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### algebra precalculus - Simplyfying each side separately and solving linear equation

I'm given an equation that first needs simplifying:$$\frac{x-2}{x-4} - \frac{x-4}{x-6} = \frac{x-1}{x-3} - \frac{x-3}{x-5}$$My next step is:$$\frac{(x-2)(x-4)(x-6)}{x-4} - \frac{(x-4)(x-4)(x-6)}{x-6} = \frac{(x-1)(x-3)(x-5)}{x-3} - \frac{(x-3)(x-3)(x-5)}{x-5}$$Tydying this up, I get:$$(x-2)(x-6) - (x-1)(x-5) - (x-3)(x-3)$$The next step:$$x^2 -6x -2x +12 - x^2 + 4x - 4x +16 = x^2 - 5x -x + 5 +3x +3x -9$$Then:$$-8x - 4 = -4$$$$-8x-5$$Coming to the final solution:$$x = \frac{5}{8}$$which I'm not sure is the correct solution. Can somebody please ta...Read more

### algebra precalculus - Silly doubt regarding onto function

Let function $f:(0,\infty)\to(0,\infty)$ be defined as $f(x)=|1-\frac{1}{x}|$. Then is it onto function?My doubt is that here the codomain doesn't include $0$ but here in the function $0$ is there on $x=1$. So, will it be an onto function? (confusion is created due to this point $(1,0)$ ).Thanks for clearing my doubt....Read more

### algebra precalculus - Condition in terms of b and a if $ax^2+bx+c=0$ has two consecutive odd positive integers as roots

The roots of the equation$$ax^2+bx+c=0$$, where $a \geq 0$, are two consecutive odd positive integers, then (A) $|b|\leq 4a$(B) $|b|\geq 4a$(C) $|b|=2a$(D) None of theseMy attemptLet p and q be the roots then if they are consecutive positive integers (q>p) then $$pq=\frac{c}{a} \geq 0$$So, $$c \geq 0$$ and $$q-p=2$$So, $$\frac{\sqrt{b^2-4ac}}{a}=2$$So,$$|b|>2a$$ $(Since, a>0,c>0)$But I know that 4ac should be taken in consideration since its not equal to zero. But I don't know how to use it. Any hints and suggestions are welcome!...Read more

### algebra precalculus - Root/Solution confusion

Suppose I have the equation $x-a=0$.So, $x-a=0\Rightarrow x=a$. The unique solution is (or root) $a$ (by definition a solution is the value for the unknown that make the equation true, $a-a=0$ ).Since $x=a$ shouldn't x(the unknown) considered a solution?Sorry if the question isn't clear but i have this stupid doubt....Read more

### algebra precalculus - How to expand this polynomial division?

My Physics teacher gave me a problem and its solution, what I have todo is to expand the solution, but when I do it I do not get to the same solution he says is the right one, here is the problem: We cut off a rectangle ABCD along the line DE through the corner D, so when we suspend the plate ABED from the point E the line DA, is on horizontal position. Calculate the magnitude EBThe image below is the solution, What I am not able to figure out is how the teacher solved the polynomial division and get to that result.Any help is appreciated...Read more