Question

What is the result of substituting for y in the bottom equation?
y =x-3
y=x2+2x-4
0y=(x-3)2+21x-31-4
0y=x2+2x-4-1x-3)
0(x-3) (x-3)=x2
0x-3=x2+2x-4

Answer 1

Let's first clarify the equations given:

1. \( y = x - 3 \)
2. \( y = x^2 + 2x - 4 \)

Next, we will substitute the first equation into the second equation.

Since both equations equal \( y \), we can set them equal to each other:

\[
x - 3 = x^2 + 2x - 4
\]

Now let's solve for \( x \) by rearranging the equation:

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

Expanding this gives us:

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

Combining like terms leads to:

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

This simplifies to:

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

To solve for \( x \), we can either use factoring (if possible) or the quadratic formula. The quadratic formula states that for an equation \( ax^2 + bx + c = 0 \), the solutions for \( x \) are given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \). Plugging in these values gives:

\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}
\]

\[
x = \frac{-1 \pm \sqrt{1 + 4}}{2}
\]

\[
x = \frac{-1 \pm \sqrt{5}}{2}
\]

Now, we have two possible \( x \) values:

\[
x = \frac{-1 + \sqrt{5}}{2}
\]
\[
x = \frac{-1 - \sqrt{5}}{2}
\]

Next, we substitute these \( x \) values back into the first equation \( y = x - 3 \) to find the corresponding \( y \) values.

1. For \( x = \frac{-1 + \sqrt{5}}{2} \):

\[
y = \frac{-1 + \sqrt{5}}{2} - 3 = \frac{-1 + \sqrt{5} - 6}{2} = \frac{\sqrt{5} - 7}{2}
\]

2. For \( x = \frac{-1 - \sqrt{5}}{2} \):

\[
y = \frac{-1 - \sqrt{5}}{2} - 3 = \frac{-1 - \sqrt{5} - 6}{2} = \frac{-7 - \sqrt{5}}{2}
\]

Thus, the solutions for the equations \( y = x - 3 \) and \( y = x^2 + 2x - 4 \) yield two points of intersection:

1. \( \left( \frac{-1 + \sqrt{5}}{2}, \frac{\sqrt{5} - 7}{2} \right) \)
2. \( \left( \frac{-1 - \sqrt{5}}{2}, \frac{-7 - \sqrt{5}}{2} \right) \)