Select the correct answer.

The sum of the digits of a three-digit number is 13. The tens digit, [tex]\( t \)[/tex], is 1 more than the hundreds digit, [tex]\( h \)[/tex]. The units digit, [tex]\( u \)[/tex], is 3 more than the sum of the tens and hundreds digits. Which system of equations can be used to find each digit?

A.
[tex]\[
\begin{aligned}
h + t + u & = 13 \\
h - t & = 1 \\
h + t - u & = 3
\end{aligned}
\][/tex]

B.
[tex]\[
\begin{aligned}
h + t + u & = 13 \\
h + u & = 1 \\
-h - t & = 3
\end{aligned}
\][/tex]

C.
[tex]\[
\begin{aligned}
h + t + u & = 13 \\
-h + u & = 1 \\
h - t + u & = 3
\end{aligned}
\][/tex]

D.
[tex]\[
\begin{aligned}
h + t + u & = 13 \\
-h + t & = 1 \\
-h - t + u & = 3
\end{aligned}
\][/tex]



Answer :

Let's break down the problem step by step:

1. We are given that the sum of the digits of a three-digit number is 13. This gives us the equation:
[tex]\[ h + t + u = 13 \][/tex]

2. The tens digit, [tex]\( t \)[/tex], is 1 more than the hundreds digit, [tex]\( h \)[/tex]. This gives us the relationship:
[tex]\[ t = h + 1 \][/tex]

3. The units digit, [tex]\( u \)[/tex], is 3 more than the sum of the tens and hundreds digits. This gives us the relationship:
[tex]\[ u = t + h + 3 \][/tex]

To formulate a system of equations, let's express all these relationships mathematically:

- We already have:
[tex]\[ h + t + u = 13 \][/tex]

- The second relationship can be rewritten as:
[tex]\[ t - h = 1 \][/tex]

- For the third relationship, let's rewrite it:
[tex]\[ u = t + h + 3 \][/tex]

By rearranging this, we get:
[tex]\[ u - t - h = 3 \][/tex]

So, our system of equations becomes:
[tex]\[ \begin{aligned} h + t + u & = 13 \\ t - h & = 1 \\ u - t - h & = 3 \end{aligned} \][/tex]

Simplifying them in a more familiar format, rearranging terms properly, we have:
[tex]\[ \begin{aligned} h + t + u & = 13 \\ -h + t & = 1 \\ -h - t + u & = 3 \end{aligned} \][/tex]

Now, let's compare this system with the provided options:

A.
[tex]\[ \begin{array}{r} h + t + u = 13 \\ h - t = 1 \\ h + t - u = 3 \end{array} \][/tex]

B.
[tex]\[ \begin{aligned} h + t + u & = 13 \\ h + u & = 1 \\ -h - t & = 3 \end{aligned} \][/tex]

C.
[tex]\[ \begin{aligned} h + t + u & = 13 \\ -h + u & = 1 \\ h - t + u & = 3 \end{aligned} \][/tex]

D.
[tex]\[ \begin{aligned} h + t + u & = 13 \\ -h + t & = 1 \\ -h - t + u & = 3 \end{aligned} \][/tex]

The system of equations perfectly matches with Option D.

Therefore, the correct answer is:
[tex]\[ \boxed{1} \][/tex] (Option D)