Solve the system by method of choice (addition or substitution)

Question

x²+2y²=324

x+y=18

Answer 1

x + y = 18

x = 18 - y

x^2 = (18 - y)^2

x^2 = 324 -36y +y^2

substitue into the other equation

x^2 +2y^2 = 324

324 -36y +y^2 = 324

solve for y

-36y +y^2 = 0

(-36 +y) y = 0

y = 36 or 0

use the original equation

x +y = 18

enter the values for y

x +36 = 18  or  x +0 = 18

solve for x

x = -18 or x = 18

Answer:  x = -18, y = 36  or  x = 18, y = 0

check

x + y = 18

-18 + 36 = 18 True

check

x + y = 18

18 + 0 = 18 True

Answer 2

Given equations are

  ---> (1)

x + y  = 18       ---> (2)

solve for y from (2).

Subtract y from each side.

x = 18 - y         ---> (3)

Substitue x = 18 - y in (1).

.

Substitute  in (3)

x = 18 - 12 = 6.

Solution  x = 6 and y = 12.

Answer 3

y^2 - 12y = 0
y(y-12) = 0
y = 0 and y = 12.
If y = 0 then x = 18 - y => x = 18.
If y = 12 then x = 18 - y => x = 18 - 12 = 6.
The two ordered-pair solutions are (x, y) = (6, 12) and (x, y) = (18, 0)