# equationsSolve for x and y if y=6/x and 2^(x+y)=32.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve for x and y given that y=6/x and 2^(x+y)=32

substitute y = 6/x in  2^(x+y)=32

=> 2^(x + 6/x) = 32

=> 2^(x + 6/x) = 2^5

=> x + 6/x = 5

=> x^2 - 5x + 6 = 0

=> x^2 - 3x - 2x + 6 = 0

=> x(x - 3) - 2(x - 3) = 0

=> (x - 2)(x - 3) = 0

=> x = 2 and x = 3

y = 3  and y = 2

The values of x and y are (2, 3) and (3,2)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll start by multiplying by x the 1st equation:

y*x = 6x/x

We'll simplify and we'll get:

x*y = 6 (1)

We'll write 32 as a power of 2.

2^(x+y) =2^5

Because of the fact that the bases are matching, we'll use one to one property:

x + y = 5 (2)

We'll note x + y = S and x*y = P.

S = 5 and P = 6

We'll create the quadratic with the sum and the product;

x^2 - Sx + P = 0

x^2 - 5x + 6 = 0

x1 = [5+sqrt(25 - 24)]/2

x1 = (5 + 1)/2

x1 = 3 => y1 = 5 - x1

y1 = 5 - 3

y1 = 2

x2 = 2

y2 = 5 - 2

y2 = 3

So, the symmetric system will have the solutions {2;3} and {3;2}.