（4x+2y）*3=（2x+5y）*2 x=?

（4x+2y）*3=（2x+5y）*2 x=?

Split, shift, merge, congener, reduce
X=Y/2
I don't know if it's the answer you want
How to find an equation and two unknowns? There must be some specific conditions

Use the plus sign, minus sign, multiply sign and bracket to form a formula for the following groups of numbers, so that the results are 21, 1, 15, 1.2 and 7

15×7×（1.2-1）=21

What is another pronunciation of "Ji"? What words can be formed?

The word "Ji" has one pronunciation: Ji, two tones. It refers to: 1. Books, books. 2. Affiliation of registration. The other "Ji" has two pronunciation: one, Ji, two tones. It refers to: 1. Trampling, insulting. It can be grouped into: dishonor; cup dishonor; reputation dishonor. 2. Surname. Two, Jie, four tones. It refers to: 1. Something under the cushion; 2. Cushion; 3. Pretense

A formula for finding the coordinates of a point with respect to a line

Set the coordinates (a, b) of the point to be solved. According to the set points (a, b) and known points (C, d), the coordinates of the symmetric point (a + C / 2, B + D / 2) can be expressed, and the symmetric point is on a straight line. So substituting this point into a straight line, a, B, the coordinates of the point to be solved can be obtained

1000 written math problems in Volume 2 of grade 3

1.4－0.5= 10÷5= 7.2－3.6= 80×2=4.4＋2.1= 30÷5= 15×3= 20×3=2.3＋0.6= 1.8＋0.7= 13×20= 59×0=90×4= 80÷4= 1.7－0.9= 2.8＋2.7=30×5= 620－420= 707÷7= 80-25=5×80= 20×40= 60×20 50×3=(40÷8)= 20...

The number of real roots of the equation (x ^ 2-1) ^ 2 - | x ^ 2-1 | + k = 0 about X is discussed

Let | X & sup2; - 1 | = y, y ≥ 0; therefore, the original formula becomes Y & sup2; - y + k = 0
(1) When k > 1 / 4, the whole equation has no solution
(2) When k = 1 / 4, the equation has two equal solutions, that is, y = 1 / 2
So x & sup2; - 1 = 1 / 2 or X & sup2; - 1 = - 1 / 2
So the equation has eight roots (four pairs of equal roots)
(3) When k = 0, that is y = 1 or y = 0 = = = > X & sup2; - 1 = 1; X & sup2; - 1 = - 1 and X & sup2; - 1 = 0
That is, x = ± 2, x = 0 and x = ± 1
So the equation has five roots
(4) . when 0

The rectangle ABCD is inscribed in the ellipse x ^ 2 + 4Y ^ 2 = 4. The length of the rectangle will make it the largest area

The elliptic equation is changed into the standard form
X^2/2^2+y^2=1
This is an ellipse with the major axis 2 on the X axis and the minor axis 1 on the Y axis
Let a point m (x, y) on an ellipse have
Y = radical (1-x ^ 2 / 4)
Take M as the vertex, X as one side, y as the other side, and s as the rectangle area
S = xy = x radical (1-x ^ 2 / 4) = radical (x ^ 2-x ^ 4 / 4)
The derivation of s leads to (2x-x ^ 3) / 2 radical (x ^ - x ^ 4 / 4)
Make it 0
Then there are two solutions: x = 0: x = radical 2
This rectangle has the largest area when x = root 2
Here y = root 2 / 2
This is a quarter of an ellipse inscribed with a rectangle
Therefore, when the length of the inscribed rectangle is 2 radical 2 (x direction) and the width is radical 2 (Y direction), the rectangle has a maximum area of 4

100 binary linear equations exercises and answers should be simple
66x+17y=3967
25x+y=1200
X = 48, y = 47, don't give it to me
1）x-y=2
Xy = 15 (x = 5, y = 3) something like this
Well, I'll give you 100

x-y=2
Xy = 15 (x = 5, y = 3) something like this
There are two solutions to this problem
The other solution is x = - 3, y = - 5

Given that the function y = f (x) is a decreasing function on [- 1,4], and f (B + 1) > F (2b-3), the value range of real number B is obtained
Online, etc!!!!!

Find the major axis length, eccentricity, focus and vertex coordinates of ellipse 25X ^ 2 + 16y ^ 2 = 400

Major axis 2A = 10, eccentricity e = 3 / 5, focus F1 (0, - 3), F2 (0,3), vertex (0,5), (0, - 5), (4,0), (- 4,0)