When a is an integer The equation for x a x 1 2x 4a 5 is solved as an integer

Solution: ax-a=2x-4a+5

a-2)x=5-3a

x=(5-3a)/(a-2)

Note: This is a fraction of a numerator and denominator are a binomial, when the number of numerators is not less than the number of times of the denominator, it can be in the form of integer + fraction, I call this method an integer).

x=(-3a+6-1)/(a-2)=[-3（a-2）-1]/(a-2)=-3- 1/(a-2)

If x is an integer (and when a is an integer), there must be 1 (a-2) as an integer.

a-2 must be an divisor of 1, i.e. a-2 = 1

a=3 or a=1

That is, when a = 3 or a = 1, the solution of the equation a(x-1) = 2x-4a+5 about x is an integer.

To make the solution of x an integer, let (5-3a) (a-2) = (a-2) (a-2) (a-2) (a----2) (a-2) (a-2) (a-2) (a-2) (a-2) (a-2) (a-2) (a-2) (a-2) (a-2 is not equal to 2, k is an integer).

Because 5-3a=-3(a-2)-1; So k = -3-1 (a-2) and only if a-2 = 1 or -1 satisfies the requirement.

So a=3,1

Solution: The original formula can be reduced to (2-a)x-3a+5=02-a)x=3a-5

3a-5 1

x= --=-3+ -

2-a a-2

It can be seen that the question is satisfied only when a=1 and 3.

Problem solving idea: Because the solution of the equation is a positive integer, Lao Huai first finds the solution of the equation x=[4 2 a], and then makes [4 2 a] a positive integer to find the value of a

Solve the equation about x 2x+1=ax+5, transfer the term to the servant and obtain: 2x-ax=5-1, combine the same terms to obtain: (2-a)x=4, and the coefficient is 1 to obtain:

x=[4/2−a]；Since [4 2 a] is a positive integer, 2-a may be , then the value of a may be , -2

2,2x+1=ax+5

2-a)x=4

x=4/(2-a)

x is a positive integer, a=1,2,0,2x+1=ax+5a-2)x=4

x=1,a=6

x=2,a=4

x=4,a=3,0,

ax+3=2x-1

2-a)x=4

x=4 (2-a) is the number of cavity socks.

Then 2-a is the divisor of 4.

So 2-a = 1, 2, 4

So. a=1,a=3,a=0,a=4,a=-2,a=6

When x 0, then x=ax+2, and the number of wax is a=-1 or 0;

When x 0, then -x=ax+2, then a=0 or a=1

Therefore, fill in 1 or 0 round file.

2ax=ax+x+3

x(a-1)=3 (this renting finch is simplified, put x aside, because it is an equation about x) 3=1 3=(which type of modulus is slow-1) (3) (because the title says an integer) a-1=1,3,-1,-3

then a=2,4,0,-2

ax+5=a^2+2a+2x

ax-2x=a^2+2a-5

a-2)x=a^2+2a-5

x=(a^2+2a-5)/(a-2)

x=(a^2-4a+4+6a-9)/(a-2)x=[(a-2)^2+6a-9]/(a-2)x=(a-2) +6a-12)+3]/(a-2)x=(a-2) +6 + 3/(a-2)

x=(a+4) +3/(a-2)

a(a≠2) is an integer.

A+4 is an integer.

The solution of a unary equation is an integer.

The value of 3 (a-2) should be an integer.

Then: a=-1 or a=1 or a=3 or a=5

So, the sum of all integer a satisfying conditions is: -1+1+3+5=8

Since a is not equal to 2, the original formula can be reduced to :

x=（a2+2a-5）/(a-2)

Further simplify the formula to:

x=/(a-2)=(a-2)+6+13/(a-2)=a+4+13/(a-2)

Since a is an integer, a+4 is an integer, as long as a-2 is divisible by 13, and the conditions are a=1,3,15,-11

So the algebraic sum is: a=1+3+15+(-11)=8.

From the original formula: x=4 (2-a).

Because x is an integer, the value of a is -2,0,1,3,4.