Solving the system of linear equations X: y = 5:3 (1) x: z = 7:2 (2) x-2y + 3Z = 4 (3) I want detailed steps ~ thank you

Solving the system of linear equations X: y = 5:3 (1) x: z = 7:2 (2) x-2y + 3Z = 4 (3) I want detailed steps ~ thank you

x：y=5：3 （1）
x：z=7：2 （2）
x-2y+3z=4 （3）
From (y) to (1 / 5)
From (2) we get z = 2x / 7 and substitute (3) to get z = 2x / 7
x-6x/5+6x/7=4
x=140/23
y=84/23
z=40/23
x=5/3y;
5/3y:z=7:2;
7z=10/3y;
5/3y-2*y+3z=4;
y=84/23;
x=140/23;
z=40/23
(a-b) &# 178; (a + b) &# 178; - (A & # 178; + B & # 178;) &# 178; factorization
(a-b)²(a+b)²-(a²+b²)²
=[(a-b)(a+b)]²-(a²+b²)²
=(a²-b²)²-(a²+b²)²
=(a²-b²+a²+b²)(a²-b²-a²-b²)
=2a²(-2b²)
=-4a²b²
(a-b)²(a+b)²-(a²+b²)²
=[(a-b)(a+b)+(a²+b²)][(a-b)(a+b)-(a²+b²)]
=(a²-b²+a²+b²)(a²-b²-a²-b²)
=2a²×(-2b²)
=4a²b²
To solve the equations: x-2y + Z = - 1, x + y + Z = 2, x + 2Y + 3Z = - 1
x-2y+z=-1①
x+y+z=2②
x+2y+3z=-1③
①+③
2x+4z=-2
x+2z=-1④
①+②×2
3x+3z=-1+4
x+z=1⑤
Substitute x = 1-z from 5 into 4
1-z+2z=-1
z=-2
∴x=1-(-2)=3
Substituting x = 3; Z = - 2 into 2
3+y-2=2
Y=1
That is: the solution of the equations is x = 3; y = 1; Z = - 2
X=3
Y=1
z=-2
Factorization of (a + b) & #178; - (a + b)
A:
(a+b)²-(a+b)=(a+b)(a+b-1)
Equation 2Y = 56
2y= 56y=28
y=28
2y=56
Divide both sides by two
y=28
Thank you.
y=28
2y=56
2y÷2=56÷2
y=28
Factoring cross multiplication ^ into square
8. A & sup2; - one sixth a - one sixth
12.x^4-13x²y²+36y^4
13.x^4+x²-2
（8）
. A & sup2; - one sixth a - one sixth
=1/6*(6a^2-a-1)
=1/6*(3a+1)(2a-1)
(12)
x^4-13x²y²+36y^4
=(x^2-4y^2)(x^2-9y^2)
=(x+2y(x-2y)(x+3y)(x-3y)
(13)
x^4+x²-2
=(x^2+2)(x^2-1)
=(x^2+2)(x+1)(x-1)
1.=a²-1/6a+1/144-1/144-1/6
=(a-1/12)²-25/144
=(a-1/12+5/12)(a-1/12-5/12)
=(a+1/3)(a-1/2)
To solve binary linear equation: 2x + 3Y = 56; 3x + 2Y = 54, to solve x, y (urgent!)
x=10.y12.
It would be better if there were fractions
1、 Which of the following polynomials contains the factor of 2x + 3? (1) 2x3 + 3 (2) 4x2-9 (3) 6x2-11x + 3 (4) 2x2 + X + 3 () 2. Which of the following polynomials is the factor of 2x2-11x-21? (1) (X-6) (2) (x + 7) (3) (2x-3) (4) (2x + 3) () 3
The solution equation: 3-1 / 5 (5-2y) = 4-1 / 10 (4-7y) + 1 / 2 (y + 2)
1. Multiply both sides by 10, that is: 30 -- 2 (5 -- 2Y) = 40 -- (4 -- 7Y) + 5 (y + 2)
2. Simplify: 30 -- 10 + 4Y = 40 -- 4 + 7Y + 5Y + 10
3. It is reduced to: - 8y = 26 y = -- 13 / 4
Factorization, application problems
Two students decompose the same quadratic trinomial into one factor. One student decomposes it into (x-1) (X-9) by misreading the coefficient of the first term, and the other student decomposes it into (X-2) (x-4) by misreading the constant term
Because the first student misread the term coefficient and decomposed it into (x-1) (X-9),
So the first student's decomposition formula is: x ^ 2-10x + 9
So the correct quadratic term is x ^ 2 and the correct constant term is 9
Because the second student misinterpreted the constant term and decomposed it into (X-2) (x-4)
So the second student's decomposition formula is: x ^ 2-6x + 8
So the straightforward term is - 6x
So the correct quadratic trinomial is x ^ 2-6x + 9
If you misread the coefficient of the first term for the first time, you can see that the constant term is right. You can see that the constant term is 9,
If you read the constant term wrong for the second time, the first term is correct, and you can see that the first term is - 6
So the original formula: x ^ 2-6 * x + 9 = 0, the factorization factor is (x-3) ^ 2
(x-1) (X-9) = x ^ 2-10x + 9. If you misread the coefficient of the first term, it means that the second term is right
(X-2) (x-4) = x ^ 2-6x + 8
To sum up, the original title is x ^ 2-6x + 9 = (x-3) ^ 2
Let the quadratic trinomial be X & sup2; - (x1 + x2) x + x1x2
If the coefficient of the first term is wrong, it means that C / A is not wrong
That is, x1x2 = 1 × 9 = 9
Read the constant term wrong, indicating that - B / A is not read wrong
That is, X1 + x2 = 2 + 4 = 6
So this quadratic trinomial is X & sup2; - 6x + 9 = (x-3) & sup2;