Question

Using the substitution method, solve the following linear quadratic system:

−x2+y=−5x−20

y=−5x+5

(1 point)

Answer 1

To solve the system of equations using the substitution method, we have the following:

1. \(-x^2 + y = -5x - 20\)
2. \(y = -5x + 5\)

We will substitute equation (2) into equation (1).

Starting with equation (1):

\[
-x^2 + y = -5x - 20
\]

We substitute \(y\) from equation (2):

\[
-x^2 + (-5x + 5) = -5x - 20
\]

Simplifying this gives:

\[
-x^2 - 5x + 5 = -5x - 20
\]

Now, we can add \(5x\) to both sides:

\[
-x^2 + 5 = -20
\]

Next, we add 20 to both sides:

\[
-x^2 + 25 = 0
\]

Now, isolate \(x^2\):

\[
-x^2 = -25
\]

Multiplying both sides by -1:

\[
x^2 = 25
\]

Taking the square root of both sides gives:

\[
x = 5 \quad \text{or} \quad x = -5
\]

Now we can substitute these values back into equation (2) to find the corresponding \(y\) values.

**For \(x = 5\):**

\[
y = -5(5) + 5 = -25 + 5 = -20
\]

**For \(x = -5\):**

\[
y = -5(-5) + 5 = 25 + 5 = 30
\]

Thus, we have the two solutions:

1. \( (5, -20) \)
2. \( (-5, 30) \)

So the solutions to the system of equations are:

\[
(5, -20) \quad \text{and} \quad (-5, 30)
\]