To answer this question, we need to translate each of the expressions into algebraic form. Then we have:
1. We have that one number is 2 less than a second number.
In this case, let x be one of the numbers, and y the second number. Now, we can write the expression as follows:
[tex]x=y-2[/tex]
2. We also have that twice the second number is 2 less than 3 times the first:
[tex]2y=3x-2[/tex]
3. And now, we have the following system of equations:
[tex]\begin{cases}x=y-2 \\ 2y=3x-2\end{cases}[/tex]
4. And we can solve by substitution as follows:
[tex]\begin{gathered} x=y-2\text{ then we have:} \\ \\ 2y=3(y-2)-2 \\ \\ 2y=3(y)+(3)(-2)-2 \\ \\ 2y=3y-6-2 \\ \\ 2y=3y-8 \end{gathered}[/tex]
5. To solve this equation, we can subtract 2y from both sides, and add 8 from both sides too:
[tex]\begin{gathered} 2y-2y=3y-2y-8 \\ 0=y-8 \\ 8=y-8+8 \\ 8=y \\ y=8 \end{gathered}[/tex]
6. Since y = 8, then we can use one of the original equations to find x as follows:
[tex]\begin{gathered} x=y-2\Rightarrow y=8 \\ x=8-2 \\ x=6 \end{gathered}[/tex]
Therefore, we have that both numbers are x = 6, and y = 8.
In summary, we have that:
• The smaller number is 6.
,
• The larger number is 8.
