The whole calculation process of 5x + 3 = 6x-12

The whole calculation process of 5x + 3 = 6x-12

Move x (i.e. primary term) to one side and constant to one side
3+12=6x-5x
The solution is as follows
x=15
6x-5x=-15
x=-15
Isn't that 15?
How do you get all - 15???
depressed
6x-5x=3+12
x=15
Despise, so simple all ask for help! Strong disdain!
5X+3=6X-12
5X-6X=-12-3
-X=-15
X=15
5x-6x=-12-3
That is - x = 15
So x = - 15
On the system of equations 3x-y = P + 1, x + y = 3 of XY, if the solution of the system of equations satisfies x > y, find the value range of P
3x-y=p+1,x+y=3,
——》x=1+p/4,y=2-p/4,
x>y,——》1+p/4>2-p/4,——》p>2,
x-y=1+p/4-(2-p/4)=p/2-1,
Four
On the system of equations 3x-y = P + 1, x + y = 3 of XY, if the solution of the system of equations satisfies
Why do I always encounter such problems today
9x & # 178; + 16-24x factorization,
9x²+16-24x
=9x²-24x+16
=(3x)²-2×(3x)×4+4²
=(3x-4) &# - 178; - - - - - (according to the complete square formula)
——————————If you don't understand, you can ask,
How to calculate 5x + 6x = 66
5x+6x=66
（5+6）x=66
11x=66
x=66÷11
X=6
11x = 66, x = 6. In fact, I'm in Grade 7. I'm working on a task
The solution of the system {x + y = 7 {xy = 10} is_____
From (1)
x=7-y (3)
Substituting (3) into (2) leads to
(7-y)y=10
y²-7y+10=0
∴y=5 y=2
Substituting y = 5 and y = 2 into (3) respectively leads to
x=2 x=5
∴x=2 x=5
y=5 y=2
Because {x + y = 7 {xy = 10, so x = 10 / y, substituting {x + y = 7 to get y + 10 / y = 7, so {x = 5, y = 2
(X & # 178; - X-2) (X & # 178; -- X-6) + 16 factorization
Factorization of X & # 178; - 10xy - + 25y & # 178; - 6x + 30y + 8
Classmate, if there is only one - sign before X in the second formula, then the original formula = (X-2) (x + 1) (x-3) (x + 2) + 16
(x2-x-2)(x2-x+6)+16=[(x2-x)-2][(x2-x)+6]+16=(x2-x)2+6(x2-x)-2(x2-x)-12+16=(x2-x)2+4(x2-x)+4=[(x2-x)+2]2=(x-2)2(x+1)2
The original formula = x & # 178; - 10xy + 25y & # 178; - 6x + 30y + 8 = (x-5y) & # 178; - 6 (x-5y) + 8 (regarding x-5y as a whole, the final result can be obtained by cross multiplication) = (x-5y-2) (x-5y-4)
Solution equation: 5x (6x + 1) - (5x + 1) (6x-1) = 7 (X-2)
5x(6x+1)-(5x+1)(6x-1)=7(x-2)
30x²+5x-30x²-x+1=7x-14
4x+1=7x-14
3x=15
X=5
X = 5: can we not imitate other people's answers!
We know the system of equations ax-y = a, X-Y = 1 about XY
We know the system of equations about XY
｛ax-y=a
x-y=1
1. When a ≠ 1, the equations are solved;
2 when a = 1, what is the solution of the equations?
3 if a = 1, what is the solution of the equation {ax-y = a?
x-y=2
Given that the equations {5x + y = 3 and {x-2y = 5 have the same solution, find the value of a and B
ax+5y=4 5x+by=1
When a is not equal to 1, y = X-1 is transferred to ax-y = a to get ax-x + 1 = a, that is, (A-1) x = A-1. Because a is not equal to 1, x = A-1 / A-1 = 1 is replaced by y = X-1 to get y = 0. When a = 1, A-1 = 0, A-1 / A-1 denominator is 0, so the equations have no solution
A
1. X is not equal to 1 or 0
2. Not established
Factorization: x4-6x & # 178; - 7x-6
Factorization: X fourth power - 6x & # - 178; - 7x-6
x^4-6x²-7x-6=x²(x²-4)-2x²-7x-6 =x²(x+2)(x-2)-(2x+3)(x+2) =(x+2)[x²(x-2)-(2x+3)] =(x+2)(x³-2x²-2x-3) =(x+2)[x²(x-3)+x²-2x-3] =(x+2)[x²(x-3)+(x+1)(...
Solve the equation 5x + 17 = 6x + 7
X=10