Question

Consider the following system of equations:
-x+y=5
y=(x-3)^2+2

A. Is (1, 6) a solution to the system of equations? Explain.
B. Solve the system of equations using substitution

Answer 1

To analyze the given system of equations:

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

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

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

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

### Solutions to the system:
The solution set is:
- \((1, 6)\)
- \((6, 11)\)