What is the largest two digit number that is divisible by three whose digits differ by two? Please show your work!

75

Step-by-step explanation:

2 digit number: tens digit (X), ones digit (X - 2)

2 digit number: 10 X + (X - 2) = 11X - 2

11x - 2 = 3 Y  (must be divisible by 3)

x = (3Y + 2) / 11

possible number of 3Y + 2 should be multiple of 11

                            11 , 22, 33, 44, 55, 66, 77, 88, 99, 110

X:                          1  ,  2 ,  3 ,   4,    5,   6,   7,   8,    9,  10

2 digit number:         20, 31,  42,  53,  64, 75, 86, 97

75 is the largest 2 digit number can be divisible by 3 and digits differs by 2

We can check if the ones digit is (X) and tens digit (X+2), will give the same number

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

step-by-step explanation: