Solve for n. A=nz+n

Question

I got 2 possible answers: 1. n=(A-z)/z

                                     2. n=A/z-1

Please solve and give details for each and every step.

Answer 1

Anna,

This question is best solved by factoring. Factor the n from the right hand size of the equation, then just use algebraic rules to solve from there.

Best of luck, Steve

Answer 2

We can use the distributive property to factor the right side of the equation for n, and we get

A = nz+n = n(z+1)

Then, we can use the division property of equality to prepare to move z+1 to the left side of the equation and to isolate n.

A/(z+1) = n×(z+1)/(z+1)

Now, we apply the identity property of division.

A/(z+1) = n×(z+1)/(z+1) = n×1 = n

So n = A/(z+1).

We can test the solutions that you found by solving them again for A.  In the first, we have the following.

n = (A-z)/z

nz = A-z

nz+z = A

Since that isn't what we started with, your solution (1) isn't correct.

For the second, we have

n = A/z-1

If that is n = (A/z)-1, then

n+1 = A/z

zn+z = A

If it is n = A/(z-1), then

n(z-1) = A

nz-n = A

That is close, but without seeing your work, it is hard to know how z+1 became z-1, which I think is where you had your error.

 

HTH.

Answer 3

Factor out n,

A = n(z+1)

Isolate n by dividing both sides by z+1,

n = A/(z+1) <==Answer