Use the substitution method to find the solutions to the linear-quadratic system y=x2−x−40, y=−3x−5?(1 point)%0D%0AResponses%0D%0A%0D%0A(−7,26)%0D%0Aleft parenthesis negative 7 comma 26 right parenthesis%0D%0A%0D%0A(5,−10)%0D%0Aleft parenthesis 5 comma negative 10 right parenthesis%0D%0A%0D%0A(5,−10) and (−7,26)%0D%0Aleft parenthesis 5 comma negative 10 right parenthesis and left parenthesis negative 7 comma 26 right parenthesis%0D%0A%0D%0A(5,−20) and (−7,16)

Answer 1

To solve the linear-quadratic system using the substitution method, we have the following equations:

1. \( y = x^2 - x - 40 \)
2. \( y = -3x - 5 \)

We can substitute the expression for \( y \) from the linear equation into the quadratic equation.

So we set \( -3x - 5 \) equal to \( x^2 - x - 40 \):

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

Now, let's rearrange the equation to form a standard polynomial equation:

\[ 0 = x^2 - x - 40 + 3x + 5 \] \[ 0 = x^2 + 2x - 35 \]

Next, we need to factor the quadratic \( x^2 + 2x - 35 \). We look for two numbers that multiply to \(-35\) and add to \(2\). These numbers are \(7\) and \(-5\).

Thus, we can factor the equation as follows:

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

Setting each factor to zero gives us the potential \(x\) values:

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

Now, we will find the corresponding \(y\) values for each \(x\):

1. For \( x = -7 \): \[ y = -3(-7) - 5 = 21 - 5 = 16 \] So, one solution is \( (-7, 16) \).

2. For \( x = 5 \): \[ y = -3(5) - 5 = -15 - 5 = -20 \] So, another solution is \( (5, -20) \).

Now we have the two solutions:

• \( (-7, 16) \)
• \( (5, -20) \)

Among your options, the closest representations are:

• \( (5, -10) \) and \( (-7, 26) \) — which are incorrect based on our calculations.
• \( (5, -20) \) and \( (-7, 16) \) — which is the accurate representation of the solutions found.

Therefore, the correct answer is:

(5, -20) and (-7, 16)