This feels like a stupid question but I'm struggling with it nevertheless. I have the following variables $$u=2x+y \qquad v=\frac{5x}{x+y}$$ I have been trying in vain to put $x$ and $y$ in terms of $u$ and $v$, but with the variables I have the method for eliminating variables is not clear to me.

Sorry if this is a little too elementary for this exchange, but is there a useful method for combining variables in situations such as this?


  • $\begingroup$ The basic idea is to eliminate one of the variables x or y. So to eliminate y for example, you need an equation without a y in it. One simple way to do this is to solve for y in one of the equations, and plug that formula into the 2nd equation. So you'll get an equation with u,v,x which you can solve for x. Then substitute that formula for x back into one of your two original equations (either one will work) and solve for y. $\endgroup$ – Ameet Sharma Nov 18 '20 at 1:17

First, note that $y = u - 2x$. Then, substituting:

$$v = \frac{5x}{u - x}$$

$$v(u - x) = 5x$$

Here, we assume $x\neq u$.

$$uv - vx = 5x$$

$$(5 + v)x = uv$$

$$x = \frac{uv}{5 + v}$$

$$y = u - 2\frac{uv}{5 + v}$$

$$y = \frac{uv + 5u}{5 + v} - \frac{2uv}{5 + v}$$

$$y = \frac{5u - uv}{5 + v}$$


$$\boxed{(x,y) = \bigg(\frac{uv}{5 + v}, \frac{5u - uv}{5 + v}\bigg)}$$


Another trick to eliminate one of the variables is to take linear combinations of the two equations to eliminate a variable. I rearrange the 2nd equation a bit

$u = 2x+y$ (1)

$\frac{5x}{v} = x+y$ (2)

I take eqn (1) and subtract (2) to eliminate y. So I take the left-side of (1) and subtract the left-side of (2). and the same with the right side. so I subtract out the y's.

$u-\frac{5x}{v} = x$ (notice there's no y)

$u = x(1+\frac{5}{v})$

multiply both sides by v to get rid of that v in the denominator

$uv = x(v+5)$


Now you can substitute this x back into either of your original two equations to solve for y.


Upon inspection, it is best to solve for $(x + y)$ as a first step (instead of solving for $x$ or $y$). $$\begin{align} u &= 2x + y\\ u-x&=x+y \end{align} $$ Substituting in to the second equation... $$\begin{align} v&=\frac{5x}{u-x} \tag{ where $u\neq x$ }\\ v(u-x)&=5x \\ uv-vx&=5x \\ uv&=5x+vx\\ uv&=x(v+5)\\ x &= \frac{uv}{v+5} \end{align}$$ We then substitute this back into the first equation to get $y$. $$\begin{align} u&=2\bigg(\frac{uv}{v+5}\bigg)+y\\ y&=u-\frac{2uv}{v+5}\end{align} $$


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