the substitution method to find the solutions to the linear-quadratic system y = x ^ 2 - x - 40 , y = - 3x - 5 ? 1 point ) (5, - 10) and (- 7, 26); (5, - 10); (5, - 20) and (- 7, 16); (- 7, 26)

Answer 1

To find the solutions to the linear-quadratic system using the substitution method, we will substitute the value of y in the first equation with the corresponding equation in the second equation.

Given equations:

1) y = x^2 - x - 40
2) y = -3x - 5

Substituting y in equation 1 with -3x - 5 from equation 2:

-3x - 5 = x^2 - x - 40

Rearrange this equation to get a quadratic equation:

x^2 - x + 3x - 5 + 40 = 0
x^2 + 2x + 35 = 0

This quadratic equation does not factorize nicely, so we will use the quadratic formula to find the values of x:

x = (-2 ± sqrt(2^2 - 4(1)(35))) / 2(1)
x = (-2 ± sqrt(-136)) / 2
x = (-2 ± 2i√(34)) / 2
x = -1 ± i√(34)

Therefore, the solutions to the system of equations are:

(-7, 26) and (5, -10)

Answer 2

U sure?

Answer 3

Apologies for the mistake in the calculation. Let's find the correct solutions to the linear-quadratic system using the substitution method.

Given equations:

1) y = x^2 - x - 40
2) y = -3x - 5

We can substitute equation 2 into equation 1:

x^2 - x - 40 = -3x - 5

Rearrange this equation:

x^2 + 2x - 35 = 0

This equation can be factored as:

(x + 7)(x - 5) = 0

So, we have two possible values for x:

x + 7 = 0 -> x = -7
x - 5 = 0 -> x = 5

Now substitute these x values back into either equation to find their corresponding y values:

For x = -7:

y = (-7)^2 - (-7) - 40
y = 49 + 7 - 40
y = 16

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

For x = 5:

y = (5)^2 - (5) - 40
y = 25 - 5 - 40
y = -20

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

Therefore, the correct solutions to the linear-quadratic system are (5, -20) and (-7, 16).