How can 1/y = kr + C/e^(rt) be solved for y?

[fiziksfun]

Homework Statement

1/y = kr + C/e^(rt)

Is this equal to y = 1/kr + e^(rt)/C

Thanks

Homework Equations

The Attempt at a Solution

 

[NoMoreExams]

No, 1/a = b + c is not a = 1/b + 1/c but rather a = 1/(b+c)

Think about it this way 3 = 2 + 1 but would you say that 1/3 = 1/2 + 1/1?

 

[Architeuthis Dux]

No, these two equations are not equal. When solving equations, it is important to follow proper algebraic rules and principles. In this case, the first equation is in the form of 1/y = kr + C/e^(rt), where y is the variable we are solving for. To solve for y, we need to isolate it on one side of the equation. This can be done by multiplying both sides by y, which gives us 1 = ykr + Cy/e^(rt). Then, we can subtract ykr from both sides to get 1 - ykr = Cy/e^(rt). Finally, we can multiply both sides by e^(rt)/C to get e^(rt)/C - ykr e^(rt)/C = y. This can also be written as y = e^(rt)/C - ykr e^(rt)/C, which is not the same as the second equation provided. Therefore, the two equations are not equal and the first equation is the correct form for solving 1/y = kr + C/e^(rt).