To analyze the given system of equations:
- \(-x + y = 5\) (Equation 1)
- \(y = (x - 3)^2 + 2\) (Equation 2)
Part A: Checking if (1, 6) is a solution
We need to check if the point \((1, 6)\) satisfies both equations.
Substituting \(x = 1\) and \(y = 6\) into Equation 1:
\[ -x + y = 5 \Rightarrow -1 + 6 = 5 \Rightarrow 5 = 5 \]
This is true.
Now substituting \(x = 1\) and \(y = 6\) into Equation 2:
\[ y = (x - 3)^2 + 2 \Rightarrow 6 = (1 - 3)^2 + 2 \] Calculating the right side: \[ (1 - 3)^2 + 2 = (-2)^2 + 2 = 4 + 2 = 6 \]
This is also true.
Since both equations hold true, the point \((1, 6)\) is indeed a solution to the system of equations.
Part B: Solving the system using substitution
We can use substitution by first isolating \(y\) in Equation 1, since it can be directly substituted into Equation 2.
From Equation 1: \[ y = x + 5 \]
Now we can substitute this expression for \(y\) into Equation 2:
\[ x + 5 = (x - 3)^2 + 2 \]
Now, let's simplify and solve for \(x\):
- Expand the right side: \[ x + 5 = (x^2 - 6x + 9) + 2 \]
- Simplify the right side: \[ x + 5 = x^2 - 6x + 11 \]
Now, rearrange the equation: \[ 0 = x^2 - 6x - x + 11 - 5 \] \[ 0 = x^2 - 7x + 6 \]
Now, factor the quadratic: \[ 0 = (x - 1)(x - 6) \]
Setting each factor to zero gives us the solutions: \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \]
Now, we need to find the corresponding \(y\) values for these \(x\) 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 the other solution is \((6, 11)\).
Solutions to the system:
The solution set is:
- \((1, 6)\)
- \((6, 11)\)