1. I think of a two-digit number. The ones digit is 6 more than the tens digit. The number is 12 more than the product of its digits. Find the original number.

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UpStudy AI · Accepted Answer

Let's denote the tens digit as $$ x $$ and the ones digit as $$ y $$.

Given:
1. The ones digit is 6 more than the tens digit: $$ y = x + 6 $$
2. The number is 12 more than the product of its digits: $$ 10x + y = xy + 12 $$

We can solve these equations to find the original number.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+6=y\10x+y=xy+12\end{array}ight.$$
- step1: Rewrite the expression:
  $$\left\{ \begin{array}{l}y=x+6\10x+y=xy+12\end{array}ight.$$
- step2: Substitute the value of $$y:$$
  $$10x+x+6=x\left(x+6ight)+12$$
- step3: Simplify:
  $$11x+6=x^{2}+6x+12$$
- step4: Move the expression to the left side:
  $$11x+6-\left(x^{2}+6x+12ight)=0$$
- step5: Calculate:
  $$5x-6-x^{2}=0$$
- step6: Factor the expression:
  $$\left(-3+xight)\left(2-xight)=0$$
- step7: Separate into possible cases:
  $$\begin{align}&-3+x=0\&2-x=0\end{align}$$
- step8: Solve the equation:
  $$\begin{align}&x=3\&x=2\end{align}$$
- step9: Calculate:
  $$x=3\cup x=2$$
- step10: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=3\y=x+6\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=x+6\end{array}ight.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=3\y=9\end{array}ight.\cup \left\{ \begin{array}{l}x=2\y=8\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=8\end{array}ight.\cup \left\{ \begin{array}{l}x=3\y=9\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=2\y=8\end{array}ight.\cup \left\{ \begin{array}{l}x=3\y=9\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(2,8ight)\cup \left(x,yight) = \left(3,9ight)$$
The solutions to the system of equations are $$ (x, y) = (2, 8) $$ or $$ (x, y) = (3, 9) $$.

Therefore, the original number can be either 28 or 39.
The original number is either 28 or 39.

I think of a two-digit number.
The ones digit is 6 more than the tens digit.
The number is 12 more than the product of its digits.
Find the original number.

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