If there is a two digit number, the number on the tenth digit is 5 times larger than the number on the single digit, and the two digit number is 5 times larger than the sum of the two digits, find the two digit number

If there is a two digit number, the number on the tenth digit is 5 times larger than the number on the single digit, and the two digit number is 5 times larger than the sum of the two digits, find the two digit number

Let X be the single digit, then x + 5 is the ten digit. From the meaning of the question, 10 (x + 5) + x = 8 [x + (x + 5)] + 5, the solution is: x = 1, then the two digit is 61

If a and B are fixed values, then the solution of the first order equation 3AK − x4-x + 2bx5 = - 1 is always 2, no matter what the value of K is

By multiplying both sides of the equation by 20 and removing the denominator, we can get 5 (3ak-x) - 4 (x + 2bx) = - 20. We can get that 15ak-8bx-9x = - 20, ∵ no matter what the value of K is, the solution of the equation is always 2, ∵ 15A = 0, - 8b × 2-9 × 2 = - 20, and the solution is a = 0, B = 18

If a and B are fixed values, the solution of the equation about X is always x = 1 no matter what the value of K is

If both sides of the equation are multiplied by 6, 4ka-x-6bx = 12, x = (4ka-12) / (6B + 1), because a and B are fixed values, then 6B + 1 is also fixed value. No matter what the value of K is, the solution is x = 1, then 4ka-12 must also be fixed value. Only a = 0 can satisfy the condition, so a = 0. If x = 1 is brought into the deformed equation, then - 1-6b = 12, B = - 13 / 6. A = 0; b = - 13 / 6