Question

Prove that
$\frac{0.09x^2+{0.4}^{2}xy+0.16y^2-0.05x^2}{0.7x+0.8y-0.5x-0.4y}=$?

Answer 1

$\frac{0.09x^2+{0.4}^{2}xy+0.16y^2-0.05x^2}{0.7x+0.8y-0.5x-0.4y}$

$=\frac{(0.02x{)}^2+(0.04y{)}^2+0.16xy}{0.2x+0.4y}$

$=\frac{(0.02x+0.04y{)}^2}{0.2x+0.4y}$

$=\frac{[2\times {10}^{-2}x+4\times {10}^{-2}y{]}^2}{2\times {10}^{-1}x+4\times {10}^{-1}y}$

$=\frac{4\times {10}^{-4}[x+2y{]}^2}{2\times {10}^{-1}[x+2y]}$

$=2\times {10}^{-3}[x+2y]$

$=0.002x+0.004y$