I have this equation  $$   4y = y $$  If I divide by y in both sides I would get this: $$ 4 = 1$$  And this does not have sense.  I know that the solution is 0 but why I get this answer when dividing by y. What's the logic behind?
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I'm currently working through an algebra book, and during the chapter about rational expressions and inequalities, the author has a side note in which he states:     Never divide both sides of the equation by a variable, even if you're   doing it to try to solve a rational equation, because there's a very   real danger that you will actually eliminate answers.   The reason I'm asking this question is because I didn't understand how answers can be eliminated when dividing both sides of an equation by a variable.  Can you show me an example of how this can happen and why?
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Frist I solved this equation $\displaystyle \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{xy}}$ by multiplying $xy \left(\frac{xy}{x}+\frac{xy}{y}\right)=y+x$ The answer is correct.  Next equation was:    $$\frac{\frac{1}{x^2}-\frac{1}{y^2}}{\frac{1}{x^2}+\frac{1}{y^2}}$$  I made the bottom multiplied and got $x^2 y^2.$ So I made the up also multiply and got     $$\frac{\frac{y^2}{x^2}-\frac{x^2}{y^2}}{\frac{1}{x^2y^2}}.$$   Now I'm stuck.  So now I doubt I did the first one the right way. I might have gotten the answer right but not the correct way to get it. Im sorry if Im confusing. Can someone help me with the two were I go wrong?
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I have seen the Pell's equation wiki page but I need to prove this from scratch without mentioning any formula. I have also seen multiple answers on this site but the answers tend to skip over and assume formulas.  This is what people do -> $x^2 - 3y^2 = 1$ has solution 1, 0.  Next they say we "observe" that $$(x',y') = (2x + 3y, x + 2y)$$ is also a solution.  What I'm asking is how they get this value. Can someone help?
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So I have the difference equation $4y_{k+1} = 2y_k$.  I need to solve this equation, first with the initial condition $y_0 = 0$ and then with the initial condition $y_0 = 2$.  The professor wrote on the answer sheet that applying iteratively the recurrence relation it is easy to check that:  $y_k= (1/2)^k y_0 $.  Since I missed the lecture where he explained this, can somebody explain how to come about with this question?  I imagine that once I get the equation, then the initial condition $y_0 = 0$ gives us the particular solution $y_k = 0$. Thus the other initial condition $y_0 = 2$ is verified by the solution $y_k = 2(1/2)^k$.  Any help on how to get to the equation $y_k= (1/2)^k y_0$ will be appreciated.
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Please guide me:  $y' + ay +b = 0$ ($a$ not zero) is supposed to be separable and has solution  $y = ce^{-ax} - \frac ba$  Here is my start to this problem:  $\frac{dy}{dx} + ay = -b$ is as far as I can go with this. How should I go about separating $x$ and
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I need help with this equation: $$100x - 23y = -19.$$  When I plug this into Wolfram|Alpha, one of the integer solutions is $x = 23n + 12$ where $n$ is a subset of all the integers, but I can't seem to figure out how they got to that answer.
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I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.  Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation.
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Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.  What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:   $(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y &gt; 1$, which can then be rewritten as:  $$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$ Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.  Do you have any suggestion on how I can continue?
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I've often heard that instead of adding up to a little less than one, 1/2 + 1/4 + 1/8... = 1. Is there any way to prove this using equations without using Sigma, or is it just an accepted fact? I need it without Sigma so I can explain it to my little sister.  It is not a duplicate because this one does not use Sigma, and the one marked as duplicate does. I want it to use variables and equations.
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if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$  putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$  i wan,t be able to  go further,could some help me with this
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I know $y=2f(x)$ doubles the $y$ value in vertical shifting, and I know that $y=(1/2)f(x)$ halves the $y$ value. But, why does $y=f(2x)$ halve the $x$ value, and why does $y=f((1/2)x)$ double the $x$ value in horizontal shifting? I looked at the graphs, and I see the difference, but I want to know the math behind why it halves instead of doubling the $x$ value for $y=f(2x)$. Maybe this is a stupid question, but could someone help me out?  Thank you.
