a(n-3)+8=b solve for n
an-3a +8=b (subtracted an from left side) I get:
3a+8 = b-an
=3a +8 = b-a(n)
=3a+8/b-a = n
but i think this is wrong...help...thank you
You are right. It is wrong
an-3a +8=b (subtracted an from left side)
Yes but
I get:
3a+8 = b-an
NO
-3a + 8 = b - an
an - 3a + 8 = b
an = 3a+b-8
n = (3a+b-8)/a
from : an-3a +8=b
an = 3a + b - 8
n = (3a + b - 8)/a
subtract 8 ... a(n-3) = b-8
divide by a ... n - 3 = (b-8)/a
add 3 ... n = [(b-8)/a] + 3
To solve for n in the equation a(n-3) + 8 = b, we can follow these steps:
1. Start with the original equation: a(n-3) + 8 = b.
2. Distribute the "a" to both terms inside the parentheses: an - 3a + 8 = b.
3. Move the 8 to the other side of the equation by subtracting 8 from both sides: an - 3a = b - 8.
4. If you want to isolate n, you need to get rid of the "-3a" term. To do that, we divide both sides of the equation by "a": (an - 3a) / a = (b - 8) / a.
5. Simplify the left side of the equation: n - 3 = (b - 8) / a.
6. Finally, add 3 to both sides of the equation to isolate n: n = (b - 8) / a + 3.
So, the solution for n in terms of a and b is: n = (b - 8) / a + 3.