complete question:
The sum of the digits of a two-digit numeral is 8. If the digits are reversed, the new number is 18 greater than the original number. How do you find the original numeral?
Answer:
The original number is 10a + b = 10 × 3 + 5 = 35
Step-by-step explanation:
Let
the number = ab
a occupies the tens place while b occupies the unit place. Therefore,
10a + b
The sum of the digits of two-digits numeral
a + b = 8..........(i)
If the digits are reversed. The reverse digit will be 10b + a. The new number is 18 greater than the original number.
Therefore,
10b + a = 18 + 10a + b
10b - b + a - 10a = 18
9b - 9a = 18
divide both sides by 9
b - a = 2...............(ii)
a + b = 8..........(i)
b - a = 2...............(ii)
b = 2 + a from equation (ii)
Insert the value of b in equation (i)
a + (2 + a) = 8
2a + 2 = 8
2a = 6
a = 6/2
a = 3
Insert the value of a in equation(ii)
b - 3 = 2
b = 2 + 3
b = 5
The original number is 10a + b = 10 × 3 + 5 = 35
Answer:
fyi b's answer has imaginary numbers in it...
Imaginary: 1 +
Imaginary: 1 -
Step-by-step explanation:
= 
... the negative root will produce imaginary solutions
Answer:
-1
Step-by-step explanation:
15x-8x-2=-9
Or, 7x=-9+2
Or, 7x=-7
Or, x=-7/7
Or, x =-1
Technically nothing times 3 get you 16 but 3 times 5 equals 15 which is the closest answer but the answer would be 5.3
