Answer:
(x, y) = (5, -3)
Step-by-step explanation:
Apparently, you're asked to solve for x first. This suggests you will be substituting for x in the quadratic. This works, but may make the problem more difficult than it has to be.
Using the found value of x, we have ...
y = (2 -y)^2 -11(2 -y) +27
y = 4 -4y +y^2 -22 +11y +27 . . . . eliminate parentheses
y^2 +6y +9 = 0 . . . . . . . . . . . . . put in standard form
(y +3)^2 = 0 . . . . . . . . . . . . . . factor
y = -3 . . . . . . . . . . . . . . . . . the value of y that makes the factor zero
x = 2 -(-3) = 5
The one solution is (x, y) = (5, -3).
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Alternate solution
The two expressions for y can be equated:
x^2 -11x +27 = -x +2
x^2 -10x +25 = 0 . . . . . rearrange to standard form
(x -5)^2 = 0 . . . . . . . . factor
x = 5 . . . . . . . . . . . x that makes the factors zero
y = -5 +2 = -3 . . . corresponding value of y
This solution removes the need for the "eliminate parentheses" step in the original solution, so cuts down on the opportunity for error.
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The attached graph shows the one solution is the point where the line is tangent to the parabola.
