How would I solve this algebra problem?

Question

2a+b=100 and a^b=1000?

"a" needs to be the same in the first equation as the next one. Likewise for "b".

Answer 1

Don't think there is a way to solve this analytically; the best one can do is solve it numerically.

I'd start by combining the equations to eliminate "b":

2a + b = 100
b = 100-2a

substitute this into the second equation:

a^(100-2a) = 1000

Now, you can start plugging numbers in to see if you can find a value of a that satisfies this equation. Plotting a^(100-2a) as a function of a will help you narrow your search (i.e., look for where this function is approximately equal to 1000.

It might be easier to work with the logarithmic form of this equation. By taking the common log of both sides, one obtains:

(100-2a)*log(a) = 3

Again, you can plot the left hand side of the equation as a function of a to see where this expression is approximately equal to 3.

I find that a = 1.0731440183655411306, b = 97.853711963269 is the solution.

Answer 2

Solve the first equation for one of the unknowns, perhaps b = ? Then plug the answer into the second equation in place of b. The second equation will now have only one unknown, a, which you can solve for. Plug that value for a back into the first equation in place of a and there will be only one unknow, b, which you can solve for. Good luck.