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the sum of the digits of a two-digit number is 7. when the digits are reversed, the number is 9les than the original number. find the numbers
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Sagot :

✒️NUMBER DIGITS

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad \Large \:»\: \rm Digits \: are \: 4 \: and \: 3 [/tex]

[tex] \qquad \Large \:»\: \rm Original \: is \: 43 [/tex]

[tex] \qquad \Large \:»\: \rm Reversed \: is \: 34 [/tex]

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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]

Let x be the first digit of the original number and y as the second digit. Represent it as a number.

  • [tex] \rm Original = \mathcal{10x + y} [/tex]
  • [tex] \rm Reversed = \mathcal{10y + x} [/tex]

Make two equations based on the given statements.

  • [tex] \begin{cases} x + y = 7 \\ 10y + x = (10x + y)-9 \end{cases} \quad \begin{align} \tt{(eq. \: 1)} \\ \tt{(eq. \: 2)} \end{align} [/tex]

Simplify the values of x and y.

  • [tex] \begin{cases} x = 7 - y \\ 10y + x = 10x + y-9 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ 10y - y = 10x -9 - x\end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ 9y = 9x -9 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ (9y)/9 = (9x -9)/9 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ y = x - 1 \end{cases} [/tex]

Substitute x in the second equation from the first equation in terms of y.

  • [tex] \begin{cases} x = 7 - y \\ y = 7-y - 1 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ y + y = 7 - 1 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ 2y = 6 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ 2y/2 = 6/2 \end{cases} [/tex]

  • [tex] \begin{cases} x = 7 - y \\ y = 3 \end{cases} [/tex]

Thus, the value of y is 3. Substitute it to the first equation to find the value of x.

  • [tex] \begin{cases} x = 7 - 3 \\ y = 3 \end{cases} [/tex]

  • [tex] \begin{cases} x = 4 \\ y = 3 \end{cases} [/tex]

Thus, the value of x is 4. Substitute the two values to find the original and the reversed digits.

  • [tex] \rm Original = \mathcal{10(4) + 3} [/tex]
  • [tex] \rm Reversed = \mathcal{10(3) + 4} [/tex]

  • [tex] \rm Original = \mathcal{40 + 3} [/tex]
  • [tex] \rm Reversed = \mathcal{30 + 4} [/tex]

  • [tex] \rm Original = \mathcal{43} [/tex]
  • [tex] \rm Reversed = \mathcal{34} [/tex]

Thus, the original number is 43 and its reversed digit is 34. Since the numbers that are asked weren't specified, I'll be putting the values of x and y along as the original and its reversed digit.

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