Question

Let x,y be real numbers and `y^2+4y+9x^2-30x+29=0`, which of the following is equal to `9x-y`? A. 17 B. 25 C. 30 D. 41

Answer 1

A

Explanation:

You may notice that it bears some similarities to a circle with the general form `(x-h)^2+(y-k)^2=r^2` where `(h,k)` is the centre and r is the radius

So first up, you need to complete the square
`y^2+4y+9x^2-30x+29=0`

`(9x^2-30x)+(y^2+4y)=-29`

`9(x^2-30/9x+(5/3)^2)+(y^2+4y+4)=-29+4+25`
In case you don't remember how to complete the square,
`ax^2+bx+(b/2)^2` is how you go about it. All you have to do to find your constant is to half the coefficient of your `x` term ie `b/2`and then square the entire thing ie `(b/2)^2`

`9(x-5/3)^2+(y+2)^2=0`

Therefore, the centre is `(5/3,-2)`

Now you have the equation `9x-y`. Sub your above point in and you will get:
`9(5/3)-(-2)=15+2=17=A`