Question

Consider the following system of equations:
-x+y=5
y=(x-3)^2+2
Is (1, 6) a solution to the system of equations?

Answer 1

Solve the system of equations using substitution.

Answer 2

To determine if the point \((1, 6)\) is a solution to the given system of equations, we need to substitute \(x = 1\) and \(y = 6\) into both equations.

1. For the first equation:
\[
-x + y = 5
\]
Substituting \(x = 1\) and \(y = 6\):
\[
-1 + 6 = 5
\]
This simplifies to:
\[
5 = 5
\]
This equation is satisfied.

2. For the second equation:
\[
y = (x - 3)^2 + 2
\]
Substituting \(x = 1\) and \(y = 6\):
\[
6 = (1 - 3)^2 + 2
\]
Calculate \((1 - 3)^2\):
\[
(1 - 3)^2 = (-2)^2 = 4
\]
Now substitute back into the equation:
\[
6 = 4 + 2
\]
This simplifies to:
\[
6 = 6
\]
This equation is also satisfied.

Since both equations are satisfied with the point \((1, 6)\), we conclude that \((1, 6)\) is indeed a solution to the system of equations.

Answer 3

To solve the system of equations using substitution, we start with the given equations:

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

**Step 1: Solve the first equation for \(y\).**

From the first equation:
\[
y = x + 5
\]

**Step 2: Substitute \(y\) into the second equation.**

Now we can substitute \(y = x + 5\) into the second equation:
\[
x + 5 = (x - 3)^2 + 2
\]

**Step 3: Simplify and solve for \(x\).**

We will first expand \((x - 3)^2\):
\[
x + 5 = (x^2 - 6x + 9) + 2
\]
\[
x + 5 = x^2 - 6x + 11
\]

Now, rearranging the equation gives:
\[
0 = x^2 - 6x + 11 - x - 5
\]
\[
0 = x^2 - 7x + 6
\]

**Step 4: Factor the quadratic.**

This can be factored as:
\[
0 = (x - 1)(x - 6)
\]

**Step 5: Solve for \(x\).**

Setting each factor to zero gives us:
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
\[
x - 6 = 0 \quad \Rightarrow \quad x = 6
\]

**Step 6: Find corresponding \(y\) values.**

Now we will find the corresponding \(y\) values using the equation \(y = x + 5\).

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

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

**Final Solutions:**
The solutions to the system of equations are:
\[
(1, 6) \quad \text{and} \quad (6, 11)
\]