Question

How do you solve the system x+2y+5z=−1, 2x−y+z=2, 3x+4y−4y=14?

Answer 1

\huge\mathfrak{Answer}Answer

65

\huge\textsf{Explanation}Explanation

Let the digits be x and y.

SO, the original number would be 10x + y (two digits numbers are of this form, when expanded)

Given, sum of digits = 11

⇒ x + y = 11

And, when the digits are interchanged, then the original number is 9 more than the new number.

So, ⇒ 10y + x + 9 = 10x + y

⇒ 9x - 9y = 9

⇒ 9(x - y) = 9

⇒ x - y = 9/9

⇒ x - y = 1

So, now we get that, x + y = 11 and x - y = 1

So, add these two equations to get the value of x

x + y + x - y = 11 + 1

⇒ 2x = 12

⇒ x = 6

Now, put the value of x in any equation to get the value of y.

⇒ x - y = 1

⇒ 6 - y = 1

⇒ - y = 1 - 6

⇒ - y = - 5

⇒ y = 5

So, the digits are 6 and 5, hence, the number would be :-

⇒ 10x + y

⇒ 10(6) + 5

⇒ 60 + 5

⇒ 65

Hence, the original number is 65.