Y=6x^2 -5x+6 ( the first option )
Answer:
x = -5, and y = -6
Step-by-step explanation:
Suppose that we have two equations:
A = B
and
C = D
combining the equations means that we will do:
First we multiply both whole equations by constants:
k*(A = B) ---> k*A = k*B
j*(C = D) ----> j*C = j*D
And then we "add" them:
k*A + j*C = k*B + j*D
Now we have the equations:
-x - y = 11
4*x - 5*y = 10
We want to add them in a given form that one of the variables cancels, so we can solve it for the other variable.
Then we can take the first equation:
-x - y = 11
and multiply both sides by 4.
4*(-x - y = 11)
Then we get:
4*(-x - y) = 4*11
-4*x - 4*y = 44
Now we have the two equations:
-4*x - 4*y = 44
4*x - 5*y = 10
(here we can think that we multiplied the second equation by 1, then we have k = 4, and j = 1)
If we add them, we get:
(-4*x - 4*y) + (4*x - 5*y) = 10 + 44
-4*x - 4*y + 4*x - 5*y = 54
-9*y = 54
So we combined the equations and now ended with an equation that is really easy to solve for y.
y = 54/-9 = -6
Now that we know the value of y, we can simply replace it in one of the two equations to get the value of x.
-x - y = 11
-x - (-6) = 11
-x + 6 = 11
-x = 11 -6 = 5
-x = 5
x = -5
Then:
x = -5, and y = -6
It would be 15 greater than or equal to 14
Answer:
the equation is : x²-x-12
Step-by-step explanation:
the quadratic equation is in the form of : y=ax²+bx+c
the product of the zeros is -12 and the sum is 1
b = - 1
c=-12 (product)
y=x²-x-12
check : factorize first (x+3)(x-4)=0
either x+3=0 then x=-3
or x-4=0 then x=4
-3*4=-12
-3+4=1
the equation is : x²-x-12
