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The sum of the digits of a two-digit number is 14. The difference between the tens digit and the units digit is 2.

If [tex]x[/tex] is the tens digit and [tex]y[/tex] is the ones digit, which system of equations represents the word problem?

A. [tex]x + y = 14[/tex] and [tex]y - x = 2[/tex]

B. [tex]x + y = 14[/tex] and [tex]x - y = 2[/tex]

C. [tex]x \cdot y = 14[/tex] and [tex]x - y = 2[/tex]

Sagot :

To solve this problem, let's define [tex]\( x \)[/tex] as the tens digit and [tex]\( y \)[/tex] as the ones digit of our two-digit number. We have two key pieces of information given in the problem:

1. The sum of the digits is 14.
2. The difference between the tens digit and the ones digit is 2.

Using this information, we can set up the respective equations.

### Step-by-Step Solution:

1. Sum of the Digits:
The sum of the digits [tex]\( x \)[/tex] and [tex]\( y \)[/tex] can be written as:
[tex]\[ x + y = 14 \][/tex]
This equation represents that the tens digit plus the ones digit equals 14.

2. Difference Between the Digits:
The difference between the tens digit [tex]\( x \)[/tex] and the ones digit [tex]\( y \)[/tex] is 2. Since the tens digit is greater than the ones digit, we can write this as:
[tex]\[ x - y = 2 \][/tex]
This equation captures the information that when you subtract the ones digit from the tens digit, the result is 2.

So, the system of equations that represents this word problem is:
[tex]\[ \begin{cases} x + y = 14 \\ x - y = 2 \end{cases} \][/tex]

Let's now solve this system of equations step-by-step:

### Solving the System of Equations

We have:
[tex]\[ \begin{cases} x + y = 14 \\ x - y = 2 \end{cases} \][/tex]

3. Add the Equations:
To eliminate [tex]\( y \)[/tex], add the two equations:
[tex]\[ (x + y) + (x - y) = 14 + 2 \][/tex]
Simplifying this, we get:
[tex]\[ 2x = 16 \][/tex]
Divide both sides by 2:
[tex]\[ x = 8 \][/tex]

4. Substitute [tex]\( x \)[/tex] back into one of the equations:
Using [tex]\( x + y = 14 \)[/tex]:
[tex]\[ 8 + y = 14 \][/tex]
Subtract 8 from both sides:
[tex]\[ y = 6 \][/tex]

Thus, the digits are:
[tex]\[ x = 8 \quad \text{(tens digit)} \][/tex]
[tex]\[ y = 6 \quad \text{(ones digit)} \][/tex]

Therefore, the correct system of equations that represents the word problem is:
[tex]\[ x + y = 14 \quad \text{and} \quad x - y = 2 \][/tex]
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