How To Solve 2x 1/y = 3 [solve For Y]

How do you solve 2(x+1)+(y-3)=4, 5(x+1)-y=0?

An ad hoc method would be noting that [math]y=5(x+1)[/math], so the other equation becomes[math]2(x+1)+5(x+1)-3=4[/math]Hence [math]7(x+1)=7[/math], so [math]x+1=1[/math] and finally [math]x=0[/math]. This implies [math]y=5[/math].A general method is to write each equation in the form [math]ax+by=h[/math]. In this case,[math]\begin{cases}2x+y=5\\[1ex]5x-y=-5\end{cases}[/math]Here several strategies are possible for removing one unknown: multiplying the first equation by 5 and the second equation by [math]-2[/math] and summing them would eliminate the [math]x[/math]. Similarly can be done with [math]y[/math], which in this case is even simpler: summing up the two equations yields [math]7x=0[/math]. Once [math]x[/math] is determined, substitute the value and solve for [math]y[/math].

How do you solve x^2 d^2y/dx^2 - 2x (1+x) dy/dx +2 (1+x) y = x^3?

On dividing the given equation by x^2, it becomes y'' - 2[(1+x)/x]y' + 2(1+x)/x^2 = x, which is a second order linear differential equation of the form : y''+f(x)y'+g(x)y = r(x), where x.f(x) and (x^2).g((x) and r(x) are analytic at x=0, i.e. x=0, is a regular singular point of the equation. Accordingly we apply the Frobenius' Extended Power series method and puty=Sigma(n=0 to inf.)(c_n)x^(n+k), in the original equation, where the coefficients c_n and the index k are to be determined by differentiating the above, twice with respect to x and by comparison of coefficients of x on the two sides of the equation.

How do I solve 2^5x = 3^2x+1?

[math]2^{5x} = 32^x[/math] ,[math]3^{2x} = 9^x[/math].Therefore, the equation can be rewritten as,[math]32^x = 9^x + 1[/math]There are two methods to take it from here,1 : Graphical Method,The point of intersection of the two curves is (0.318, 3.012)So, the values of x is 0.318 .2 : Analytical Method.Let [math]f(x) = 32^x - 9^x - 1[/math]We wish to find the roots of this function.The method I will be using is called Newton’s method, where we start off with an approximation [math]x_0[/math]Let [math]x_0 = 0[/math]Then [math]x_1 = x_0 - \frac {f(x_0)}{f'(x_0)}[/math][math]f(x_0) = -1 and f'(x_0) = ln(32) - ln(9) \approx 1.26[/math][math]x_1 = 0 - \frac {-1}{1.26} = 0 + 0.79 = 0.79[/math][math]x_2 = 0.79 - \frac {f(0.79)}{f'(0.79)} = 0.58 [/math][math]x_3 = 0.58 - \frac {f(0.58)}{f'(0.58)} = 0.43[/math][math]x_4 = 0.43 - \frac {f(0.43)}{f'(0.43)} = 0.30[/math]You can iterate a couple more times using more accurate values, but I think you get the idea.[math]x \approx 0.30[/math]FINAL ANSWER : [math]x = 0.318 \approx 0.300[/math]

How would you solve [math]1+2^x+2^ {2x+1} =y^2[/math] over the integers?

First note that for [math]x\leq -2[/math], the LHS is not an integer. When [math]x=-1[/math], the LHS is [math]2[/math] which is not a perfect square. When [math]x=0[/math], we have [math]y=\pm 2[/math]. We can try [math]x=1,2[/math] and see that these do not work.So consider [math]x\geq 3[/math]. We have [math](y-1)(y+1)=2^x (2^{x+1}+1)[/math]. Now certainly [math](y-1)[/math] and [math](y+1)[/math] are both even, and since they differ by 2 at most one of them can be a multiple of 4.But the RHS has a factor of [math]2^x[/math], so what does this mean? It means this power [math]2^x[/math] can only be distributed among [math](y-1)[/math] and [math](y+1)[/math] as [math]2 [/math]and [math]2^{x-1}[/math] or [math]2^{x-1} [/math]and [math]2[/math] respectively. For [math]x\geq 3[/math], any other distribution leaves either one of [math](y-1),(y+1)[/math] to be odd or both of them a multiple of 4.In other words, it means either [math]y-1=2^{x-1}k[/math] or [math]y+1=2^{x-1}k[/math] for some odd [math]k[/math]. (Note [math]k[/math] is odd because [math]2^{x-1}[/math] should already account for all the powers of 2 which divide [math]y\pm 1[/math].)Going back to the original equation, we have [math](2^{x-1}k\pm 1)^2 = 2^{2x+1}+2^x+1[/math]. Now if [math]k\geq 5[/math], we have [math]y\geq 4\times 2^{x-1}=2^{x+1}[/math] so [math]y^2\geq 2^{2x+2}[/math]. This means [math]2^{2x+1}+2^x+1\geq 2^{2x+2}[/math], or [math]2^x+1\geq 2^{2x+1}[/math], which is evidently impossible.Thus [math]k=1[/math] or [math]3[/math]. Let [math]2^{x-1}=t[/math] for convenience. When [math]k=1[/math], we have [math](t\pm 1)^2 = 8t^2+2t+1[/math], or [math]t\pm 2[/math][math]=[/math][math]8t+2[/math], which has no possible solutions for [math]t[/math].When [math]k=3[/math], we have [math](3t\pm 1)^2 = 8t^2+2t+1[/math], or [math]9t\pm 6=8t+2[/math]. For [math]9t-6=8t+2[/math] this gives [math]t=8[/math] while for [math]9t+6=8t+2[/math] there are no possible [math]t[/math].So [math]t=8[/math] which gives [math]x=4[/math] and then [math]y=\pm 23[/math].In conclusion, the only solutions are [math](x,y)=(0, \pm 2), (4, \pm 23)[/math].

How do I solve the equation 2y + 1 = y + 3?

2y+1 = y+3 ,2y-y = 3–1 ,Y(2–1) = 2= Y is 2 .