If x = 2a² + 3b² - 5ab, y = b² - 3a² + 7ab and z = 6a² - b² + ab, find :

(i) x + y - z

(ii) x - y + z

Given:

x = 2a² + 3b² - 5ab

y = b² - 3a² + 7ab

z = 6a² - b² + ab

(i) x + y - z

First, let's calculate x + y:

Arranging the terms to match (ab, b², a²):

$\begin{array}{rcccc} -5ab & + & 3b^2 & + & 2a^2 \\ +7ab & + & b^2 & - & 3a^2 \\ \hline 2ab & + & 4b^2 & - & a^2 \\ \hline \end{array}$

The sum is 2ab + 4b² - a² .

Now, subtract z from the above sum:

$\begin{array}{rcccc} 2ab & + & 4b^2 & - & a^2 \\ +ab & - & b^2 & + & 6a^2 \\ -\phantom{ab} & + & & - \\ \hline ab & + & 5b^2 & - & 7a^2 \\ \hline \end{array}$

∴ x + y - z = ab + 5b² - 7a²

(ii) x - y + z

First, let's calculate x - y:

Arranging the terms to match (a², b², ab):

$\begin{array}{rcccc} 2a^2 & + & 3b^2 & - & 5ab \\ -3a^2 & + & b^2 & + & 7ab \\ +\phantom{3a^2} & - & & - \\ \hline 5a^2 & + & 2b^2 & - & 12ab \\ \hline \end{array}$

The result is 5a² + 2b² - 12ab.

Now, add z to the above result:

$\begin{array}{rcccc} 5a^2 & + & 2b^2 & - & 12ab \\ +6a^2 & - & b^2 & + & ab \\ \hline 11a^2 & + & b^2 & - & 11ab \\ \hline \end{array}$

∴ x - y + z = 11a² + b² - 11ab