Question

Question #13424

Answer 1

We have: `x = (15 +- 3sqrt(15))/10` and `y = (-5 +- 9sqrt(15))/10`

Explanation:

Since `y` is already isolated, we can just substitute the second equation into the first.

`x^2 + (3x- 5)^2 = 16`

`x^2 + 9x^2 -30x + 25 = 16`

`10x^2 - 30x + 9 = 0`

If we use the discriminant, `b^2 - 4ac`, to check the nature of the solutions, we find that

`30^2 - 4 * 10 * 9 = 540`

There will be two solutions, but since `sqrt(540)` isn't an integer, factoring will not work.

By the quadratic formula:

`x= (-(-30) +- sqrt((-30)^2 - 4 * 10 * 9))/(2 * 10)`

`x= (30 +- sqrt(540))/20`

`x = (30 +- 6sqrt(15))/20`

`x = (15 +- 3sqrt(15))/10`

So the value of `y` is

`y = 3x -5`

`y = 3((15 +- 3sqrt(15))/10) - 5`

`y = (45 +- 9sqrt(15))/10 - 50/10`

`y = (-5 +- 9sqrt(15))/10`

We can confirm graphically:

Hopefully this helps!