a certain two-digit number is 6 less than the sum of its tens digit and 7 times its units digit. if the digits are reversed the number is increase by 27. fin the original two-digit numbers
let the tens digit be x and the unit digit be y
then the number is 10x + y
10x+y = (x+7y) - 6
9x - 6y = -6
3x - 2y = -2 , #1
the number reversed would be 10y + x
so 10y+x = 10x+y + 27
9x - 9y = -27
3x - 3y = -9 , #2
#1 - #2
y = 7
then from #1, x = 4
the number is 47
check:
reverse it --- 74
is 74 -47 = 27 , so far so good
what is the sum of its tens digit and 7 times its unit digit
4 + 7(7) = 53
is 53 greater than 47 by 6 ? YEAHHHH
Let's solve this step by step.
Let's assume the tens digit of the original two-digit number is represented by "x", and the units digit is represented by "y". Thus, the original two-digit number can be written as 10x + y.
According to the given information, the original number is 6 less than the sum of its tens digit and 7 times its units digit. Mathematically, we can express this as:
10x + y = x + 7y - 6
Next, we are told that if the digits are reversed, the number is increased by 27. This means the new number is 27 greater than the original number. Mathematically, we can write this as:
10y + x = 10x + y + 27
Now we have a system of two equations:
1) 10x + y = x + 7y - 6
2) 10y + x = 10x + y + 27
To solve this system, we can use the method of substitution or elimination.
Let's solve it using substitution:
From equation 1, we can isolate x in terms of y:
10x - x = 7y - y -6
9x = 6y - 6
x = (6y - 6) / 9
Substituting this value of x in equation 2, we get:
10y + (6y - 6) / 9 = 10((6y - 6) / 9) + y + 27
Now, solve for y:
90y + 6y - 6 = 60y - 60 + 9y + 243
96y - 6 = 69y + 183
96y - 69y = 183 + 6
27y = 189
y = 189 / 27
y = 7
Now substitute this value of y back into equation 1 to find x:
10x + 7 = x + 49 - 6
10x - x = 49 - 6 - 7
9x = 36
x = 4
Therefore, the original two-digit number is 47.