Question

Find two consecutive odd integers such that two-fifths of the smaller exceeds two-ninth of the greater by $4$.

Answer 1

Let the two consecutive odd integers be $x$ and $x+2.$

$x$ is the smaller one, and $x+2$ is the greater one.

According to the question, we have

$\Rightarrow \frac{2}{5}x=\frac{2}{9}(x+2)+4$

$\Rightarrow \frac{2x}{5}=\frac{2x}{9}+\frac{4}{9}+4$

$\Rightarrow \frac{2x}{5}-\frac{2x}{9}=\frac{4+36}{9}$
$\Rightarrow \frac{(18x-10x)}{45}=\frac{40}{9}$

$\Rightarrow 8x=\frac{40}{9}\times 45$

$\Rightarrow 8x=200$

$\therefore x=\frac{200}{8}=25$

Therefore, the two consecutive positive odd integers are $x=25$ and $x+2=25+2=27$.