Question

Simplify the expression $3(x^2+y^2+xy)+xyz-2xy-30$ and find its value when $x=6$, $y=-7$, and $z=4$.

• A. $3x^2$+$3y^2$+$xyz$+$xy-30$ and $15$
• B. $3x^2$+$3y^2$+$xyz$-$xy-30$ and $15$
• C. $3x^2$+$3y^2$+$xyz$+$5xy-30$ and $30$
• D. $3x^2$+$3y^2$+$xyz$+$6xy-30$ and $7$

Answer 1

The correct option is A $3x^2$+$3y^2$+$xyz$+$xy-30$ and $15$

$3(x^2+y^2+xy)+xyz-2xy-30$
=$3x^2+3y^2+3xy+xyz-2xy-30$
=$3x^2+3y^2+xyz+xy-30$ ------(1)

Substituting $x=6$, $y=-7$, $z=4$ in the equation (1),

$3(6{)}^2+3(-7{)}^2+\left(6\right)(-7)\left(4\right)+\left(6\right)(-7)-30$
=$3\left(36\right)+3\left(49\right)-168-42-30$
=$108+147-168-42-30$
=$255-240$
=$15$