Ask some equation questions, equation questions?

Question 1 3 (x-60)-20=x

x=100 Question 2 5x-8066=x

x=3 3x=y

y-22=z

x-6=zz=2 x=8

y=24

For brevity, write a1 as a

By Eq. Q=(6-A) a

Substitution and finishing:

a(6-a)=(6-a)²/a

a²=6-a

a+3)(a-2)=0

a=-3, or a=2

q=[6-(-3)] 3)=-3 or.

q=(6-2)/2=2

It's OK. <>

<>q=-2 detailed steps are in the figure above, I hope it will be helpful to you.

- Satisfied with the ---

q^2 + 4q + 4 = 0

q + 2)^2 = 0

q + 2 = 0q = 2

With the cross multiplication, the result is the same.

q^2+4q+4=0

q + 2)(q + 2) =0

q + 2 = 0

q = 2

The original formula is equal to (x-4)(x+2)>0

Solve x<-2 or x>4

If you want to multiply two parentheses greater than zero, they must each be positive or negative.

This should be done: -2(2-x)*(x-1)=-2(2-x) to solve the equation, x=2, or x=1,

If you understand a little bit of math official website, you will know what is going on.

Solve the lack of filial piety in two steps. Let's start with cos as a variable.

First, solve a quadratic equation to examine the limbs. In the manuscript of the topic, the cross multiplication method is used. After finding the cos value. To solve is how much he is for.

Solution: Suppose there are students in the second year of junior high school to complete the remaining parts, which will take a total of x hours to complete.

2+3+3x=15

3x=10x=10/3

Answer: Zheng Shisheng in the second year of junior high school needs to complete the remaining part 10 3 when shouting pins.

I hope the above will be useful to you.

The elimination method solves the limb group, and represents the a2 in the first formula as 15 - a3, and brings it into the second formula. i.e. (15 - a3)a3 = 54, and then solve a3 = 6....

The root 2 equation is eliminated on both sides, p square = 4, p = plus or minus 2