If the number a is divided by the number B, the quotient 4 is more than 5. If the number a is increased by 4 times, the number B remains unchanged. Then the quotient 18 has no remainder. The original number a is ()

Are you sure it's quotient 4 + 5? It's impossible. The remainder must be less than divisor

The sum of number a, number B, quotient and remainder is 2008, and number B is ()

（2008-10-16-10）÷（16+1）
=1972÷17
=116
(1) 2008-10-16 find the sum of a and B
(2) Because "number a divided by number B, quotient 16 is more than 10", so number a is 10 times more than number B. therefore, number a minus 10 is just 16 times of number B, that is, the sum of number a and number B minus 10, that is, (2008-10-16-10) is just (16 + 1) times of number B

The sum of a, B, quotient and remainder is 163. What are the two numbers?

A △ B = 3 So a = 3 B + 10 A + B + 3 + 10 = 163 3 B + 10 + B + 3 + 10 = 163 4 B = 140 B = 35 A = 3 B + 10 = 105 + 10 = 115 satisfied, please adopt it in time

If the number a is divided by the number B, the quotient 4 is more than 5. If the number a is increased by 4 times, the number B remains unchanged. Then the quotient 18 has no remainder. What is the original number a
Thank you

Let B be X
4(4x+5)=18x
16x+20=18x
2x=20
x=10
4x+5=40+5=45
The number a is 45

The quotient of a divided by B is 20 and there is no remainder. The sum of a and B is 315. What are the two numbers?
Use the ordinary method, not the equation

In this problem, a divided by B equals 20. It shows that a is 20 times of B, then B can be regarded as a share, a can be 20 times multiplied by 1, so that the total number of a and B is 21. 315 / 1 + 20 = 15 is the result of B. 15x20 shows that a is 300

X & # 178; + ax-2a & # 178; = 0 ABX & # 178; - (A & # 178; - B & # 178;) x-ab = 0 (a, B are constants and ab is not equal to 0) to solve these two equations

x²+ax-2a²=0
(x-2a)(x+a)=0
a1=2a x2=-a
abx²-(a²-b²)x-ab
(ax+b)(bx-a)=0
x1=-b/a x2=a/b

Given that two of the equations X & # 178; + ax + B = 0 are exactly a and B, then AB is equal to?

Using Vader's theorem: a + B = - A
ab＝b
The solution is a = 1, B = - 2
So AB = - 2

For X, the equation ax-b = bx-a (a is not equal to b),

AX-B=BX-A
AX-BX=B-A
X(A-B)=B-A
X=(B-A)/(A-B)=-1

On the equation of X: ax-m = BX + n (a is not equal to b)
On the equation 2x - [2 - (2b-1) x] = A-2 of X
On the equation of X | x | + | X-2 | = 6
On the equation of X | x | - x = 5
On the equation B (a + 2x) - a = (B + 2) x + ab of X
Please complete the steps

On the equations of X and Y {3x-4y = - 6, ax = 3By = - 4 and {3bx = 2ay = 0, 2x-y = 1, we have the same solution, find a, B

Solution: it is known that: there is the same solution, so two equations can be combined with a quaternion linear equation system as follows: 3x-4y = - 6 ① ax-3by = - 4 ② 3bx-2ay = 0 ③ 2x-y = 1 ④ from ④? 4 - ① get: x = 2 substitute x = 2 into ④ get: y = 3 substitute x = 2, y = 3 substitute x = 2 into ② ③ get: 2a-9b = - 4 ⑤ 6b-6a = 0 ⑥ substitute 6 into

If the number a is divided by the number B, the quotient 4 is more than 5. If the number a is increased by 4 times, the number B remains unchanged. Then the quotient 18 has no remainder. The original number a is ()