It depends. If I had only this specific set of equations to solve, and I wanted an answer in the next few minutes, I might solve the second equation to give X in terms of Y , substitute the result in the first, then plot the result over a range that seemed plausible.

fun2 <- function(Y){
X <- (7-4*(Y^2))/3
2*exp(X)+X^2+3*Y
}
plot(fun2)
fun2.0 <- 2*exp(1)+4
abline(h=fun2.0)

From here, one can read the solution from the plot: Y is approximately 1.

If you want more than this, see "?optim", work the examples, etc. If you want more than this, check the references given with "?optim".

hope this helps. spencer graves

??? wrote:

Hello:

I have a question in Math.
If we want to get X's and Y's solution, X>0 and Y>0
We have two equation :

2*exp(X)+X^2+3*Y=2*exp(1)+4
3*X+4*(Y^2)=7

How I use R-project to  solve above question??

THANKS  YOU !!!!
                                  HLC
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