The solution of the system of ternary quadratic equations, the solution of the system of ternary qua

Guo Dunyun: The original system of equations is.

y =x +1 (1)36+x =r (2)(r y) 2=5 y (3) is obtained by (1), x =y 1 (4), substituting (2) obtains, y +35=r (5), substituting (3) obtains, y y+35) 2=5 y, y(y y+35)=10,y3

y²+35y－10=0 （6）

Try-step approximation method to solve the unary cubic equation of equation (6) - when y=, y3

y²+35y－10=；

When y=, y3

y²+35y－10=；

When y=, y3

y²+35y－10=；

When y=, y3

y²+35y－10=

When y=, y3

y²+35y－10=

Substituting y= into (4), x = y 1= , x= imaginary root) substituting y= into (5), r = y +35=

The solution of the equation is: x= imaginary root), y=, r=, and it is tested to be correct.

ry=10+y^2

y^2=x^2+1

36+x^2=r^2

ry=10+y^2

ry=x^2+11

36+x^2=r^2

ry=10+y^2

36+ry=r^2+11

r-y)y=10

35+y^2=r^2

r-y)y=10

35=(r-y)(r+y)

r+y)/y=7/2

r/y=5/2

2r=5y2ry=20+2y^2

2r=5y2ry=20+2y^2

5y^2=20+2y^2

y = positive (2 3) times root number 15

r = positive (5 3) times root number 15

x = plus or minus (1 3) times the root number 51

Or. y = minus (2 3) times root number 15

r = minus (5 3) times root number 15

x = plus or minus (1 3) times the root number 51

There is a problem with the first two equations, are you looking at the problem correctly? If this is the equation, there is no solution.

The solution of the system of ternary quadratic equations is as follows:

The solution of the ternary quadratic equation system is the substitution elimination method, and its basic methods are the substitution method and addition and subtraction.

1. Formula: Carry out a ternary formula, so that two of the unknowns are parameters, and the remaining one is formulated like a one-dimensional quadratic equation.

2. Elimination: merge similar terms and merge the coefficient into one.

Specific steps: 1. Use the substitution method or addition and subtraction method to eliminate an unknown number and obtain a binary system of equations.

2. Solve this system of binary equations and find the values of two unknowns.

3. Substitute the stupid values of these two unknown stall pairs into the simpler equation in the original equation, find the value of the third unknown, and put these three.

The numbers written together are the solutions to the system of ternary equations that are sought.

When solving a system of equations, we follow four steps: one look, two variations, three matches, and four solutions.

One look: that is, to observe the coefficients of each unknown number in the equation system, whether there is a 1 or 1, and whether there is a relationship between multiples of each other; It is easy to solve after it is determined.

Match-3: From ternary to binary, and then to unelemental, find the value of an unknown number; i.e. the process of 3-2-1.

Four solutions: The process of bringing in the value of one unknown number and finding the value of the other two unknowns respectively, that is, 1-2-3.

The solution of the system of ternary quadratic equations is as follows:The main solution method is the addition and subtraction elimination method and the substitution elimination method, usually using the addition and subtraction elimination method, if the equation is rough and difficult to solve, use the substitution elimination method, which varies from problem to problem.

1. Use the substitution method or the stool beat subtraction method to eliminate an unknown number and obtain a binary system of equations.

2. Solve this system of binary equations and find the values of two unknowns.

3. Substituting the values of these two unknowns into one of the simpler equations in the original equation, finding the value of the third unknown, and writing these three numbers together is the solution of the ternary linear equation system found.

a+b=5ab

a+c=6ac

b+c=7bc

a+b=5ab, a+b-5ab=0, a(1-5b)=-b, a=b/(5b-1)

b+c=7bc, 7bc-c=b, c(7b-1)=b, c=b/(7b-1)

Substitute a=b (5b-1), c=b (7b-1) with: a+c=6ac, b (5b-1)+b (7b-1)=6[b (5b-1)]*b (7b-1)].

1/(5b-1)+1/(7b-1)=6b/[(5b-1)(7b-1)]

12b-2=6b

6b=2b=1/3

a=b/(5b-1)=(1/3)/(5/3-1)=(1/3)/(2/3)=1/2

c=b/(7b-1)=(1/3)/(7/3-1)=(1/3)/(4/3)=1/4

Problem solving ideas: a, b, c three unknowns, two of the unknowns with the remaining one unknown to express, substitute into the original test question, it becomes an unknown equation to solve.

From 1: a(1+b)(1-b+b2)=18 from 2:ab(1+b)=12

Divide the two formulas to yield: (1-b+b 2) b=3 22-2b+2b 2=3b

2b^2-5b+2=0

2b-1)(b-2)=0

b=1/2,b=2

Replace b = 1 2 into 1.

a+1/8a=18

A = 16 and b = 2 into 1.

a+8a=18

a=2 so the solution of the system of equations is.

a=16,b=1/2

or {a=2, b=2.}

Both 1 and 2 have a common factor a(b+1) on the left

That is, (1) a(b+1)(b 2-b+1)=18(2) a(b+1)b=12

It is clear that both A and B+1 are not zero.

So (1) and (2) are divided left and right, (b 2-b + 1) b = 18 12

This is a quadratic equation with solutions of 1 2 and 2

Subscribing (1) yields a = 16 and 2

Its solution is (a,b) = (16,1 2) and (2,2).

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If i is a known number, three unknowns and four equations are too many, and they will usually be unsolvable.

Considering only the first 3 equations, it can be solved like this:

Looking at f1-f2, we can see that the quadratic term x +y +z of the unknown number is just eliminated.

That is, we get a one-time equation for x, y, z.

In the same way, f1-f3 gives another one-time equation for x, y, z.

Simultaneous two linear equations, in general, can eliminate one unknown to represent the other two.

Something like y = ax+b, z = cx+d.

Substituting it into f1 to get a quadratic equation about x, solve it (there may be two real roots), and then calculate the corresponding y, z on behalf of it.

If i is also an unknown number, and r1(i) is the meaning of r1·i, then it can also be solved.

Firstly, the i term is eliminated by f1·r2-f2·r1, f1·r3-f3·r1, f1·r4-f4·r1, and the three-element quadratic equations about x, y, and z are obtained.

Noting that the quadratic terms of the three equations are (r2-r1) (x +y +z), r3-r1) (x +y +z), r4-r1) (x +y +z), and that the quadratic terms can still be eliminated with appropriate elimination, and the two quadratic equations for x, y, z can still be obtained by eliminating the quadratic terms.

After that, do the same as above, solve x, y, z and finally replace f1 to find i.

If i is an unknown and r1(i) is a function about i, then it depends on the form of the function.

If the function is complex, it is very likely that it will not be solved.