Question

Solve the system by substitution.

y=2x^2−3x−1


y=x−3


(1 point)
Responses

(1,−2)
open paren 1 comma negative 2 close paren

(3,0),(5,2)
open paren 3 comma 0 close paren comma open paren 5 comma 2 close paren

(4,1)
open paren 4 comma 1 close paren

no solution

Answer 1

To solve the system of equations by substitution, we have:

1) \( y = 2x^2 - 3x - 1 \)

2) \( y = x - 3 \)

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

\[
2x^2 - 3x - 1 = x - 3
\]

Now, let's rearrange this equation to set it to zero:

\[
2x^2 - 3x - 1 - x + 3 = 0
\]

Combining like terms gives us:

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

Now we can simplify this equation by dividing all terms by 2:

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

This is a perfect square trinomial:

\[
(x - 1)^2 = 0
\]

So, solving for \( x \), we get:

\[
x - 1 = 0 \implies x = 1
\]

Now we substitute \( x = 1 \) back into one of the original equations to find \( y \). We'll use the second equation:

\[
y = x - 3 \implies y = 1 - 3 = -2
\]

Thus, the solution to the system of equations is:

\[
(1, -2)
\]

Therefore, the answer is:

**(1, −2)**