How do you find two consecutive odd integers such that 3 times the first integer is 5 more than twice the second?

Question

3364 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-01T18:06:35

The consecutive odd integers are $9$ and $11$

Let the two consecutive odd integers $= x$ and $x+2$

Three times first integer $= 3x$

Five more than twice second $=5 + 2(x+2)$

As per data provided :
$3x = 5 + 2(x+2)$

$3x = 5 + 2(x) + 2 (2)$

$3x = 5 + 2x + 4$

$3x = 2x +9$

$3x -2x = 9$

$x=9$
The next odd integer $= 9+2 = 11$

The consecutive odd integers are $9$ and $11$