Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40, y=−3x−5

Question

?(1 point)
Responses

(−7,26)

left parenthesis negative 7 comma 26 right parenthesis
(5,−10)
left parenthesis 5 comma negative 10 right parenthesis
(5,−20)
and (−7,16)
left parenthesis negative 5 comma 20 right parenthesis and left parenthesis 7 comma negative 16 right parenthesis
(5,−10)
and (−7,26)

Answer 1

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

1. \( y = x^2 - x - 40 \) (let's call this Equation 1)
2. \( y = -3x - 5 \) (let's call this Equation 2)

We can substitute Equation 2 into Equation 1. This means we'll replace \( y \) in Equation 1 with the expression for \( y \) from Equation 2.

So, we set the two equations equal to each other:

\[ x^2 - x - 40 = -3x - 5 \]

Now, let's move all terms to one side to rearrange the equation:

\[ x^2 - x + 3x - 40 + 5 = 0 \]

Simplifying this gives:

\[ x^2 + 2x - 35 = 0 \]

Next, we'll factor this quadratic equation:

\[ (x + 7)(x - 5) = 0 \]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x + 7 = 0 \) → \( x = -7 \)
2. \( x - 5 = 0 \) → \( x = 5 \)

Now we can find the corresponding \( y \) values for each \( x \) value by plugging them back into either of the original equations. We'll use Equation 2 for convenience.

For \( x = -7 \):

\[ y = -3(-7) - 5 = 21 - 5 = 16 \]

So, one solution is \( (-7, 16) \).

For \( x = 5 \):

\[ y = -3(5) - 5 = -15 - 5 = -20 \]

So, the other solution is \( (5, -20) \).

Thus, the solutions to the system are:

1. \( (-7, 16) \)
2. \( (5, -20) \)

Among the options provided, we can see the solutions correspond to:

• (−7,16) and (5,−20).