These are the problems i have been trying to solve.
Solve for n.
1.) na + n= cd
2.) n/x + 1= x - d
My teacher gave me no explanation for these types of problems. Can someone help me figure them out?
I see a number of mistakes here.
This isn't right. If you want to divide both sides by a here, you have to divide the cd term by a too, like this:
n/a = cd/a - n
This isn't right either, for the same reason. If you want to multiply both sides by a, you have to multiply ALL terms by a, like this:n + n/a = cd
n+ n = cda
na + n = cda
Here's what you should do: one way of factoring is to take out the greatest common monomial factor -- in a sense, you are reversing the distributive property. Since on the left side there is a common factor of n, you "take it out":
$\displaystyle \begin{aligned}
na + n &= cd \\
na + n(1) &= cd \\
n(a + 1) &= cd \\
\end{aligned}$
You can see that this is true by going backwards and distributing the n. Now that you have n times something, you can divide both sides by that other factor:
$\displaystyle \begin{aligned}
n(a + 1) &= cd \\
\frac{n(a + 1)}{(a + 1)} &= \frac{cd}{(a + 1)} \\
n &= \frac{cd}{a + 1}
\end{aligned}$
It's the first one. Here's how to get it, step by step:
$\displaystyle \begin{aligned}
\frac{n}{x} + 1 &= x - d \\
x \left( \frac{n}{x} + 1 \right) &= x(x - d) \\
x \left( \frac{n}{x} \right) + x(1) &= x(x) - x(d) \\
n + x &= x^2 - dx
\end{aligned}$
Then subtract x from both sides, and you can clean up along the way:
$\displaystyle \begin{aligned}
n + x &= x^2 - dx \\
n &= x^2 - dx - x \\
n &= x^2 - x(d + 1) \\
n &= x^2 - (d + 1)x \\
\end{aligned}$
You should learn some LaTeX in typing these equations, because I'm not sure what your equation is. I could read it in one of two ways.
(A)$\displaystyle n + \frac{2}{c} = n + \frac{c}{b}$
(B)$\displaystyle \frac{n + 2}{c} = \frac{n + c}{b}$
So which one is it? (A) or (B)? Because if it is (A), the n's would cancel from both sides.
LaTeX is the use of math tabs. You put the square brackets around 'math' like this [tex] and you close by inserting a similar tab but with '/' in front of 'math'. This allow the ease of display of some functions and fractions
If it's B, then
$\displaystyle \frac{n+2}{c} = \frac{n+c}{b}$
Multiply throughout by bc. Some b and c will cancel out, giving;
$\displaystyle b(n+2) = c(n+c)$
Expand;
$\displaystyle bn + 2b = nc + c^2$
Remove nc from both sides, and remove 2 b too from both sides;
$\displaystyle bn + 2b -2b - nc = nc + c^2 - 2b - nc$
$\displaystyle bn - nc = c^2 - 2b$
Factor n,
$\displaystyle n(b - c) = c^2 - 2b$
Divide by (b-c);
$\displaystyle n= \frac{c^2 - 2b}{b-c}$
Looks like you did well!