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)
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)