Solving equations in MAPLE

Question

I am using maple for the first time. I want to invert the equations using any maths tool. I tried it using MAPLE.

I have given two equations in this form:

$a=x*b^r + y*(b^s)*(c^t) + z*c^u$

Now I want to flip this equation (invert this equation)

I now want two equations of b and c in terms of a

e.g

$b= 0.98989a^2 + 78a + 87$

$c= 56a + 89$

These are not my results. I just gave an example.

Any advice?

My two equations are:

Equation#1 is:

$a= 1.093*(b^0.002939) - 0.1887*(b^0.7637)*(c^0.2306) - 0.04425 *(c^0.9143)$

Equation#2 is:

$a= 1.088*(b^0.003058) -0.1813*(b^0.7627)*(c^0.2501) -0.03958 *(c^0.9365)$

From Equation#1, i can find b equation. I put the bequation in Equation#2 . In this way equation#2 has only a and c variables. i can get c equation. and then by putting this equation can get b equation ( with only a variables)

But these equations are very complex and it takes a lot of time to evaluate in MAPLE. How to simplify it? I tried to use natural log. But this doesn't seem to be working.

Please guide me. How should I get b and c equations from these two equations?

Answer 1

You can eliminate $a$ from the two equations by subtraction, leaving just one equation involving powers of $b$ and $c$, but this will involve some ugly fractional powers, and it's very unlikely that you could get a closed-form solution. On the other hand, for any $c$ for which real solutions exist you might get numerical values of the solutions using fsolve. For example, taking $c=1$:

 eq1:= a= 1.093*(b^0.002939) - 0.1887*(b^0.7637)*(c^0.2306) 
     - 0.04425 *(c^0.9143);
 eq2:= a= 1.088*(b^0.003058) -0.1813*(b^0.7627)*(c^0.2501) 
     -0.03958 *(c^0.9365);
 b1:= fsolve(eval(rhs(eq1)=rhs(eq2), c = 1));

$$b1 := 0.04810865707$$

eval(eq1,{c=1,b=b1});

$$a = 1.020451636$$