Solve the system below.
y=x2-2x-6
y=4x+10
A. (-2, 2)
B. (8, 42)
C. (-2, 2) (8, 42)
D. There is no real solution
do you mean
y = x^2 - 2 x - 6 ????
if so then
4 x + 10 = x^2 - 2 x - 6
x^2 - 6 x - 16 = 0
(x-8)(x+2) = 0
x = +8 or x = -2
C is looking good but check it
if x = -2 does y = +2 work?
yes
if x = +8 does 42 work?
looks good so C
thanks Damon:)
Knock, knock.
Who's there?
Solve.
Solve who?
Solve this system, silly!
To find the solution, we need to set the two equations equal to each other:
x^2 - 2x - 6 = 4x + 10
Now, let's simplify:
x^2 - 2x - 6 - 4x - 10 = 0
x^2 - 6x - 16 = 0
Now, we can use the quadratic formula to solve for x:
x = (-(-6) ± √((-6)^2 - 4(1)(-16))) / 2(1)
x = (6 ± √(36 + 64)) / 2
x = (6 ± √100) / 2
x = (6 ± 10) / 2
So we have two possible values for x: x = 8 and x = -2.
Now let's substitute each value of x back into one of the original equations to find the y-coordinate:
For x = 8:
y = (8)^2 - 2(8) - 6
y = 64 - 16 - 6
y = 42
For x = -2:
y = (-2)^2 - 2(-2) - 6
y = 4 + 4 - 6
y = 2
Therefore, the solution to the system is (8, 42) and (-2, 2), which is answer option C: (-2, 2) (8, 42).
Looks like those two points were having a secret meeting!
To solve the system of equations, we can set y in both equations equal to each other:
x^2 - 2x - 6 = 4x + 10
Rearranging the equation, we get:
x^2 - 6x - 16 = 0
Factoring or using the quadratic formula, we find:
(x - 8)(x + 2) = 0
So, x = 8 or x = -2.
To find the corresponding values of y, we substitute these values of x back into either equation.
For x = 8:
y = (8)^2 - 2(8) - 6
= 64 - 16 - 6
= 42
For x = -2:
y = (-2)^2 - 2(-2) - 6
= 4 + 4 - 6
= 2
Therefore, the solutions are (-2, 2) and (8, 42).
The correct answer is C. (-2, 2) (8, 42).
To solve the system of equations, we can set the two equations equal to each other and solve for x.
x^2 - 2x - 6 = 4x + 10
Start by combining like terms:
x^2 - 2x - 4x - 6 - 10 = 0
x^2 - 6x - 16 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = -6, and c = -16. Plugging these values into the quadratic formula:
x = (-(-6) ± √((-6)^2 - 4(1)(-16))) / (2(1))
x = (6 ± √(36 + 64)) / 2
x = (6 ± √100) / 2
x = (6 ± 10) / 2
Now we solve for x using both the positive and negative roots:
x₁ = (6 + 10) / 2 = 16 / 2 = 8
x₂ = (6 - 10) / 2 = -4 / 2 = -2
So the solutions for x are x = 8 and x = -2.
Now we can substitute these values back into either equation to solve for y. Let's use the first equation y = x^2 - 2x - 6:
When x = 8:
y = 8^2 - 2(8) - 6 = 64 - 16 - 6 = 42
When x = -2:
y = (-2)^2 - 2(-2) - 6 = 4 + 4 - 6 = 2
Therefore, the solutions to the system of equations are (-2, 2) and (8, 42). So the answer is option C: (-2, 2) (8, 42).