Question

what is the result of substituting for y in the bottom equation?

y=x-7
y=x^2-2x-4

Answer 1

To find the result of substituting for \( y \) in the bottom equation, we will start with the two equations you provided:

1. \( y = x - 7 \)
2. \( y = x^2 - 2x - 4 \)

We will substitute the expression for \( y \) from the first equation into the second equation.

Starting with the second equation:

\[
x - 7 = x^2 - 2x - 4
\]

Now, we will rearrange this equation to isolate all terms on one side:

\[
0 = x^2 - 2x - 4 - (x - 7)
\]
\[
0 = x^2 - 2x - 4 - x + 7
\]
\[
0 = x^2 - 3x + 3
\]

Now, we have a quadratic equation:

\[
x^2 - 3x + 3 = 0
\]

To find the values of \( x \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = -3, c = 3 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-3)^2 - 4(1)(3) = 9 - 12 = -3
\]

Since the discriminant is negative, it indicates that there are no real solutions for \( x \). Instead, we will have two complex solutions.

Now, plugging the values into the quadratic formula:

\[
x = \frac{3 \pm \sqrt{-3}}{2} = \frac{3 \pm i\sqrt{3}}{2}
\]

Thus, the two complex solutions for \( x \) are:

\[
x_1 = \frac{3 + i\sqrt{3}}{2}, \quad x_2 = \frac{3 - i\sqrt{3}}{2}
\]

To find the corresponding values of \( y \), we can substitute these \( x \) values back into the first equation \( y = x - 7 \):

1. For \( x_1 = \frac{3 + i\sqrt{3}}{2} \):
\[
y_1 = \frac{3 + i\sqrt{3}}{2} - 7 = \frac{3 + i\sqrt{3} - 14}{2} = \frac{-11 + i\sqrt{3}}{2}
\]

2. For \( x_2 = \frac{3 - i\sqrt{3}}{2} \):
\[
y_2 = \frac{3 - i\sqrt{3}}{2} - 7 = \frac{3 - i\sqrt{3} - 14}{2} = \frac{-11 - i\sqrt{3}}{2}
\]

In summary, the results of the substitutions lead to complex solutions for \( x \) and corresponding \( y \) values:

\[
x_1 = \frac{3 + i\sqrt{3}}{2}, \quad y_1 = \frac{-11 + i\sqrt{3}}{2}
\]

\[
x_2 = \frac{3 - i\sqrt{3}}{2}, \quad y_2 = \frac{-11 - i\sqrt{3}}{2}
\]