# Math Help - Algebra Subtraction

1. ## Algebra Subtraction

Subtract $1 / 2[12y^2 + 1 / 4x(16x-8y)]$ from $12y^2 + 7xy + 5x^2$

And this is how I done it...

$12y^2 + 7xy + 5x^2 - 1/2[12y^2 - 1/4x(16x + 8y)]$
$= 12y^2 + 7xy + 5x^2 - 6y^2 - 2y + 2xy$
$= 6y^2 + 8xy + 5x^2 - 2x$

However, it turns out wrong. If you could point out my mistake while solving this sum, I would appreciate your kindness. Thank you!

2. $12y^2+7xy+5x^2-\frac{1}{2}(12y^2+\frac{1}{4x}(16x-8y)) = 12y^2+7xy+5x^2-\frac{1}{2}12y^2-\frac{1}{2}\frac{1}{4x}(16x-8y) =$

$= 12y^2+7xy+5x^2-6y^2-\frac{1}{8x}(16x-8y) = 12y^2+7xy+5x^2-6y^2-2+\frac{y}{x}$

This is gonna be without the parentheses

3. Originally Posted by DengXiang
$12y^2 + 7xy + 5x^2 - 1/2[12y^2 - 1/4x(16x + 8y)]$
This is wrong... The right is $12y^2+7xy+5x^2-1/2[12y^2+1/4x(16x-8y)]$