What is the largest of three consecutive odd integers if the product of the first and third integers is 6 more than three times the second integer?
Let the largest of the odd integers be $\displaystyle 2x+1$, then the others are $\displaystyle 2(x-1)+1$ and $\displaystyle 2(x-2)+1$ (where $\displaystyle x$ is an integer).
So we are told that: $\displaystyle [2(x-2)+1][2x+1]=3[2(x-1)+1]+6$
This is a quadratic in $\displaystyle x$, so solve it and choose the root (or roots) that makes sense.
CB
Hello, magentarita!
Another approach . . .
Consecutive odd integers "go up by two's".What is the largest of three consecutive odd integers if the product
of the first and third integers is 6 more than three times the second integer?
Let $\displaystyle x$ = the smallest odd integer.
Then $\displaystyle (x+2)\text{ and }(x+4)$ are the next two odd integers.
. . $\displaystyle \underbrace{\text{Product of first and third}} _{x(x+4)}\; \underbrace{\text{is}}_{=} \; \underbrace{\text{6 more than}}_{6\;\; +} \; \underbrace{\text{3 times second}}_{3(x+2)}$
And we have: .$\displaystyle x(x+4) \;=\;3(x+2)+6 \quad\Rightarrow\quad x^2 + x - 12 \:=\:0$
Solve the quadratic equation (and take the odd root).
And remember, they asked for the largest of the three.