Answer:
uh what is this address for or am i not supposed to be in this
Step-by-step explanation:
sorry if im getting in your buisness
Notice that the 2 expressions have 2 common terms.
(r-s) is just (s-r) times (-1)
similarly
(t-s) is just (s-t) times (-1)
this means that :
(r-s) (t-s) + (s-r) (s-t)=-(s-r)[-(s-t)]+(s-r) (s-t)
the 2 minuses in the first multiplication cancel each other so we have:
-(s-r)[-(s-t)]+(s-r) (s-t)=(s-r) (s-t)+(s-r) (s-t)=2(s-r) (s-t)
Answer:
d)<span>2(s-r) (t-s) </span>
Answer:
Ok, we have a system of equations:
6*x + 3*y = 6*x*y
2*x + 4*y = 5*x*y
First, we want to isolate one of the variables,
As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:
(6*x + 3*y)/(2*x + 4*y) = 6/5
now we isolate one off the variables:
6*x + 3*y = (6/5)*(2*x + 4*y) = (12/5)*x + (24/5)*y
x*(6 - 12/5) = y*(24/5 - 3)
x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y
Now we can replace it in the first equation:
6*x + 3*y = 6*x*y
6*(0.5*y) + 3*y = 6*(0.5*y)*y
3*y + 3*y = 3*y^2
3*y^2 - 6*y = 0
Now we can find the solutions of that quadratic equation as:

So we have two solutions
y = 0
y = 2.
Suppose that we select the solution y = 0
Then, using one of the equations we can find the value of x:
2*x + 4*0 = 5*x*0
2*x = 0
x = 0
(0, 0) is a solution
if we select the other solution, y = 2.
2*x + 4*2 = 5*x*2
2*x + 8 = 10*x
8 = (10 - 2)*x = 8x
x = 1.
(1, 2) is other solution
If we add the equations it looks like
-5y + 8x + 5y + 2x = -18+58
so 10x=40
so x=40/10=4
now let's replace x by 4 in the second equation
5y +2*4=58
5y=58-2*4=58-8=50
so y=50/5=10
so (x, y) = (2, 10)
I will help you if you show me what questions 1,2,3,4 are.