Find the ordered pair solutions for (x,y)

Question

Use the method shown in the introduction to this section to solve the system of equations. (Enter your answers as a comma-separated list of ordered pairs.)

x+y=6

xy=18

Answer 1

Just as the method of substitution can be used to solve a system of 2 linear equations in 2-unknowns, this method can also be applied to the case shown above ( one linear and one non-linear) equations in two unknowns.

x + y = 6 (1)

x.y = 18 (2)

solving (1) for y obtains

y = 6 - x (3)

substitute this expression for y into equation 2.

x. (6-x) = 18 (4)

expanding the equation and setting it equal to zero obtains (5)

6x -x^{2 _}18 = 0

multiply by minus one obtains

x² -6x +18 = 0

The quadratic formula yields complex roots:

--6 +/- SQRT( (-6)² _- 4.1.18)/2 = 6 +/- SQRT(36 -72)/2 =[ 6 +/- SQRT(-36)]/2

which simplifies to two complex solutions for x:

x = (3 +i.3) and ( 3 - i.3) (6)

now solve for y by plugging x into equation (3)

y = 6 -x

y = 6 - [ 3+ i 3} = 3 -i.3 when x = 3 + i.3

y = 6 - [ 3 - i 3] = 3 + i 3 when x = 3 - i.3 (7)

so x and y are complex conjugates !

lets verify they satisfy (2)

x.y = (3+i.3)(3-i.3) = 3^{2 --} 3^{2 =} = 18

expressing the solution(s) as ordered pairs we have solutions:

( x = 3+ i 3, y = 3 -i.3) is one solution

( x = 3- i 3, y = 3 +i.3) is the other solution

Answer 2

There are no real solutions, but we can find complex ones.

y = 6 - x

Then, xy = 18, so x(6 - x) = 18

6x - x^2 = 18

x^2 - 6x + 18 = 0

x = 3 ± 3i

y = 6 - (3 ± 3i) = 3 -± 3i

(3 + 3i, 3 - 3i), (3 - 3i, 3 + 3i)