(5a + 4A + 7a) + (5c-3b-6a), (the square of 8xy-x + the square of Y) - (the square of X - the square of one + 8xy)?

(5a + 4A + 7a) + (5c-3b-6a), (the square of 8xy-x + the square of Y) - (the square of X - the square of one + 8xy)?

（1/3a-3/4b+5c)-（3/4a-1/3b-3c） =（1/3a-3/4a)+(-3/4b+1/3b）+(5c+3c) =1/12a-5/12b+8c

If 7a-8b = 5. Find the value of (3a-4b) (7a-8b) - (11a-12b) (8b-7a)

(3a-4b)(7a-8b)-(11a-12b)(8b-7a)
=(7a-8b)(3a-4b+11a-12b)
=(7a-8b)(14a-16b)
=2(7a-8b)²
=2×5²
=50

It is known that the solution 3x minus ay equals 5, x plus by equals 6 is x = 5, y = 6, then the solution of the system 3 (x + y) - A (X-Y) = 5, x + y + B (X-Y) = 11 is

3x ay = 5, x + by = 6, the solution is x = 5, y = 6, then 15-6a = 5, a = 10 / 6 = 5 / 35 + 6B = 6, B = 1 / 63 (x + y) - A (X-Y) = 5 （1）x+y+b(x-y)=11 …… （2）（2）×3-（1）3b(x-y)+a(x-y)=33-5 x-y=28/(a+3b)=28/(5/3+3×1/6)=28×6/13=168…… （3）x+y=11-1...

Given that the equations 3x + y = 3,4x + ay = 2 have integer solutions (a is an integer), find the value of A
I solved the equations 3x + y = 3 (1) 4x + ay + 2 (2)
From (1), y = 3-3x (3)
Substituting (3) into (2), we get 4x + a (3-3x) = 2

3x+y=3,4x+ay=2
Eliminate X and get X
y=6/(4+3a)
With integer solution (a is an integer)
So the number divisible by 6 is 1,2,3,6, - 1, - 2, - 3,6
So 4A + 3 must be 1,2,3,6, - 1, - 2, - 3,6
Because a is an integer
So a = 0, - 1
When a = 0, y = 2, x = 1 / 2 does not conform to rounding
When a = - 1, y = 6x = - 1
So a = - 1

Given that {x = 3, y = 2 is a solution of the bivariate linear equation 3x ay = 6, then a = what?

3*3-a*2=6
9-2a=6
2a=3
a=3/2

Given that x = 1, y = 2 is a solution of the bivariate linear equation 3x ay = 2, then a = ()

Substituting x = 1, y = 2 into 3x ay = 2
3-2a=2
3-2=2a
1=2a
a=1÷2
a=1/2

If the solution of the quadratic equation 3x + ay = 7 is x = 3, y = - 1, then the value of a is?

Just take X and Y in,
3*3+a*（-1）=7
a=2
PS: it's too late. Let's go to bed early and take it

It is known that three sides a, B and C of a triangle satisfy a + B + C = AB + BC + AC. the shape of the triangle is determined by the multiplication formula

2A + 2B + 2C = 2Ab + 2BC + 2Ac
The formula is as follows
(a-b) square + (B-C) square + (A-C) square = 0
So A-B = 0
b-c=0
a-c=0
So a = b = C
Then this is an equilateral triangle

The three sides of a triangle satisfy A.A + B.B + C.C = AB + BC + AC

The two sides of the equation are multiplied by 2, the term is shifted, the formula, the miracle happens. Equilateral triangle

The square of a - 2 ab + 2 times the square of B + 4A + 8 = 0
A and B are rational numbers,

The square of a - 2Ab + 2 times the square of B + 4A + 8 = 0
That is, 2 (A / 2-B) ^ 2 + A ^ 2 / 2 + 4A + 8 = 0
2(a/2-b)^2+(a+4)^2/2=0
That is, a / 2-B = 0, a + 4 = 0
So a = - 4, B = - 2
ab=8