Question

Question #03033

Answer 1

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)