Question #03033

1 Answer
Oct 8, 2015

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 x+y=3 and set it equal to either x or y.
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)