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Possible Duplicate:          I obtained a question asking for how to solve $\large x^y = y^x$. The given restraints was that $x$ and $y$ were both positive integers. By a bit of error an trial we quickly see that $x=2$ and $y=4$ is one solution.   My question is: How do one show that $(2\,,\,4)$ is the only non trivial, positive solution to the equation?   Now my initial approach was as follows: We have   $$\large x^y = y^x$$  The trivial solution is obviously when $y=x$, so let us focus on when $y \neq x$.  Let us make a more general statement. Firstly I take the log of both sides  $$\large y \log x = x \log y $$  Let us divide by x and \log x (We now assume $x\neq 0$ and $x\neq 1$ since 0 is not a positive number, and 1 gives us a trivial solution)  $$\large \frac{y}{x} = \frac{\log y }{\log x}$$  For these sides to be equal, we must remove the logarithms on the right hand side, this is achived if $y$ is on the form $x^a$. Now This gives  $$\large \frac{x^a}{x} = \frac{\log \left(x^a\right) }{\log(x)}$$  $$\large x^{a-1} = a$$  So finaly we obtain that $ \displaystyle \large x=\sqrt[ a-1]{a}$ and $\displaystyle \large y = \sqrt[a-1]{a^a}$  Now setting $a=2$ gives us $x = 2$ and $y=4$ as desired.  My question is, how do we prove that $x=2$ and $y=4$ is the only integer solutions? My thought was to show that $ \displaystyle \large \sqrt[ a-1]{a}$ and $\displaystyle \large \sqrt[a-1]{a^a}$ are both irrational when a>2, but I have not been able to show this.   Any help is greatly appreciated, cheers =)
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This may be an incredibly stupid question, but why does partial fraction decomposition avoid division by zero?  Let me give an example:  $$\frac{3x+2}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$$  Multiplying both sides by $x(x+1)$ we have:  $$3x+2=A(x+1)+Bx$$  when $x \neq -1$ and $x \neq 0$.  What is traditionally done here is $x$ is set to $-1$ and $0$ to reveal: $$-3+2=-B \implies 1=B$$ and  $$2=A$$  so we find that   $$\frac{3x+2}{x(x+1)}=\frac{2}{x}+\frac{1}{x+1}$$  Why can $x$ be set equal to the roots of the denominator (in this case, $0$ and $-1$) without creating a division by zero problem?
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Given  If $|x -4| + |y - 4| =4$ then how many integers value the sets $(x,y)$ have? $x$ and $y$ are both positive and negative numbers.  options a)infinite b)$3$ c)$5$ d)$12$ e)$16$ Solution My Approach: i have just put all the values satisfying the equation Hence, Possible values are $(0,4),(4,0)$ $ (2,6),(6,2)$ $ (1,3),(3,1)$ $ (3,7),(7,3)$ $ (5,7),(7,5)$ $ (8,0),(0,8)$ I am getting wrong ans as the Ans is $16$ Is there any better approach to this problem and why the my ans is wrong.
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I am so confused,  how to get a solution of following differential equation:  $$(y')^2 -xy' + y = 0$$  My Thought:  Try solving $y'$ to get $y' = \dfrac{ x \pm \sqrt{x^2 - 4y}}{2}$ and now its unsolvable. Kindly help, appreciated.
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I was working my way through some number theoretic proofs and being a newbie am stuck on this problem :     If $(x, 4) = 2$ and $(y, 4) =2$, then $(x + y, 4) = 4$,     where $(a,b)$ denotes the greatest common divisor of $a$ and $b$.   My Solution  (Incorrect)   $x-4 = 2t$ $y-4 = 2p$ $x + y - 8 = 2(t+p) \Rightarrow x + y - 4 = 2f + 4 = 2(f+2)\Rightarrow 2$ divides $(x+y , 4)$   My Question:  My solution is definitely inadequate. Can someone help me out?
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Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.