Solving the equation of X △ 40 = (x-40) △ 24-3

Solving the equation of X △ 40 = (x-40) △ 24-3

x÷40＝（x－40)÷24－3
Multiply by 120 at the same time
3x=5（x－40)-360
3x=5x-200-360
560=2x
x=280

The solution equation: 4 of X + (10 of X + 24) = 14 of X + 8

About an X on both sides
The original equation is equivalent to 4 / x + 10 / 24 = 14 / 8
4/X=32/24
8x=24 X=3

14 / 25 △ x is equal to 7 / 10. What is x by solving the equation

X = four fifths
14 out of 25 is equal to 7 out of 10
Remove 14 out of 25 times 7 out of 10 = x
14 out of 25 * 10 out of 7 = x
X = four fifths
Like it, too

Equation: 14x + 5-2x = 125
How to do it?

14x+5-2x=125
14x-2x=125-5
12x=120
x=120/12
x=10

Let a be a set of roots of equation x ^ 3-7x ^ 2 + 14x-8 = 0, and B be a set of roots of equation x ^ 3 + 2x ^ 2-xc ^ 2-2c ^ 2 = 0, where C ≥ 0
Now take the elements of set a and B as the two roots of the quadratic equation x ^ 2 + PX + q = 0, and note that the minimum value of F (x) = x ^ 2 + PX + Q is m. find the maximum value of M and the minimum value of M,

X ^ 3-7x ^ 2 + 14x-8 = (x ^ 3-8) - (7x ^ 2-14x) = (X-2) (x ^ 2 + 2x + 4) - 7 (x - 2) = (X-2) (x ^ 2 + 2x-3) = (X-2) (x-3) (x + 1) = 0: X1 = 2, X2 = 3, X3 = - 1, a = {2,3, - 1} x ^ 3 + 2x ^ 2-xc ^ 2-2c ^ 2 = x ^ 2 (x + 2) - C ^ 2 (x + 2) = (x + 2) (x ^ 2-C ^ 2) = (x + 2) (x + C) (x-C)

Solve the equation 2 / x + 3 / (12-x) = 20%

2/x－3/﹙x－12﹚＝1/5
x²－7x＋120＝0
b²－4ac＝49－480＝﹣431＜0
The equation has no real solution

How to solve the following equation? Please use two methods to connect the following equation: 24 △ x = 20
This is the fifth question on page 7 of three exercises in Lesson 1 of Su Education Press!
Fifth grade
What's the meaning of '/'?

The first one is the first one
Transfer: x = 24 △ 20 = 24 / 20 = 6 / 5
Second:
Divisor: 24 △ x = 20
24÷X÷4=20÷4
24÷4÷X=5
6÷X=5
X=6/5

Given that x = 2 is a solution of the equation 3x + a = 0 about X, then the value of a is ()
A. -6B. -3C. -4D. -5

Substituting x = 2 into the equation: 6 + a = 0, the solution: a = - 6

Collocation method of quadratic equation with one variable x (5x + 4) = 5x + 4

(5x+4)(5x+4-x)=0
(5x+4)(4x+4)=0
x1=-4/5,x2=-1

Estimation method of quadratic equation with one variable
Ax ^ 2 + BX + C "0 and ax ^ 2 + BX + C" 0 how to find these two numbers

To be constant or greater than or equal to 0, the opening of the parabola must be upward, that is, a > 0. Moreover, there can only be one or no intersection point between the parabola and the x-axis, so B ^ 2-4ac