May 3, 2020 · 1 Here is one way to use elimination to solve your system of linear equations. However we first must put the equations into standard form.

Dec 17, 2016 · I spent hours trying to solve this system of equations, they're two equations with two variables, and both are 2nd degree equations, I don't know, both equal zero, so each one equals the …

Conversely, you can solve for y in terms of x x, and then use (any? or just integer x x so that y y is an integer -- you filed this under "diophantine equations") x x choices to get y y s for (x,y) (x, y) solution …

Jul 2, 2018 · If $x^2 - 4yx + 3y^2 = 0$, we can view this as a quadratic equation in standard form with $a=1$, $b=-4y$, and $c=3y^2$; so, we can use the quadratic formula just as in the single variable case:

Sep 2, 2019 · Equations that have only integer solutions (or, in this case, positive integer solutions) are called Diophantine equations. A linear Diophantine equation like $7w+3d=44$ is relatively easy to …

May 10, 2016 · It depends on what you mean by "solve", and also whether the "two variables each" come in distinct pairs. For example we can easily demonstrate a system of three equations, each …

Jan 30, 2018 · $$3y^2-63+12x=0$$ Actually they are partial derivatives of this equation $f (x,y)=x^3+y^3-63 (x+y)+12xy$ I am doing maxima minima questions of two variables and i am really …

Feb 25, 2016 · 2 This system of equations can be solved, it just doesn't have a unique solution. We can try to solve it as if we had just 2 variables, x x and y y. Subtract the first equation from the second to …

Jan 8, 2017 · How to solve a differential equation of the following form: $\dfrac {ax} {\dfrac {\partial {f (x,y)}} {\partial x}} + \dfrac {by} {\dfrac {\partial {f (x,y)}} {\partial y}}=-1 $

May 9, 2014 · 2 I know how to solve quadratic equations when there's only one unknown, but I'm a bit confused on what I do if I have $2$ unknowns e.g: $$x^2 + 2kx + 81 =0$$ With just $x^2 + 2kx + …