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 - 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 - 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

algebra precalculus - Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?

Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?I am learning trigonometric identities one identity I have to proof is the next:$ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$so I tried to resolve the identity for the left:$ 1 + \sin^2 \alpha+ \cos^2\alpha - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $$= 1 + 1 - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $ $= 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$And I got stuck, I did not know what to do, so I went to ...Read more

education - Free introductory resources for learning algebra?

I have a non-mathematician friend who is interested in re-learning algebra. I am more than happy to help, but I am in no position to judge what is a good introductory text --- only to identify when a text is a very bad one.My friend is interested in starting from "basics" --- he's comfortable with order-of-operations, minus-times-minus-is-plus, and positive integer exponents. He's shaky with negative exponents, logarithms, and roots.Any recommendations for good, freely-available resources?...Read more

algebra precalculus - How do you deal with Math Frustration?

So I'm covering Algebra 1 and 2 atm and I'm wondering how you guys deal with frustration regarding math problems. Sometimes I find myself watching explanations and just can't figure out how it makes sense, so I get frustrated, leave for a couple hours. When I return I usually sit with it for 5 min and it turns out to make sense. I get this quite a lot and it interrupts my study sessions. Tips/Tricks?...Read more

algebra precalculus - Good books for building number/math intuition

I'm wondering if there are some good book/textbooks that were designed with algebraic logic in mind (ie. building intuition rather than rote learning).As an example of what I mean, consider this question I found on Khan Academy: Given that $c<d<b$, is $\frac{b}{b+c+d}$ greater than, less than, or equal to $1/3$? Solving this problem myself really helped with my math intuition (sometimes I'm appalled with how bad my math intuition is) and I'd love to work through more things like this.Edit: I have read How to Solve It and I'm looking for s...Read more

algebra precalculus - Common volume of three cylinders with unequal radii

I would like to solve it please for the case where the radii can be a similar size - so the case where this statement is NOT true: $$ \mathbf r_1^2 \mathbf \geq \mathbf r_2^2 \mathbf + \mathbf r_3^2 $$ How do you solve for the common volume of 3 cylinders with unequal radii? (If you could please include the integral needed - I think it might need a cartesian equation system? Or if you have any good idea of what direction or things I could read up on to please learn to solve this. Thank you so much for your help!$$x^2 + y^2 = r_1^2$$$$x^2 + z^...Read more

algebra precalculus - If the radius of the smallest circle is x cm form an equation in x and show that it reduces to a quadratic equation

What i triedI tried finding the area of the rectangle ABCD which is $5*7=35$ and then i tried to find the area of the three quadrants in terms of $x$ and add those areas up, however, i still couldn't find the area of the two region that arent under the quadrants. I have tried various ways to find the area of these two regions outside the quadrants but i couldn't find them. Could anyone explain please. Thanks...Read more

algebra precalculus - Collinearity when $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$

Let $\mathbf{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$. Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}.$Since the cross product of two vectors gives an area, and for two vectors to give an area of $0$ they need to be on the same line (or they can be a point, but I'm assu...Read more