Solving Simple System of Equations: Step-by-Step Guide

[DrummingAtom]

Homework Statement

2^x + 2^y = 10

(2^x)² + (2^y)² = 68

Homework Equations

The Attempt at a Solution

If I divide them, I'm right back where I started to:

2^x+ 2^y = 6.8

I can't subtract or substitute either because that turns into a mess. Thanks for any help.

 

[mege]

Try solving for 2^x in the top expression then insert that in the bottom expression and expand. Then solve the quadratic replacing 2^y with z (or whatever) and then find the appropriate result.

 

[Mark44]

Homework Statement

2^x + 2^y = 10

(2^x)² + (2^y)² = 68

Homework Equations

The Attempt at a Solution

If I divide them, I'm right back where I started to:

2^x+ 2^y = 6.8

I can't subtract or substitute either because that turns into a mess. Thanks for any help.

I can see that 68/10 = 6.8. Are you thinking that
$$\frac{(2^x)^2 + (2^y)^2}{2^x + 2^y} = 2^x + 2^y \text{?}$$

That's not true at all. That's like saying that (3 + 8)/(1 + 4) = 3 + 2 = 5. The true value of the expression on the left is 11/5 = 2.2.

 

[Ray Vickson]

Homework Statement

2^x + 2^y = 10

(2^x)² + (2^y)² = 68

Homework Equations

The Attempt at a Solution

If I divide them, I'm right back where I started to:

2^x+ 2^y = 6.8

I can't subtract or substitute either because that turns into a mess. Thanks for any help.

Write a = 2^x and b = 2^y, so your equations become a + b = 10 and a^2 + b^2 = 68. Note that if we square both sides of the first equation we get 100 = (a+b)^2, and we can write this as 100 = a^2 + b^2 + 2*a*b. Since we already know a^2 + b^2 from the second equation, we can get a*b. So now we have a simpler system in which we know a + b and a*b. From these, we can get 'a' by solving a quadratic equation, although you need to worry about which of the two quadratic roots you should use. Once you have 'a' you can get 'b', and then you can get x and y from those.

RGV

 

[mege]

...although you need to worry about which of the two quadratic roots you should use. Once you have 'a' you can get 'b', and then you can get x and y from those.

Little note: there is not a single solution to these equations (when I worked it, I got two symmetrical solutions). So both roots should be valid. (multiple solutions make sense since y and x are equally weighted in both systems)