Finding minimum for an equation with two variables

[Hivoyer]

Homework Statement

I have the equation: x^2 + 2*x*y + 5*y^2 - 4*x - 6*y +7 and I have to find the minimum value
I'm getting something that looks half like the correct answer, but not quite right...

Homework Equations

The answer from the answer book is:

[x + 2*(y - 1)]^2 + (y + 1)^2 + 2

The Attempt at a Solution

Ok first I took 2*x*y and -4*x and turned them into 4*x*(y - 1), so I got:

x^2 + 4*x*(y - 1) + 5*y^2 - 6*y + 7

Then I turned x^2 + 4*x*(y - 1) into a square: [(x + 2*(y - 1)]^2 and subtracted [2*(y - 1)]^2, which is 4*(y - 1)^2 to balance it out, so I got:

[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*y^2 - 6*y + 7

However when I complete the square for the other part I get:

[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*[(y - 3)^2 - 9] + 7

when then gives me:

[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*(y - 3)^2 - 38

and this is not what the answer in the answer book I've written above is.Where did I go wrong?

 

[LCKurtz]

Homework Statement

I have the equation: x^2 + 2*x*y + 5*y^2 - 4*x - 6*y +7 and I have to find the minimum value
I'm getting something that looks half like the correct answer, but not quite right...

Homework Equations

The answer from the answer book is:

[x + 2*(y - 1)]^2 + (y + 1)^2 + 2

That doesn't look like a "minimum value" of anything. Please give the exact wording of the problem from your text.

 

[Hivoyer]

That doesn't look like a "minimum value" of anything. Please give the exact wording of the problem from your text.

Here is the problem from the book(part (c) in the red rectangle):

Here is the solution of (c) from the answer book(again surrounded in red):

 

[LCKurtz]

Homework Statement

I have the equation: x^2 + 2*x*y + 5*y^2 - 4*x - 6*y +7 and I have to find the minimum value
I'm getting something that looks half like the correct answer, but not quite right...

Homework Equations

The answer from the answer book is:

[x + 2*(y - 1)]^2 + (y + 1)^2 + 2

The Attempt at a Solution

Ok first I took 2*x*y and -4*x and turned them into 4*x*(y - 1),

$2xy-4x\ne 4x(y-1)$