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10. the sum of the digits of a certain two - digit number is 7. reversi…

Question

  1. the sum of the digits of a certain two - digit number is 7. reversing its digits increases the number by 9. what is the number?

Explanation:

Step1: Define variables

Let the tens digit be \( x \) and the units digit be \( y \). The two - digit number can be expressed as \( 10x + y \), and the number with reversed digits is \( 10y+x \).

We know two conditions:

  1. The sum of the digits is 7, so \( x + y=7 \), which can be rewritten as \( y = 7 - x \).
  2. Reversing the digits increases the number by 9, so \( (10y + x)-(10x + y)=9 \).

Step2: Simplify the second equation

Simplify \( (10y + x)-(10x + y)=9 \):
\[

$$\begin{align*} 10y+x - 10x - y&=9\\ 9y-9x&=9\\ y - x&=1 \end{align*}$$

\]

Step3: Substitute \( y = 7 - x \) into \( y - x=1 \)

Substitute \( y = 7 - x \) into \( y - x = 1 \):
\[

$$\begin{align*} (7 - x)-x&=1\\ 7-2x&=1\\ - 2x&=1 - 7\\ -2x&=-6\\ x&=3 \end{align*}$$

\]

Step4: Find the value of \( y \)

Since \( y=7 - x \) and \( x = 3 \), then \( y=7 - 3=4 \).

Step5: Find the two - digit number

The two - digit number is \( 10x + y \), substituting \( x = 3 \) and \( y = 4 \), we get \( 10\times3+4 = 34 \).

Answer:

34