Solution:
Given:
A word problem.
To solve the question, we develop the statements into mathematical equations.
Hence,
Let the two numbers be represented by x and y
where;
x is one number
y is another number
[tex]\begin{gathered} \text{One number is 2 less than 3 times another.} \\ \text{Mathematically means;} \\ x=3y-2 \\ R\text{earranging, th}e\text{ equation becomes; }x-3y=-2\ldots\ldots\ldots\ldots.(1) \\ \text{The sum of the two numbers is 14.} \\ \text{Mathematically means;} \\ x+y=14\ldots\ldots\ldots\ldots\ldots(2) \end{gathered}[/tex]
Solving both equations generated simultaneously, using the elimination method, we subtract equation (1) from equation (2).
That is equation (2) - equation (1);
[tex]\begin{gathered} x-3y=-2\ldots\ldots\ldots\ldots\text{.}\mathrm{}(1) \\ x+y=14\ldots\ldots\ldots\ldots\ldots\text{.}(2) \\ \text{equation (2)-(1);} \\ x-x+y-(-3y)=14-(-2)_{} \\ y+3y=14+2 \\ 4y=16 \\ \text{Dividing both sides by 4,} \\ y=\frac{16}{4} \\ y=4 \end{gathered}[/tex]
Substituting the value of y gotten into equation (2) to solve for x,
[tex]\begin{gathered} x+y=14 \\ x+4=14 \\ x=14-4 \\ x=10 \end{gathered}[/tex]
Hence, the solution to the equations is;
[tex](x,y)=(10,4)[/tex]Therefore, the numbers are 10 and 4.