# Please solve this both 6&7 question in step by step. ?

### 3 Answers

- 冷眼旁觀Lv 64 weeks agoBest Answer
6.

In the question, it should be "18" instead of '8".

Let u be the units digit and t be the tens digit.

t + u = 6 …… [1]

(10t + u) + 18 = 10u + t …… [2]

From [2]:

10t + u + 18 = 10u + t

9t - 9u = 18

t - u = 2 …… [3]

[1] + [3]:

(t + u) + (t - u) = 6 + 2

2t = 8

t = 2

Substitute t = 2 into [1]:

2 + u = 6

u = 4

Hence, the number is 24.

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7.

Let x and y be the two numbers, and x > y.

4(x + y) = 140 …… [1]

x - y = 13 …… [2]

[1] ÷ 4:

x + y = 35 …… [3]

[2] + [3]:

(x - y) + (x + y) = 13 + 35

2x = 48

x = 24

Substitute x = 24 into [2]:

24 - y = 13

y = 11

Hence, the two numbers are 11 and 24.

- Φ² = Φ+1Lv 74 weeks ago
6. This has no solution. The starting numbers are 60, 51, 42, 33, 24, 15, and maybe even 06. Adding 8 to these yields 68, 59, 50, 41, 32, 23, and 14. None of these is the reverse of the matching number in the first group.

If this had been 18, then you could start with 24 and finish with 42.

7. We can restate this as: The sum of two numbers is 35 and their difference is 13.

The numbers are then (35+13)/2 = 24 and (35-13)/2 = 11.

I needed the 6th solution plz solve this in step by step