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Part 1
Solve the system by substitution. Check the​ answer(s).
left brace Start 2 By 1 Matrix 1st Row 1st Column y equals 2 x squared minus 27 x minus 1 2nd Row 1st Column y equals x minus 99 EndMatrix

Question content area bottom
Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
The​ solution(s) is/are

enter your response here.
​(Type an ordered pair. Use a comma to separate answers as​ needed.)
B.
There is no solution.
left brace Start 2 By 1 Matrix 1st Row 1st Column y equals 2 x squared minus 27 x minus 1 2nd Row 1st Column y equals x minus 99 EndMatrixSelect the correct choice below and, if necessary, fill in the answer box to complete your choice.
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To solve the system of equations by substitution, let's proceed step by step. The system is:

\[
\begin{aligned}
1) & \quad y = 2x^2 - 27x - 1 \\
2) & \quad y = x - 99
\end{aligned}
\]

### Step 1: Substitute the second equation into the first.
Since both equations equal \( y \), we can set them equal to each other:

\[
2x^2 - 27x - 1 = x - 99
\]

### Step 2: Simplify and solve for \( x \).
Move all terms to one side to set the equation equal to zero:

\[
2x^2 - 27x - 1 - x + 99 = 0
\]

Simplify the equation:

\[
2x^2 - 28x + 98 = 0
\]

Now, divide the entire equation by 2 to make it simpler:

\[
x^2 - 14x + 49 = 0
\]

### Step 3: Factor the quadratic equation.
The quadratic equation can be factored as:

\[
(x - 7)^2 = 0
\]

So, \( x - 7 = 0 \) leads to:

\[
x = 7
\]

### Step 4: Solve for \( y \) using the value of \( x \).
Substitute \( x = 7 \) back into the second equation to find \( y \):

\[
y = 7 - 99 = -92
\]

### Step 5: Check the solution.
Substitute \( x = 7 \) and \( y = -92 \) into the first equation:

\[
y = 2(7)^2 - 27(7) - 1 = 2(49) - 189 - 1 = 98 - 190 = -92
\]

Since both equations are satisfied, the solution is correct.

### Final Answer:
The solution is \((7, -92)\).

Would you like to explore the steps in more detail, or do you have any questions?

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Here are five related questions you might consider:

1. How would the solution change if the second equation was \( y = x - 50 \)?
2. What if the first equation was \( y = 2x^2 - 27x + 1 \) instead?
3. How do you solve the system if both equations were linear?
4. Can the quadratic equation \( x^2 - 14x + 49 = 0 \) have any other solutions?
5. How do you determine if a system of equations has no solution?

**Tip:** When solving by substitution, always ensure that after substituting, the equation is simplified properly before solving for the variable.

Learn how to solve a system of equations using the substitution method with a detailed step-by-step guide. This problem involves quadratic equations and linear equations, showcasing key algebraic concepts. Suitable for Grades 9-12.

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