 \] This is true.
For the second equation \( y = (x - 3)^2 + 2 \): \[ y = (1 - 3)^2 + 2 = (-2)^2 + 2 = 4 + 2 = 6 \] This is also true.
Since both equations are satisfied by the point (1, 6), we conclude that (1, 6) is indeed a solution to the system of equations.
Solving the System using the Substitution Method
We already have the equations:
- \( -x + y = 5 \)
- \( y = (x - 3)^2 + 2 \)
From the first equation, we can express \( y \) in terms of \( x \): \[ y = x + 5 \]
Now, we can substitute this expression for \( y \) into the second equation: \[ x + 5 = (x - 3)^2 + 2 \]
Next, we simplify and solve for \( x \): \[ x + 5 = (x - 3)(x - 3) + 2 \] \[ x + 5 = x^2 - 6x + 9 + 2 \] \[ x + 5 = x^2 - 6x + 11 \]
Rearranging the equation gives: \[ 0 = x^2 - 7x + 6 \]
Factoring the quadratic: \[ 0 = (x - 6)(x - 1) \]
Thus, the solutions for \( x \) are: \[ x = 6 \quad \text{or} \quad x = 1 \]
Now we can find the corresponding \( y \) values using \( y = x + 5 \):
- For \( x = 1 \): \( y = 1 + 5 = 6 \), so one solution is \( (1, 6) \).
- For \( x = 6 \): \( y = 6 + 5 = 11 \), so another solution is \( (6, 11) \).
Final Solutions
The system of equations has two solutions:
- \( (1, 6) \)
- \( (6, 11) \)
Infinite Solutions in the System of Equations
A system of two equations that includes a linear equation and a quadratic equation could potentially have an infinite number of solutions, but this typically occurs in very specific scenarios.
For example, if the quadratic equation is such that it represents a parabola that is tangent to the line of the linear equation or if the line lies entirely on the parabola, then it can have infinitely many solutions.
From the graph: If you were to graph the two equations and find that the line touches the parabola at an infinite number of points (i.e., the linear equation perfectly coincides with a part of the quadratic's curve), this would indicate an infinite number of solutions.
However, in the case where the linear equation is not tangent to the quadratic, or intersects the parabola at points, you will typically find a finite number of solutions (0, 1, or 2 solutions).
Thus, to answer your final question, yes, it can have infinitely many solutions if the linear equation coincides with the quadratic curve over some interval, which can be observed graphically. Otherwise, such systems typically have a finite number of intersections.