Question

{x + y = 2

Answer 1

1) I put them in two separate brakets.

2) I solved it by equating both of them.

Answer 2

Answer:

  1. Substitute for y in the first equation. Solve for x by completing the square. Substitute for x in the first equation to find y.

  2. (x, y) = (2+2√2, -2√2) and (2-2√2, 2√2)

Step-by-step explanation:

1.

The first equation is a linear equation in standard form. The second equation is a quadratic equation giving an expression for y. The system can be solved by using the second equation to substitute for y in the first equation. After the resulting equation is put in suitable form, the irrational solutions can be found by completing the square. Then the y-values can be found by substituting the x-values into either of the given equations. Usually, it is easiest to substitute into the linear equation.

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2.

Substituting for y in the first equation:

  x +(-1/4x^2 +3) = 2

Multiplying by -4 to eliminate the fraction and make the leading coefficient 1, we have ...

  -4x +x^2 -12 = -8

  x^2 -4x = 4 . . . . . . add 12

  x^2 -4x +4 = 8 . . . . add 4 to complete the square

  (x -2)^2 = 8 . . . . . . write as a square

  x -2 = ±√8 = ±2√2 . . . . take the square root

  x = 2 ±2√2 . . . . . . add 2

Solving the first equation for y, we get ...

  y = 2 -x

Substituting the values of x we found gives ...

  y = 2 -(2 ±2√2) = ∓2√2

Solutions are ...

  (x, y) = (2+2√2, -2√2) and (2-2√2, 2√2)