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

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justaguide's profile pic

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.

Now 2^(x+y)=32

=> 2^(x+ y) = 2^5

As the base is the same we can equate the exponents

=> x+ y = 5

=> x = 5 - y

Substitute this inĀ  y = 6/x

=> y = 6/ (5 - y)

=> 5y - y^2 = 6

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

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

So y can be 2 and 3.

x = 6/y can be 3 and 2.

The values for x and y are x = 2, y = 3 and x = 3, y = 2.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll multiply 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

Since 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 form the quadratic equation with the sum and the product;

x^2 - Sx + P = 0

x^2 - 5x + 6 = 0

We'll apply the quadratic formula:

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}.

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