When B = 1, the equation a (3x-2) + B (2x-3) = 8x-7 has innumerable solutions, then what is a equal to?

When B = 1, the equation a (3x-2) + B (2x-3) = 8x-7 has innumerable solutions, then what is a equal to?

3 (2a-b) square - 3A (4a-3b) + (2a + b) (2a-b) - B (a + b) where a = 1, B = 2, hurry up~~~
The original formula = 12a & # 178; - 12ab-3b & # 178; - 12a & # 178; + 9ab + 4A & # 178; - B & # 178; - ab-b & # 178;
=4a²-4ab-5b²
=4-8-20
=-24
Figure (1) is a square with side length of (a + b). The shadow part in figure (1) is put together into the shape of figure (2). The formula that can be verified is ()
A. （a+b）（a-b）=a2-b2B. （a+b）2-（a2+b2）=2abC. （a+b）2-（a-b）2=4abD. （a-b）2+2ab=a2+b2
According to the graph, we can get: ∵ AB = A2 + B2, ∵ s shadow = (a + b) 2 - (A2 + B2) = 2Ab
Equation: when B = 1, equation (2x-3) B + (3x-2) a = 8x-7 has infinite solutions, then a is equal to?
Substituting B = 1 into the equation, the equation is (2x-3) + (3x-2) a = 8x-7,
(3x-2) a = 6x-4 because X has infinitely many solutions, that is, both sides of the equation are equal. When a = 2, 2 (3x-2) = 6x-4
So a = 2
(3x-2)(a-2)=0 a=2
Given the square of a = - 4A + 3a and the square of B = - 2A - A-1, find the value of 3b-2a when (1) A-B (2) a = - 2
More details. Thank you!
We must pay attention to the same base, such as "a". Secondly, we have the same number of times. Then A-B = - 4A ^ 2 + 3A - (- 2A ^ 2-a-1) = - 2A ^ 2 + 4A + 13b-2a = 3 (- 2A ^ 2-a-1) - 2 (- 4A ^ 2 + 3a) = 2A ^ 2-11a-3, so when a = - 2, the above formula = 2 (- 2) ^
A-B=-2a^2+4a+1
3B-2A=2a^2-9a-3 =23
Ask a few math problems about factorization in junior high school
1. Decomposition factor: 4m * 3 + M-1
2. Given x2-x-1 = 0, find x * 5-x * 4-3x * 3 + 3x * 2 + X
3. Given a2-9x2 + 6xy-y2 / (a + 3x) 2 - (ay + 3xy) = 1, we prove that y = 6x
You can do whatever you want
It's m3, because it can't be typed. It's not m times 3. It's all cubic or square. It's just a process that you can understand,
4m^3+m-1=(8m^3-1)-(4m^3-m)=(2m-1)(4m^2+2m+1)-m(4m^2-1)=(2m-1)(4m^2+2m+1)-m(2m+1)(2m-1)=(2m-1)[4m^2+2m+1- m(2m+1)]=(2m-1)[4m^2+2m+1- 2m^2-m]=(2m-1)(2m^2+m+1)x^2-x-1=0,x^2-x=1x^2-1=xx^5-x^4-3x^3+3x^2+x=...
If we know what conditions a (3x-2) + B (2x-3) = 8x-7 A. B satisfies, the equation has innumerable solutions?
Urgent (need the whole process)
a(3x－2)＋b(2x－3)=8x－7a
(3a＋2b－8)x＋(5a－3b)=0
Then, if 3A + 2b-8 = 0 and 5a-3b = 0, the equation has innumerable solutions
3A + 2b-8 = 0 and 5a-3b = 0
The solution is as follows
a=24/19,b=40/19
.（2a+3b-4c)^2-(2a-3b+4c)^2
.（2a+3b-4c)^2-(2a-3b+4c)^2
=(2a+3b-4c+2a-3b+4c)(2a+3b-4c-2a+3b-4c)
=4a(6b-8c)
=8a(3b-4c)
=24ab-32ac
Take 50 points to adjust people's appetite? What about the question? It's not finished, is it
（2a+3b-4c)^2-(2a-3b+4c)^2
=[(2a+3b-4c)+(2a-3b+4c)][(2a+3b-4c)-(2a-3b+4c)]
=4a(6b-8c)
=8a(3b-4c)
=24ab-32ac
Application of square difference formula
（2a+3b-4c)^2-(2a-3b+4c)^2
=[2a+(3b-4c)]^2-[2a-(3b-4c)]^2
=[2a+(3b-4c)+2a-(3b-4c)][2a+(3b-4c)-2a+(3b-4c)]
=(2a+3b-4c+2a-3b+4c)(2a+3b-4c-2a+3b-4c)
=4a(6b-8c)
=24ab-32ac
（2a+3b-4c)^2-(2a-3b+4c)^2
=(2a+3b-4c+2a-3b+4c)(2a+3b-4c-2a+3b-4c)
=8a(3b-4c)
A mathematical problem of factorization
We know that a + B + C = 0, and prove: a cubic + A & # 178; C + B & # 178; c-abc + B cubic = 0
The process of factorization···
A "'+ a" C + B "C - ABC + B"' = (a "'+ B"' + (a "C - ABC + B" C) = (a + b) (a "- AB + B") + C (a "- AB + B") = (a + B + C) (a "- AB + B") because a + B + C = 0, the formula is = 0 (a "- AB + B") = 0
The third power of ｛ + A + B
=a³+b²+a²c+b²c-abc
=(a+b)(a²-ab+b²)+c(a²+b²-ab)
=(a+b+c)(a²-ab+b²)
=0*(a²-ab+b²)
=0
Immediate evidence
Given the equation a (3x-2) + B (2x-3) = 8x-7 (a, B are constants) about X, the conditions that a and B satisfy are obtained respectively?
① Infinite solution, unique solution, no solution