The equations X / 2-y + 1 / 3 = 1,3x + 2Y = 10

The equations X / 2-y + 1 / 3 = 1,3x + 2Y = 10

X/2-(Y+1)/3=1
Double six on both sides
3X-2(Y+1)=6
3X-2Y=8 (1)
3X+2Y=10 (2)
(1)+(2)
6X=18
X=3
Y=(10-3X)/2=1/2

Given X of y = 1 of 2, find the value of x ^ 2-xy + y ^ 2 of x ^ 2-2xy + y ^ 2

By dividing the numerator and denominator by Y & # 178;, (X & # 178; - 2XY + Y & # 178;) / (X & # 178; - XY + Y & # 178;) = [(x / y) & # 178; - 2 (x / y) + 1] / [(x / y) & # 178; - (x / y) + 1] = [(1 / 2) & # 178; - 2 (1 / 2) + 1] / [(1 / 2) & # 178; - (1 / 2) + 1] = 1 / 3 (i.e. 1 / 3) = [(1 / 2) & # 178; - (1 / 2) + 1] = [(1 / 2) / [(1 / 2) & # 178; - (1 / 2) + 1] = 1 / 3 (i.e

Solving a quadratic equation of one variable
-1 / 2x square + 3x = 10
Not yet,

Root formula: two solutions are respectively: [- B + radical (bsquare-4ac)] / 2a and [- b-radical (bsquare-4ac)] / 2A
Because B squared - 4ac

The teacher assigned an assignment; when a is what, the fraction A-1 / A & # 178; - 1 is meaningless?
Xiao Gang's solution is as follows: A-1 / A & # 178; - 1 = A-1 / (A-1) (a + 1). From a + 1 = 0, a = - 1, so when a = - 1, the fraction is meaningless. Is Xiao Gang's solution correct? If there is any mistake, please find out the reason for the mistake and correct it

No
The denominator should not be 0 before making an appointment
a²-1≠0
a²≠1
So a ≠ - 1 and a ≠ 1
If you don't understand, I wish you a happy study!

I'm a freshman in junior high school

It should be, if a digit is x, then ten digit is 4x + 1, so 10 (4x + 1) + X - (10x + 4x + 1) = 63 gets 27x = 54, so x = 2, then original number is 92, then ten digit is x, then ten digit is x + 5, so 10 (x + 5) + x = 8 (x + X + 5) + 5 gets 11x + 50 = 16x + 45, so x = 1, then original number is 61, then ten digit and one digit

How to solve the equations with three unknowns
When m is an integer, the solution of the equations 2x my = 6, x-3y = 0 is a positive integer?
Man. Can you make it easier to understand? I don't understand

By solving the system of equations 2x my = 6, x-3y = 0 with respect to XY, we obtain:
x=18/(6-m)
y=6/(6-m)
If the solutions of the equations 2x my = 6 and x-3y = 0 are positive integers, then
18/（6-m ）>0
6/（6-m ）>0
That is, 6-m > 0
Therefore, when m is 0,1,2,3,4,5, the solutions of the equations 2x my = 6, x-3y = 0 are positive integers

-7x+﹙6x²－5xy ﹚－﹙3y²+xy－x²﹚

-7x+﹙6x²－5xy ﹚－﹙3y²+xy－x²﹚
=-7x+6x²-5xy-3y²-xy+x²
=7x²-7x-6xy-3y²

It is known that the real number A.B.C satisfies | a + 1 | + (3b + 1) ^ 2 + | C-2 | = 0, so that the value of (3AB) ^ 2 times (- A ^ 2C) ^ 3 times (6ab ^ 2) ^ 2 can be obtained

Because the [a + 1] is ≥ 0, (3b + 1) ^ 2 ≥ 0, the [C-2] is ≥ 0, and [a + 1 + (3b + 1) ^ 2 + C-2 = 0, so [a + 1 = 0, (3b + 1) ^ 2 = 0, (3b + 1) ^ 2 = 0, (3b + 1) ^ 2 = 0, (3b + 1) ^ 2 = 0, (3b + 1) ^ 2 = 0, that is, a = -1, B = - (1 / 3), C = 2, so (3AB) ^ 2 * (- A ^ 2 * (- A ^ 2C ^ 2C) ^ 3 * (6ab ^ 2) ^ 2 * (6ab ^ 2 * (6ab ^ 2) (6ab ^ 2) ^ 2) 2 = (3 * (3 * (3 * (3 * - 1 * - (1 * - (1 / 3 / 3)) (1 / 3)) (2 * (2 * (2 * (6 * (6 * (6 * (6 * (6 * - 2) ^ 2 = 1 * 1 (- 2 / 3) ^ 2 = 4 / 9

(x + 3Y) (x-3y) (x ^ 2 + 9y ^ 2) and (- 3) ^ 2n + 1 + 3 (- 3) ^ 2n

1（x+3y)(x-3y)(x^2+9y^2) =(x^2-9y^2) (x^2+9y^2) =(x^4-81y^4)
（-3）^2n+1+3(-3)^2n
Is 2 (- 3) ^ (2n + 1) + 3 (- 3) ^ 2n = (- 3) ^ (2n + 1) - (- 3) ^ 2n=
（-3）^(2n+1)-(-3)^（2n+1）=0

(x square plus 5x plus 3) (x square plus 5x minus 2) minus 6

Original form
=(x²+5x)²+(x²+5x)-6-6
=(x²+5x)²+(x²+5x)-12
=(x²+5x+4)(x²+5x-3)
=(x+1)(x+4)(x²+5x-3)