Question

the sum of the digits of a two digit number is 7 when the digits are reversed then the number is deacresed by 9 find the number

Answer 1

Let the ones digit no y.
tens digit no.x.
atq 10x+y=7 - 1st
10x+y-10y-x= 9
9x-9y=9
x-y=1 -2nd
equate 1and 2
x=11/8
y= 3/8

Answer 2

$\boxed{\boxed{\boxed{\boxed{\boxed{\large\mathbf{43}}}}}}$

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Every 2-digit number can be made using a formula 10x + y

where, x = tens place and

           y = ones place

For example,

56 = 10(5) + 6

x = 5 ; y = 6

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So, let the number be 10x + y

Hence, the number obtained on reversing will be 10y + x

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A.T.Q,

x + y = 7 ---- (1)

10x + y - (10y + x) = 9

=> 9x - 9y = 9

=> 9(x - y) = 9

=> x - y = $\frac{9}{9}$

=> x - y = 1 ----(2)

Now,

By Elimination Method,

Adding equation(1) and equation(2) we get,

2x = 8

=> x = $\frac{8}{2}$

=> $\boxed{\boxed{\mathbf{x = 4}}}$

Now,

Substituting value of x in equation(1) we get,

4 - y = 1

=> -y = 1 - 4

=> -y = -3

=> $\boxed{\boxed{\mathbf{y = 3}}}$

Hence, the number is 10x + y.

Substitutimg the values of x and y, we get

10(4) + y

=> 40 + 3

=> 43

Hence, the number is 43.