Question

Please help me solve this​

Answer 1

Given :

• The unit digit of 2digit number is 6 more than the digits at tens place
• Product of tens and unit digit is 12 less than the 2-digit number

To find :

Unit digit of 2-digit number.

Let :

• Digit at unit place = x
• Digit at tens place = y
• Therefore, 2-digit number = 10y + x

Solution :

According to question,

Condition 1)

The unit digit of 2digit number is 6 more than the digits at tens place

➝ x = 6 + y

➝ y = x - 6 equation 1

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Condition 2)

Product of tens and unit digit is 12 less than the 2-digit number

➝ (x)(y) = [ 10y + x ] - 12 equation 2

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Put value of y(= x - 6 ) from equation 1 into equation 2,

➝ (x)( x-6) = [ 10(x-6) + x ] - 12

➝ x² - 6x = 10x - 60 + x - 12

➝ x² - 6x = 11x - 72

➝ x² - 6x - 11x + 72 = 0

➝ x² - 17x + 72 = 0

[ Note - Splitting middle term. ]

➝ x² - (8+9)x + 72 = 0

➝ x² - 8x - 9x + 72 = 0

➝ x(x-8) - 9(x-8) = 0

➝ (x-8)(x-9) = 0

Either. (x-8) = 0 or (x-9) = 0

Either ( x = 8 ) or. ( x = 9 )

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Therefore, unit place for given 2 digit number can be either 8 or 9

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ANSWER :

Possible Unit digit = Either 8 or 9

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

• Let digit at tens place be 'x'

So,

• Digit at unit place be (x + 6).

So,

• Two digit number formed is

$\rm :\implies\:Two \: digit \: number \: = 10 \times x + (x + 6) \times 1$

$\rm :\implies\: = 10x + x + 6$

$\rm :\implies\: = 11x + 6$

Now,

According to statement,

• Product of 2 digits is 12 less than two digit number.

$\rm :\implies\:x(6 + x) = 11x + 6 - 12$

$\rm :\implies\:6x + {x}^{2} = 11x - 6$

$\rm :\implies\: {x}^{2} - 5x + 6 = 0$

$\rm :\implies\: {x}^{2} - 3x - 2x + 6 = 0$

$\rm :\implies\:x(x - 3) - 2(x - 3) = 0$

$\rm :\implies\:(x - 2)(x - 3) = 0$

$\rm :\implies\: \red{x \: = \: 2} \: \: or \: \: \red{x \: = \: 3}$

So,

$\begin{gathered}\begin{gathered}\bf \:Possible \: unit \: digits \: are - \begin{cases} &\sf{x + 6 = 2 + 6 = 8} \\ &\sf{x + 6 = 3 + 6 = 9} \end{cases}\end{gathered}\end{gathered}$