Algebra

To solve this, let's first isolate a.

 

$(a+b)^2 = 10^2$

$a^2+2ab+b^2 = 100$

$64 + 2ab = 100$

$2ab = 36$

$ab = 18$

$b = 10 - a$

$a(10-a) = 18$

$-a^2+10a-18 = 0$

$a^2-10a+18 = 0$

 

Now, we can solve for a

 

$a^2 - 10a + 25 = 7$

$(a-5)^2 = 7$

$a-5 = \pm \sqrt{7}$

$a = 5 \pm \sqrt{7}$

 

We can now finally solve for b by plugging in each value of a.

 

a = (5 + sqrt(7)):

$(5 + \sqrt{7}) + b = 10$

$b = 5 - \sqrt{7}$

 

a = (5 - sqrt(7)):

$(5 - \sqrt7) + b = 10$

$b = 5 + \sqrt7$

 

Therefore, the solutions are (5 + sqrt(7), 5 - sqrt(7)) and (5 - sqrt(7), 5 + sqrt(7))