Question

If $x=4a^2+b^2-6ab,y=3b^2-2a^2+8ab$ and $z=6a^2+8b^2-6ab,$ find: (ii) 𝑥−𝑦−𝑧

Answer 1

Step: Find the value of 𝑥−𝑦−𝑧 $x-y-z=(4a^2+b^2-6ab)-(3b^2-2a^2+8ab)-(6a^2+8b^2-6ab)$$=4a^2+b^2-6ab-3b^2+2a^2-8ab-6a^2-8b^2+6ab$$=(4a^2+2a^2-6a^2)+(b^2-3b^2-8b^2)+(-6ab-8ab+6ab)$
$=(6a^2-6a^2)+(b^2-11b^2)+(-8ab)=0-10b^2-8ab$
$=-10b^2-8ab$
Hence, the answer is $-10b^2-8ab.$