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Solve for n. A=nz+n

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3 Answers


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

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.




Apologies to Robert; my computer displayed his answer as though it had no content.