Question

The sum of the digits of a two-digit number is 11. The digit at unit's place is 2 more than twice the digits at ten's place. Find the number.​

Answer 1

Answer:

Please find the attachment

Answer 2

Answer:

the number is 38

Step-by-step explanation:

if a is the number at the ten's place and b is the number at the unit's place, then it means a + b = 11.

the number in the unit's place or b is 2 + 2a (twice the number in the ten's place which is a)

so a + 2 + 2a = 11

combine like terms, which are a and 2a:

a + 2 + 2a = 11 becomes 3a + 2 = 11

then get rid of 2:

3a + 2 = 11

3a + 2 - 2 = 11 - 2

3a = 9

divide both sides by 3 to get the value of a

3a ÷ 3 = 9 ÷ 3

a = 3.

now that we know that a = 3, we can now solve for b which is 2 + 2a.

2 + 2a becomes 2 + (2×3)

2 + 6 = 8. so b is 8