Solving Simultaneous Equations Using Substitution Practice

Question

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Question
Use the substitution method to determine whether the linear-quadratic system y=x2−9x+39
, y=9x−42
has 0, 1, or 2 solutions.(1 point)
This system has
solution(s).
Remaining Attempts : 3
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Answer 1

To determine the number of solutions for the system of equations

1. \( y = x^2 - 9x + 39 \)
2. \( y = 9x - 42 \)

we can use the substitution method. Since both equations equal \( y \), we can set them equal to each other:

\[ x^2 - 9x + 39 = 9x - 42 \]

Now, we'll rearrange this equation to form a standard quadratic equation:

\[ x^2 - 9x - 9x + 39 + 42 = 0 \]

This simplifies to:

\[ x^2 - 18x + 81 = 0 \]

Next, we can factor this quadratic equation:

\[ (x - 9)(x - 9) = 0 \]

This gives us:

\[ (x - 9)^2 = 0 \]

The solution to this equation is:

\[ x - 9 = 0 \implies x = 9 \]

Now, substituting \( x = 9 \) back into one of the original equations to find \( y \):

Using \( y = 9x - 42 \):

\[ y = 9(9) - 42 = 81 - 42 = 39 \]

So the solution to the system is \( (9, 39) \).

Since we arrived at a single value of \( x \) which gives a corresponding value of \( y \), the system has 1 solution.

Thus, the answer is:

This system has 1 solution.