Question

How do you solve the system `x^2+y^2=17` and `y=x+3`?

Answer 1

The Soln. Set `={(1,4),(-4,-1)}`.

Explanation:

Sub.ing `y=x+3` in the first eqn. to get.

`x^2+(x+3)^2=17, or, 2x^2+6x+9-17=0`, i.e.,

`x^2+3x-4=0`

`rArr (x-1)(x+4)=0`

`rArr x=1, x=-4`. Then, from `y=x+3`,

`y=4, y=-1`.

These roots satisfy the eqns.

Thus, the Soln. Set `={(1,4),(-4,-1)}`.

Answer 2

`(x, y) = (-4, -1)` or `(1, 4)`

Explanation:

Here's an alternative method that tries to use symmetry to help...

Let `t = (x+y)/2`

Then `x = t-3/2` and `y = t+3/2`

So we have:

`17 = x^2+y^2`

`= (t-3/2)^2+(t+3/2)^2`

`= t^2-color(red)(cancel(color(black)(3t)))+9/4+t^2+color(red)(cancel(color(black)(3t)))+9/4`

`= 2t^2+9/2`

Subtract `9/2` from both ends and transpose to get:

`2t^2 = 17-9/2 = 25/2`

Divide both ends by `2` to get:

`t^2 = 25/4 = 5^2/2^2`

Hence:

`t = +-5/2`

If `t = -5/2` then `x = -5/2-3/2 = -4` and `y = -5/2+3/2 = -1`

If `t = 5/2` then `x = 5/2-3/2 = 1` and `y = 5/2+3/2 = 4`