Math: (a/3)+(b/4)=13/12. Find a^2+b^2?

Answer 1

Put the two terms on the left as a common denominator:
(4a+3b)/12 = 13/12

Multiply by 12 in both sides:
4a+3b = 13

At this point, solve for a and b in terms of each other:
a = (13-3b)/4
b = (13-4a)/3

Now, square these quantities individually and add them up.

a^2 = 9/16*b^2 - 39/8*b + 169/16
b^2 = 16/9*a^2 - 104/9*a + 169/9


a^2 + b^2 = 9/16*b^2 - 39/8*b + 4225/144 + 16/9*a^2 - 104/9*a

Answer 2

First find the the lowest common denominator: 12, and change your fractions into that:

(a/3)*(4/4) + (b/4)*(3/3) = 13/12
(4a/12) + (3b/12) = 13/12
(4a+ 3b) / 12 = 13 / 12
4a + 3b = 13

a = (13 - 3b) / 4
b = (13 - 4a) / 3

Therefore: a^2 + b^2 =
(13 - 3b)^2 / 16 + (13 - 4a)^2 / 9
(169 - 78b + 9b^2) / 16 + (169 - 104a + 16a^2) / 9
169/16 - 78b/16 + 9b^2/16 + 169/9 - 104a/9 + 16a^2/9
(1521 - 702b + 81b^2 + 2704 - 1664a + 256a^2) / 144
(4225 - 702b + 81b^2 - 1664a + 256a^2) / 144
29.3403 - 4.875b + 0.5625b^2 - 11.5556a + 1.7778a^2

It's an ugly solution, accurate to the 4th decimal place (ten-thousandths), and can be written a handful different ways.

Answer 3

Let's solve this in integers.
a/3 + b/4 = 13/12
yields 4a + 3b = 13
and this yields the solution a = 1, b = 3, a² + b² = 10.
However, there are infinitely many other solutions.
Since we have the solution a = 1, b = 3
all other solutions are given by
a = 1-3t, b = 3 + 4t, a² + b² = 10 + 18t + 25t².
where t is any integer.
So, for example, let t = 1.
Then a = -2, b = 7, a² + b² = 53.
If t = -1 we get a = 4, b = -1, a² + b² = 17.
Hope that helps!

Answer 4

(a/3)+(b/4)=13/12
4a+3b=13
for: a=1=>3b=9=>b=3=>correct solution
a=2=>3b=5=>incorrect solution
a=3=>3b=11=>incorrect solution
for: b=1=>4a=10=>2a=5=> incorrect solution
b=2=>4a=7=> incorrect solution
b=3=>4a=4=>a=1=>correct solution
b=4=>4a=1=>incorrect solution
so the solution is: a=1 & b=3
and the final answer is:10
In what grade are you??? Just curious.