Question

9. The digits of a 2-digit number differ by 5. If the digits are interchanged and
the resulting number is added to the original number, we get99. Find the
original number.
10. Solve : 5x - 3 = 3x + 7

Answer 1

Let tens place be x and ones be y.
So, x-y = 5. -> (1).
The number = 10x+y and when interchanged, it’ll be 10y+x.
On adding, we get 11x + 11y = 99.
Therefore x+y = 9 (dividing whole equation by 11).

Now, add .

=> 2x = 14. X = 14/2 = 7.
And 7-y = 5. Hence y = 2.

Hope this answer helps you. Please mark as brainliest! :)

Answer 2

Answer:

9) Let the digit in ten's. Place be x and the digit in the one's place be y.

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+y

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+x

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14⇒x=7

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14⇒x=7Putting the value of x in equation (i)

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14⇒x=7Putting the value of x in equation (i)x−y=5

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14⇒x=7Putting the value of x in equation (i)x−y=5⇒y=2

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14⇒x=7Putting the value of x in equation (i)x−y=5⇒y=2∴ Number=10x+y

Let the digit in ten's. Place be x and the digit in the one's place be y.∴x−y=5⟶(i)∴ Two digit number=10x+yTwo digit after reversing the digits=10y+xAccording to question ,∴10x+y+10y+x=99⇒11x+11y=99⇒x+y=9⟶(ii)On adding (i) and (ii),2x=14⇒x=7Putting the value of x in equation (i)x−y=5⇒y=2∴ Number=10x+y                    =72.