x> Why can numbers after 18 be continuous? = 0, Y > = 0, 4 * x + 7 * y = Z; Z can be 4,7,8,11,12,14,15,16,18,19,20,21

x> Why can numbers after 18 be continuous? = 0, Y > = 0, 4 * x + 7 * y = Z; Z can be 4,7,8,11,12,14,15,16,18,19,20,21

I don't know where your problem is? This is an indefinite equation. The condition of having integer solution is (4,7) | Z, where (4,7) is the greatest common factor of 4 and 7. In fact, all integers Z can make the equation have integer solution x, y, but your problem requires x ≥ 0, y ≥ 0, that is to say, it needs non negative integer solution

18%: 3 / 20 = 6

6.5 x = x / 6.5
x/6.5=1.2
x=1.2×6.5
x=7.8

The solution equation: (X-2) (x + 1) (x + 4) (x + 7) = 19

Let y = (x2 + 5x − 14) + & nbsp; (x2 + 5x + 4) & nbsp; & nbsp; 2 = x2 + 5x-5, then (Y-9) (y + 9) = 19, that is, y2-81 = 19. The solution is y1.2 = ± 10. Substituting the values of Y1 and Y2 into formula ①, x1.2 = − 5 ± 852, x3.4 = − 5 ± 52,

How to solve the equation; (A-2) (a + 1) (a + 4) (a + 7) = 19

a-2)(a+7)(a+1)(a+4)=19
(a^2+5a-14)(a^2+5a+4)=19
(a^2+5a)^2-10(a^2+5a)-56=19
(a^2+5a)^2-10(a^2+5a)-75=0
(a^2+5a-15)(a^2+5a+5)=0
A ^ 2 + 5a-15 = 0 or a ^ 2 + 5A + 5 = 0
I don't need to solve it next

How to solve the equation x-5.3 + 4.7 = 19

x-5.3+4.7=19
x-0.6=19
x=19.6

X-4 = 2 * [1.5 (x-4) - 24] how to solve the equation?

x-4=2*[1.5(x-4)-24]
x-4=3(x-4)-48
x-4=3x-12-48
2x=56
x=28

Solving equation X-1 / 4 = 24

x－1／4=24
x=24+1／4
X = 24 1 / 4

The solution equation: (x + 3) 4 + (x + 1) 4 = 82

Suppose y = x + 3 + X + 12 = x + 2, then the original equation becomes (y + 1) 4 + (Y-1) 4 = 82 {[(y + 1) 2 - (Y-1) 2] 2 + 2 (y + 1) 2 × (Y-1) 2-82 = 0 {16y2 + 2 (y2-1) 2-82 = 0 {Y4 + 6y2-40 = 0} (Y2 + 10) (y2-4) = 0. ∵ Y2 + 10 ≠ 0 ∵ only y2-4 = 0, then y = ± 2

82-4 (x-1) = 70

82-4(x-1)=70
82-4x+4=70
86-4x=70
-4x=-16
x=4

The reduction ratio is 1 / 3:1 / 5, 3 / 4:1 / 9, 1 / 20:1 / 4, 15:12, 17:51 and 5.7:19

1/3:1/5 =5：3
3/4:1/9=27：4
1/20:1/4=1：5
15:12=5：4
17:51 =1：3
5.7：19=3：10
If you don't understand, please ask. If you understand, please adopt it in time! (*^__ ^*)