The sum of solutions of the system of quadratic equations x + y = 2x-8 is equal

The sum of solutions of the system of quadratic equations x + y = 2x-8 is equal

That is, x = y is substituted by x = y
So - y = 3, y = - 3
x=y=-3
So - 6-3k = 8
-3k=14
k=-14/3
Is that right
I don't know how to ask
It is known that the factorization result of the quadratic trinomial 2x2 + MX + n is (2x − 1) (x + 14), and the values of M and N are obtained
(2x − 1) (x + 14) = 2x2 + 12x-x-14 = 2x2-12x-14. Then M = - 12, n = - 14
The factorization factor (6x & # 178; - 3x) - 2 (7x-5) is used,
(6x²-3x)-2(7x-5)
=6x²-3x－14x＋10
=6x²－17x＋10
=(X-2) (6x-5) (cross multiplication)
x ﹣2
X
6x ﹣5
（6x²-3x)-2(7x-5)
=6x²-3x-14x+10
=6x²-17x+10
=(x-2)(6x-5)
Bivariate linear equation, 7x + 2Y equals 49
Positive integer solution
7X+2Y=49
X=1,Y=21
X=3,Y=14
X=5,Y=7
7X+2Y=49
X=1,Y=21
X=3,Y=14
X=5,Y=7
x=1 y=21
On the quadratic trinomial of X MX & # 178; - (2m-1) x + m + 1 when m is a value, the quadratic trinomial can be factorized in the range of real number. When m is a value, it can be factorized into a complete square. What is the complete square?
The quadratic trinomial MX & # 178; - (2m-1) x + m + 1 of X
∴m≠0
In the range of real number, energy factorization → △ = (2m-1) ^ 2-4m (M + 1) = - 8m + 1 ≥ 0
M ≤ 1 / 8 and m ≠ 0
The energy factor is decomposed into a complete square formula → △ = - 8m + 1 = 0
m=1/8 fgndkjguiofghdiodrjgjfiodgjiro
1) M is not equal to 0 and deta
Decomposition factor: 2.6x & # 178; + 7x + 2 = 3.4m & # 178; - 12m + 9 = 4.5 + 7x-6x & # 178; = 5.12x & # 178; + xy-6y & # 178=
6x²+7x+2
=（3x+2）（2x+1）
4m²-12m+9
=(2m-3)²
5+7x-6x²
=-(6x²-7x-5)
=-(3x-5)(2x+1)
12x²+xy-6y²
=(3x-2y)(4x+3y)
6x²+7x+2= (2x+1)(3x+2)
4m²-12m+9=(2m-3)^2
5+7x-6x²= (5-3x)(1+2x)
12x²+xy-6y²=(3x-2y)(4x+3y)
Given the solution of the system of equations {7x + 3Y = 4,5x-2y = 7, the equation 12x + y = 7 holds and the value of M is obtained
Given the solution of the system of equations {7x + 3Y = 4,5x-2y = M-1, the equation 12x + y = 7 holds and the value of M is obtained
The answer is 4
Three unknowns and three equations can be solved
ASA
What about M?
From 7x + 3Y = 4
And 12x + y = 7
① - 2
2y=5x-3 ③
From 5x-2y = M-1
M = 5x-2y + 1, substituting 3, we get
m=5x-(5x-3)+1=4
Find out the integer value a that can factorize the quadratic trinomial x2 + ax-6 (in the range of integers), and factorize it
x2+x-6=（x+3）（x-2）；x2-x-6=（x-3）（x+2）；x2+5x-6=（x+6）（x-1）；x2-5x-6=（x-6）（x+1）．
Find the range of function y = √ X & # 178; + 6x + 10
√(x²+6x+9+1)
=√[(x+3)²+1]
>=1
The range of the function y = √ X & # 178; + 6x + 10 is [1, + ∞)
3x + 4Y = 5 -- 7x = 9y = 2 / 5 substitution method
3x + 4Y = 5 -- 7x + 9y = 2 / 5 substitution method
3x+4y=5
7x+9y=2/5
3x+4y=5
3x=5-4y
x=(5-4y)/3
7x+9y2=2/5
7(5-4y)/3+9y=2/5
Multiply 15 on both sides
35(5-4y)+135y=6
175-140y+135y=6
-5y=6-175
5y=169
y=169/5
x=(5-4y)/3
x=(5-169*4/5)/3
x=-217/5