In a 2-digit number, the tens digit is 5 less than the units digit. If you reverse the number, the result is 7 greater than double the original number. Find the original number.

The original number is 38

A 2-digit number can be written as:

N = a*10 + b*1

Where a is the tens digit, and b is the units digit, these two are single-digit numbers.

We know that:

"the tens digit is 5 less than the units digit."

This means that:

a = b - 5

(notice that a must be larger than zero and smaller than 10, from this, we can conclude that b is a number in the range {6, 7, 8, 9})

"If you reverse the number, the result is 7 greater than double the original number"

The reverse number is:

b*10 + a

and this is 7 greater than 2 times the original number, then:

b*10 + a = 7 + 2*(a*10 + b)

Then we found two equations:

a = b - 5

b*10 + a = 7 + 2*(a*10 + b)

Replacing the first equation in the second, we get:

b*10 + (b - 5) = 7 + 2*((b - 5)*10 + b)

Now let's solve that:

b*10 + b - 5 = 7 + 2*(11*b - 50)

11*b - 5 = 7 + 22*b - 100

-5 - 7 + 100 = 22*b - 11*b

88 = 11*b

88/11 = b = 8

Now that we know that b = 8, we can use the equation:

a= b - 5

a = 8 -5 = 3

Then the original number is:

a*10 +  b = 3*10 + 8 = 38

The original number is 38

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step-by-step explanation:

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.