Answer:

✏️DIGITS

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Problem: The product of a two-digit number and its tens digit is 105. Find the numbers if its ones digit is 2 more than its tens digit.

Solution: Represent xx and yy as the tens and ones digit respectively. Make equations on the given statements.

\red{(eq. \: 1)}(eq.1) We can represent the two-digit number as (10x + y)(10x+y) since xx is a face value of tens which is multiplied to 10, then add yy which is the ones. This two-digit will be multiplied to xx to get a product of 105 as it was said to the statement.

\red{(eq. \: 2)}(eq.2) yy is said to be the sum of xx and 2.

\begin{gathered} \begin{cases} (10x + y)x = 105 \\ y = x + 2 \end{cases} \: \begin{aligned} \red{(eq. \: 1)} \\ \red{(eq. \: 2)} \end{aligned} \end{gathered}

{

(10x+y)x=105

y=x+2

(eq.1)

(eq.2)

» Substitute yy from the second equation to the first equation in terms of x.x.

\begin{gathered} \begin{cases} (10x + x + 2)x = 105 \\ y = x + 2 \end{cases} \end{gathered}

{

(10x+x+2)x=105

y=x+2

\begin{gathered} \begin{cases} (11x + 2)x = 105 \\ y = x + 2 \end{cases} \end{gathered}

{

(11x+2)x=105

y=x+2

\begin{gathered} \begin{cases} 11x^2 + 2x = 105 \\ y = x + 2 \end{cases} \end{gathered}

{

11x

2

+2x=105

y=x+2

\begin{gathered} \begin{cases} 11x^2 + 2x - 105 = 0 \\ y = x + 2 \end{cases} \end{gathered}

{

11x

2

+2x−105=0

y=x+2

» Solve the quadratic equation in the first equation using the quadratic formula. Use only the positive solution.

\begin{gathered} x = \frac{\text-2 + \sqrt{2^2 - 4(11)(\text-105)} }{2(11)} \\ \end{gathered}

x=

2(11)

-2+

2

2

−4(11)(-105)

\begin{gathered} x = \frac{\text-2 + \sqrt{4 - 4(11)(\text-105)} }{2(11)} \\ \end{gathered}

x=

2(11)

-2+

4−4(11)(-105)

\begin{gathered} x = \frac{\text-2 + \sqrt{4 + 4620} }{22} \\ \end{gathered}

x=

22

-2+

4+4620

\begin{gathered} x = \frac{\text-2 + \sqrt{4624} }{22} \\ \end{gathered}

x=

22

-2+

4624

\begin{gathered} x = \frac{\text-2 + 68 }{22} \\ \end{gathered}

x=

22

-2+68

\begin{gathered} x = \frac{66}{22} \\ \end{gathered}

x=

22

66

x = 3x=3

\begin{gathered} \begin{cases} x = 3 \\ y = x + 2 \end{cases} \end{gathered}

{

x=3

y=x+2

» Therefore the tens digit of the two-digit number is 3. Substitute it to the second equation to find the ones digit.

\begin{gathered} \begin{cases} x = 3 \\ y = 3 + 2 \end{cases} \end{gathered}

{

x=3

y=3+2

\begin{gathered} \begin{cases} x = 3 \\ y = 5 \end{cases} \end{gathered}

{

x=3

y=5

» Thus, the ones digit of the two-digit number is 5. Now find the two-digit number.

2\text{-}digit \: number = 10x + y2-digitnumber=10x+y

2\text{-}digit \: number = 10(3) + 52-digitnumber=10(3)+5

2\text{-}digit \: number = 30 + 52-digitnumber=30+5

2\text{-}digit \: number = 352-digitnumber=35

\therefore∴ The two-digit number and its tens digit that has a product of 105 is...

\Large \underline{\boxed{\tt \purple{35 \:\: and \:\: 3}}}

35and3

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