Question

if a(y+z)=b(z+x)=c(x+y) and a≠b≠c then show that y–z/a(b-c) = z-x/b(c-a) = x-y/c(a-b)​

Answer 1

$\huge\star{\red{Q}{uestion}}\star\:$

Sum of the digits of a two-digit number is 12. The given number exceeds the number obtained by interchanging the digits by 36. Find the given

number.

$\huge\star{\red {A}{nswer}}\star\:$

$\huge\underline {Let,}$

The tens digit of the required number be x

and the units digit be y

$\huge\underline {Then,}$

x + y = 12 ......... eq. (1)

Required number = (10x + y)

Number obtained on reversing the digits = (10y + x)

$\huge\underline {Therefore,}$

(10y + x) - (10x + y) = 18

9y - 9x = 18

x - y = 12 ......... eq. (2)<br>

On adding eq. (1) and eq. (2)

$\huge\underline {We\: get}$

x + y + y - x = 12 +2

2y = 14

y = 2

$\huge\underline {Therefore}$

x = 5

Hence, the required number is 57

$\huge\green { Hope\: this\: helps\: you}$

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