Question

If $x^2+xy+x=12$ and $y^2+xy+y=18$, then the value of $x+y$ is __________?

• A. $5$ or $-6$
• B. $3$ or $4$
• C. $5$ or $3$
• D. $6$ or $-3$

Answer 1

The correct option is A $5$ or $-6$

Let $x^2+xy+x=12$ ...(i)

and $y^2+xy+y=18$ ...(ii)

Adding (i) and (ii), we get

$(x^2+y^2+2xy)+x+y=30$

$\Rightarrow (x+y{)}^2+(x+y)-30=0$

Let $(x+y)=t$

Thus $t^2+t-30=0$

$\Rightarrow t^2+(6-5)t-30=0$

$\Rightarrow t^2+6t-5t-30=0$

$\Rightarrow t(t+6)-5(t+6)=0$

$\Rightarrow (t+6)(t-5)=0$

which gives $t=-6$or $t=5$

So, $t=x+y=-6$ or $5$

Hence, option A is correct.