Help solve two equations 5, help solve two equations

16-2x)(12-2x) 1 2·16·12 Simplification: (8-x)(6-x)=24

x²-14x+24=0

x-2)(x-12)=0

x1=2,x2=12

1/x+1/(x-30)=1/20

Both sides of the equation are multiplied by 20x (x-30) at the same time

20x-600+20x=x²-30x

x²-70x+600=0

x-10)(x-60)=0

x1=10,x2=60

After testing, all of them conform to the meaning of the topic, and they are all solutions to the original equations.

1) 1+2: 5x+2z=3

1-3 gets: 2z-2x=0

From above, we get: z=x=3 7

Substituting 1 yields: y=-6 7

The solution of the system of equations is: x=3 7, y=-6 7, z=3 72)1-2+3 yields: -c=0, i.e.: c=0, so c can be omitted.

Equation 2 - Equation 3*3: 7b-7d = -28, that is: b = d - 4 after substituting 1: can obtain:

2a-3(d-4)+5d=6, that is: 2a+2d+12=6, i.e., a+d=-3, a=-(d+3).

Substituting 3 yields: -2(d+3)-3(d-4)+5d=5, obtains: d=arbitrary number, a=-(d+3), b=d-4, c=0

Note: The 1+2, 1-3, and so on in front of the word are the equation numbers of your equations from top to bottom.

2x-y+3z=3 ……1）

3x+y-z=0 ……2）

4x-y+z=3 ……3）

x+3y-13z=-6………4）

2x-(y-z)+2z=3 ……5）

3x+(y-z)=0 ……6）

4x-(y-z)=3 ……7）

x+3(y-z)-10z=-6………8）

From (6), (7):

3x+(y-z)=0

4x-(y-z)=3

Additive: x=3 7

So. (y-z)+2z=15/7 ……9）y-z=-9/7 ……10）

3(y-z)-10z=-45/7 ……11) 10) substitution (9), (11):

2z=24/7

y-z=-9/7

10z=18/7

z=12/7

y-z=-9/7

z=9/35

Contradiction. No solution.

y-z=-9/7

1) becomes. 2x-(y-z)+2z=3

2*3/7+9/7+2z=3

z=3/7

The title is wrong! The first question has 3 unknowns and 4 equations, and the second problem has 4 unknowns and 3 equations???

Categories: Education, Science, >> Learning Aid.

Problem description: 1. A plane flies between two cities, and the plane flies 552 kilometers per hour when there is no wind, and in a round-trip flight, the plane flies downwind for hours, and flies against the wind for 6 hours, and the wind speed of this flight is found.

2. When the steamer sails between two wharves, it takes 4 hours to sail along the water, 5 hours to sail against the current, and when the speed of the water is 2 kilometers, and the speed of the steamer in still water is x kilometers, then the equation can be solved to obtain the distance between the two wharves of x

Resolution: 552+x)*

x=24km/h

4*(x+2)=5*(x-2)

x=18km/h

s=4*(18+2)=80km

Let's find a convenient solution, use the algebraic calculator in the Yili Zhishi software on your mobile phone to solve it, and the result is as follows:

The detailed solution process is available in the "Algebra Problem Solving Steps", which is too long to take screenshots here.

Substituting the value of y into the binary quadratic equation of x and y is sufficient.

Please feel free to use it, but if you have any questions, please feel free to ask.

Your adoption will be the strongest motivation for me to continue to work hard to help others!

Square on both sides.

1）3/(m^2+3)=4×39/13×131/(m^2+3)=4/13

4(m^2+3)=13

4m^2+12=13

4m^2=1

m^2=1/4

m=±1/2

2）m^2/(m^2+3)=2m^2/161/(m^2+3)=1/8

m^2+3=8

m^2=5m=±√5

If you don't understand, please ask, and if you understand, please adopt it in time! (*Thank you.)

Solution x -3x-6 = 0

Use the formula method. =b²-4ac=9+24=33

x=(3+√33)/2

or x=(3-33)2

2x-1)²-64=0

2x-1)²=64

2x-1=8 or 2x-1=-8

x = 9 2 or x = -7 2

<> is like a trembling stool, a blind eggplant, and an empty picture is coarse.

This is summarized in the chapter "Systems of Linear Equations"! (Hehe, have you read the book?) 1) Homogeneous (linear branches) equations must have zero solutions;

2) The system of homogeneous linear equations has a non-zero solution of the sufficient and important clauses, and the determinant of the coefficient matrix is zero;

3) In general, the sufficient and necessary condition for the solution of the system of fierce linear equations is that the rank of the coefficient matrix is equal to the rank of the augmented matrix;

4) a system of non-secondary linear equations whose number of equations is equal to an unknown number of equations, and when the determinant of the coefficient matrix is not equal to zero, the system of equations has a unique solution; When the determinant of the coefficient matrix is equal to zero, but the rank of the augmentation matrix is not equal to the rank of the coefficient matrix, the system of equations has no solution; When the coefficient determinant is zero and the coefficient matrix is of equal rank to the augmentation matrix, the system of equations has an infinite number of solutions.