Question

repare simultaneous equations as per given instructions:
The number
is 4 times
The number formed
The digit in tens place x by interchanging the
The digit in units place y digits = 2 x original
the sum of
number-9
the digits.
... (1)
ese equations and find the original number.​

Answer 1

Answer:

The number is four times the sum of its digits => 10*x + y = 4(x+y)

if interchanged, the number is 9 less then 4 times the original number

interchanged number = y*10 +x

original number = x*10 + y

=> y*10 + x = 4(x*10+y) - 9

So, we got the equations

10x + y = 4x + 4y —————————-(1)

10y + x = 40x + 4y - 9 ————————(2)

=>

(10–4)x = (4–1)y => 6x = 3y => 2x = y —————(3)

(40–1)x = (10–4)y +9 => 39x -6y = 9 —————-(4)

substitute eq3 in eq4

39(x) - 6(2x) = 9

(39–12)x = 9

27x = 9

x = 9/27 = 1/3

y = 2(x) = 2(1/3) =2/3

so the value of x and y is 1/3 and 2/3 respectively

and

The 2-digit number is 10*x + y = (10*1/3) + 2/3 = 12/3 = 4.

Step-by-step explanation:

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