Question

Solve the following system of equations algebraically:
y, equals, x, squared, plus, 11, x, minus, 10
y=
x
2
+11x−10
y, equals, 2, x, minus, 10
y=
2x−10
Answer

Answer 1

To solve the system of equations given by:

1. \( y = x^2 + 11x - 10 \)
2. \( y = 2x - 10 \)

We can set the equations equal to each other since they both equal \( y \):

\[
x^2 + 11x - 10 = 2x - 10
\]

Now, we will move all terms to one side of the equation:

\[
x^2 + 11x - 10 - 2x + 10 = 0
\]

Combine like terms:

\[
x^2 + 9x + 0 = 0
\]

This simplifies to:

\[
x^2 + 9x = 0
\]

We can factor out \( x \):

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

Setting each factor equal to zero gives us:

\[
x = 0 \quad \text{or} \quad x + 9 = 0
\]
\[
x = 0 \quad \text{or} \quad x = -9
\]

Now, we can find the corresponding \( y \) values for each \( x \).

1. For \( x = 0 \):
\[
y = 2(0) - 10 = -10
\]
So one solution is \( (0, -10) \).

2. For \( x = -9 \):
\[
y = 2(-9) - 10 = -18 - 10 = -28
\]
So the other solution is \( (-9, -28) \).

Thus, the solutions to the system of equations are:

\[
(0, -10) \quad \text{and} \quad (-9, -28)
\]