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)
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)
U sure?
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).