Question

Frame an equation : A two digit number having x at ten’s digit and y at units digit is 4 times the sum of the digit.​

Answer 1

Answer:

yx = 4(y+x) is your answerrrr

Answer 2

Answer:

• The required two digit number be 83

Step-by-step explanation:

$\underline{ \rule{80mm}{1mm}}$

Given information

• A two digit number having tens digit be x and units digit be y
• Y at units digit is 4 times of the sum of the digit.

Need to find out

• Two digit number?

Explanation

• we can take two digit number be 10x + y where x is ten's place & y is units place from the above nentioned data .Inter - changing the digit means writing the number be 10y + x where y is ten's place & x is units place

Solution

Tow digit number ten's digit exceeds twice the unit's digit by 2 , the number obtained by inter - changing the digit is 5 more then three times the sum of the digit .

❍Suppose the two-digit number is given by 10x + y

$\underline{ \bigstar \pmb{ \sf \:According \: to \: the \: \: question }}$

The relation between x&y

$\dashrightarrow \: \text{x = 2y + 2}\: \: \dots \dots (\bf{eq}^{n} - 1)$

$\pmb{ \tt{After \: inter - \: changing}}$

$\dashrightarrow \text{10y + x = 3x + 3y + 5} \: \: \dots \dots (\bf{eq}^{n} - 2)$

$\pmb{ \sf{Now \: simplifying \: equation \: (ii) \: we \: get , }}$

$\\ \dashrightarrow \tt7y - 2(2y + 2) = 5 \\ \\ \dashrightarrow \tt7y - 4y - 4 = 5 \\ \\ \dashrightarrow \tt3y = 5 + 4 \\ \\\dashrightarrow \tt3y = 9 \\ \\ \dashrightarrow \tt \: y = \cancel\frac{9}{3} \\ \\ \pink{ \dashrightarrow \tt\boxed{\boxed{ \frak{y = 3}}}} \\ \\ \\ \\ \\ \pmb {\sf \: Putting \: y=3 \: in \: equation \: (i) \: we \: get, } \\ \\ \\ \dashrightarrow \tt \: x = 2y + 2 \\ \\ \dashrightarrow \tt \: x = 2 \times 3 + 2 \\ \\\dashrightarrow \tt \: x = 6 + 2 \\ \\ \pink{\dashrightarrow \tt \: x = 8}$

$\pmb{ \sf{So, \: the \: digit \: formed \: is ,}}$

$\\ \sf: \longmapsto \: 10x + y \\ \\ \sf: \longmapsto10(8) + 3 \\ \\ \sf: \longmapsto80 + 3 \\ \\ \large \pink{: \longmapsto} \boxed{ \boxed{83}} \bigstar$

$\fbox{ \boxed{ \text{Hence the required two digit number is 83}}} \bigstar$

$\underline{ \rule{30mm}{1mm}} \underbrace{ \boxed{ \pink{ \frak{ItzDipti }}}}_ {\sf{ \boxed{ \sf \: XxMrZombiexX}}} \underline{\rule{30mm}{1mm}}$