Question

Find $64x^2+81y^2+\frac{144xy}{48}$ when $x=11$ and $y=\frac{4}{3}$

Answer 1

We have, $64x^2+81y^2+14xy$

When $x=11,y=\frac{4}{3}$, then

$=64(11{)}^2+(81\times \frac{16}{9})+(\frac{144}{48}\times 11\times \frac{4}{3})$

$=(64\times 121)+(9\times 16)+44$

$=7744+144+44$

$=7932$

Hence, $64x^2+81y^2+14xy=7932$ when $x=11$ and $y=\frac{4}{3}$