# Thread: Solving for one of the variables help

1. ## Solving for one of the variables help

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?

2. The first problem, factor out the n. The second problem, first multiply both sides by x. Does this help?

3. 1.) n= cda/2 ?

2.) n= x^2 -dx -1 ?

Are these right? or am i completely off.

5. My work says that they are both wrong. I need a math book! I'm on summer break.

6. Can you help me factor the first one for n?

7. Originally Posted by Youngice11
Can you help me factor the first one for n?
Have you studied factoring? Please show all your work you've done so far.

8. We did not use factoring in Geometry. Obviously. This is my summer assignment for physics. I don't remember how to factor.

na + n= cd
n= cd -na
n/a = cd- n
n + n/a = cd
n+ n = cda
2n=cda
n= cda/2

9. I see a number of mistakes here.
Originally Posted by Youngice11
na + n= cd
n= cd -na
n/a = cd- n
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

n + n/a = cd
n+ n = cda
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:
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}

10. OK. Thank you i understand the distributing and how you got the answer. The second problem....

n/x + 1 =x -d
first you want to multiply times x.

n + x = x^2 -dx or n + 1 = x^2 -dx?

What is next?

11. Originally Posted by Youngice11
OK. Thank you i understand the distributing and how you got the answer. The second problem....

n/x + 1 =x -d
first you want to multiply times x.

n + x = x^2 -dx or n + 1 = x^2 -dx?

What is next?
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}

12. OK. On the third line once you subtracted x, you factored the x. correct? Also.. Can you tell me if this answer is correct to this problem?

solve for n.

n+2/c = n+c/b

n= c^2 -2b/ b-c

13. 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.

14. B)
How do i use LaTeX?

15. 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!

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