What are variables x and y if 3^(x+y)-243=0 and x=6/y ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We need to find x and y given that 3^(x + y) - 243 = 0 and x = 6/y

3^(x + y) - 243 = 0 can be written as

3^(x + y) = 243

=> 3^(x + y) = 3^5

As the base 3 is common equate the exponent

=> x + y = 5

=> x = 5 - y

We also have x = 6/y

=> 5 - y = 6/y

=> 5y - y^2 = 6

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

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

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

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

=> y = 2 and y = 3

x = 3 and x = 2

The value of x and y are (3, 2) and (2, 3).

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write 243 as a power of 3.

3^(x+y) - 3^5 = 0

We'll move 3^5 to the right side:

3^(x+y) = 3^5

Since the bases are matching, we'll use one to one property of exponentials:

x + y = 5 (1)

We'll multiply by y the second equation:

x*y = 6*y/y

We'll simplify and we'll get:

x*y = 6 (2)

To ease the work, we'll use the following terms instead of the variables x and y:

x + y = S and x*y = P.

S = 5 and P = 6

We'll create the quadratic equation  with the sum and the product above:

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 variables x and y are represented by the following pairs: {2; 3} and {3; 2}.

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