I'm sorry I can't see it clearly.
Answer:
(B) 3n - 11
Step-by-step explanation:
(2n + 5) - (3n + 7) + (4n - 9) = 2n + 5 - 3n - 7 +4n - 9 = 3n - 11
Answer:
(1, 3)
Step-by-step explanation:
x - 3y = -8
3x + y = 6
Isolate a variable in one of the equations:
y = 6 - 3x
Substitute the value of y into the other equation:
x - 3(6 - 3x) = -8
Use distributive property:
x - 18 + 9x = -8
Combine like terms:
10x - 18 = -8
Isolate the variable:
10x = 10
x = 1
Substitute the value of x into any equation:
3(1) + y = 6
3 + y = 6
Isolate the variable:
y = 3
No
3÷4 = .75
4÷5 = .8
5÷6 ≈ .8333
6÷7 ≈ .8571
These ratios are not the same.
Note that if

, then

, and so we can collapse the system of ODEs into a linear ODE:


which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

so that the characteristic solution is

Now let's suppose the particular solution is

. Then

and so

Thus the general solution for

is

and you can find the solution

by simply differentiating

.