find the solutions to the system
y=x^2+5x=6
y=4x=12
A( 2,20) and (-3,0)
B (2,20) and (-3,18)
C(-2,-10) and (-3,-18)
D no solutions
some of those = signs are probably +
which ones?
then, show us your work at solving the system ...
Or heck, just try plugging in the 3 points to see which on works, if any.
Y = x^2 + 5x + 6.
Y = 4x + 12.
4x + 12 = x^2 + 5x + 6.
x^2 + x - 6 = 0. -6 = -2 * 3. sum = -2 + 3 = 1 = B.
(x-2)(x+3) = 0.
x - 2 = 0, X = 2.
x + 3 = 0, X = -3.
When X = 2, Y = 4x + 12 = 4*2 + 12 = 20. (2, 20).
When X = -3, Y = 4x + 12 = 4*(-3) + 12 = 0. (-3, 0).
y = x^2+5x+6 = (x+2)(x+3)
y = 4x+12 = 4(x+3)
so, 4(x+3) = (x+2)(x+3)
4 = x+2 or 0 = x+3
...
To find the solutions to the system of equations, let's solve each equation separately and then compare the solutions.
Given system of equations:
1. y = x^2 + 5x - 6
2. y = 4x + 12
1. Solving equation (1):
y = x^2 + 5x - 6
To solve this equation, we can set it equal to zero and factor it or use the quadratic formula. Let's factor it:
0 = x^2 + 5x - 6
The factors of -6 that add up to +5 are +6 and -1:
0 = (x + 6)(x - 1)
Setting each factor equal to zero:
x + 6 = 0 or x - 1 = 0
Solving for x:
x = -6 or x = 1
So the solutions for equation (1) are x = -6 and x = 1. Now let's substitute these values into equation (1) to find the corresponding y-values.
When x = -6, substituting into equation (1):
y = (-6)^2 + 5(-6) - 6
y = 36 - 30 - 6
y = 0
So one solution is (-6, 0).
When x = 1, substituting into equation (1):
y = (1)^2 + 5(1) - 6
y = 1 + 5 - 6
y = 0
So another solution is (1, 0).
2. Solving equation (2):
y = 4x + 12
This equation represents a linear equation with a slope of 4 and a y-intercept of 12. We can choose any x-value and substitute it into the equation to find the corresponding y-value.
For example, let's substitute x = 2 into equation (2):
y = 4(2) + 12
y = 8 + 12
y = 20
So one solution is (2, 20).
Similarly, let's substitute x = -3 into equation (2):
y = 4(-3) + 12
y = -12 + 12
y = 0
So another solution is (-3, 0).
Comparing the solutions of both equations, we find that the common solution is (-3, 0).
Therefore, option D, which states "no solutions," is incorrect.
The correct answer is option A, which states "A(2,20) and (-3,0)."