Question #03033
1 Answer
Two different points satisfy the equation, (1,2) and (2,1)
Explanation:
This, without using a graphing calculator, is a substitution problem.
Take the equation
Setting it equal to y:
#y=3-x#
Setting it equal to x:
#x=3-y#
I will demonstrate using the y equation, but you can decide which equation to substitute in. Now we want to substitute the equation into the other one:
#x^2 + (3-x)^2 = 5#
#x^2 + 9 - 6x + x^2 = 5#
#2x^2 - 6x + 9 = 5#
We then need to get the equation equal to 0 so we can factor and set each part of the factor equal to 0.
#2x^2 - 6x + 4 = 0#
#2(x^2 - 3x + 2) = 0#
Now factor the polynomial and set each part equal to 0:
#2(x-2)(x-1) = 0#
#x-1 = 0#
#x-2 = 0#
#x = 1,2#
Now set x into the original y equation:
#y=3-1#
#y=3-2#
#y=2,1#
Therefore we can put the points together and get the answer, (1,2) and (2,1)