When what is the value of a, the solution of the fractional equation (4ax + 3) / (a + 2x) = 3 is x = 1?

When what is the value of a, the solution of the fractional equation (4ax + 3) / (a + 2x) = 3 is x = 1?

(4ax + 3) / (a + 2x) = 3, x = 1
（4a+3）/(a+2)=3
4a+3=3a+6
a=3

The solution of the equation (x + m) divided by (X-2) = 1 of X is greater than zero, and the value range of M is obtained
It's better to have a detailed explanation

（x+m)/(x-2)=1
m=(x-2)-x=-2

The fractional equation x + 3 / 4 = 2x + 2 / 3 is transformed into 1-ary 1-th power

2X + 3 / 4
3（2x+3）=2(x+4)
6x+9 = 2x+8
4x +1 = 0
x = - 1/4

If the solution of the fractional equation MX − 1 + 31 − x = 1 of X is positive, then the value range of M is ()
A. M ＞ 2B. M ＞ 2 and m ≠ 3C. M ＜ 2D. M ＞ 3 and m ≠ 2

The denominator of the fractional equation is removed to get M-3 = X-1, and the solution is x = m-2. According to the meaning of the question, we get m-2 ＞ 0 and m-2 ≠ 1, and the solution is m ＞ 2 and m ≠ 3

If there is no solution to the fractional equation X-5 parts x + 1 = 10-2x parts m of X, find the value of M

(x+1)/(x-5)=-m/2(x-5)
Double 2 (X-5)
2x+2=-m
x=(-m-2)/2
If there is no solution, then this is root adding
So the denominator X-5 = 0
(-m-2)/2-5=0
-m-2=10
m=-12

It is known that the fractional equation x / (x-3) - 2 = m / (x-3) about X has a positive solution. Try to find the value range of M

Multiply both sides of the equation by x-3 to get X-2 (x-3) = M
x=6-m
If there is a positive solution, then there is 6-m > 0

Solve the equation (1) 2x + 3Y = 40x − y = − 5 (2) x + 13 − y + 24 = 0x − 34 − y − 33 = 112

(1) 2X + 3Y = 40, ① x − y = − 5, ②, y = x + 5, ③ is substituted into ①, 2x + 3 (x + 5) = 40, the solution is x = 5, and x = 5 is substituted into ③, y = 5 + 5 = 10, so the solution of the equations is x = 5Y = 10; (2) the equations can be reduced to 4x − 3Y = 2, ① 3x − 4Y = − 2, ②, ① × 4, 16x-12y = 8

If (A-2) x + (B + 1) y = 13 is a quadratic equation with respect to X and y, then a and B satisfy______ Conditions

∵ (A-2) x + (B + 1) y = 13 is a bivariate linear equation about X, y, ∵ a − 2 ≠ 0b + 1 ≠ 0, the solution is a ≠ 2, B ≠ - 1

If x (exponent 2a-b-2) + y (exponent a + 5B + 1) = 13 is a bivariate linear equation about X and y, then what conditions do a and B satisfy?

Bivariate linear equation means that there are two unknowns in the equation, and the index of the unknowns is one. According to this definition, 2a-b-2 = 1 and a + 5B + 1 = 1

If (A-2) x + (B + 1) y = 7 is a quadratic equation with respect to X and y, then the condition A.B should satisfy is

Well, it's as follows:
Because it's binary. X and y are called binary, two unknowns
So a should not be equal to 2, otherwise there will be no X
Of course, B cannot be equal to - 1, otherwise there will be no y
So: A is not equal to 2; B is not equal to - 1