Junior high school practical solution of unary equations, junior high school unary equation solution

1. Solution: 1) Set up a total of x potted flowers purchased by the school, and the number of flowers assigned to each class is equal.

5+(x-5) 10=10+[5+(x-5) 10-10] 10 solution: x=45

A: The school purchased a total of 45 potted flowers.

2) Take 5 + (45-5) 10 = 9 pots in one shift.

A: Each class is assigned 9 pots of flowers.

3) Because all classes take the same flowers, there are class digits.

45 9 = 5 classes.

A: There are 5 classes in this school.

2. Solution: The speed of the armor is xkm h

The sum of the velocities of A and B is: 18 9 5 = 10 km h, then: B's velocity is: (10 x) km h

Inspired by the title: 2 3 3 2) x 3 2 (10 x) 1813 6 x 15 3 2 x 18

2/3 x＝3

x A: A's velocity is, B's velocity is.

3. Set the specified time to x

15*(x-15 60)=9*(x+15 60)x=1, that is, the distance is: 15*[1-15 60]=kilometers.

A: The distance from home to school is kilometers.

4. If the speed of the slowest athlete is x kilometers, then the speed of the fastest athlete is kilometers per hour.

List equations: (

Solve the equation to obtain: x=30

The fastest athlete travels at 42 kilometres in 35 minutes.

The slowest athlete travels 30 kilometers in 35 minutes and travels in kilometers.

5, (1) 550t+250t=400, t= minutes.

A: A and B set off in opposite directions at the same time and in the same place, and they met again 1 2 minutes later.

2) 550t=250t+400, t=4 3 minutes.

A: A and B set off in the same place and direction at the same time, and they met 4 3 minutes later.

3) If B is 100 places ahead of A at the start, and A and B start at the same time and in the same place, ask how many minutes later will they meet for the second time?

The question is wrong, since B is 100 places in front of A, it is impossible for A and B to set off at the same time and place at the same time

550-250)t=100+400

t=5 3 A5 3 minutes later they met for the second time.

Hope it works for you, thanks.

The school has enough to get x pots of flowers.

5+(x-5)*1 10=15+-15}*1 10 gives x=855

.No problem with the question?

855 90 is not an integer.

Keep dividing until you finish taking it, and each class gets the same number of flower pots?

Agree with Anko Sakura's Haha to learn it.

1.Denominator: Multiply by the least common multiple of each denominator on both sides of the equation;

2.Remove braces: first go to the small braces, then the middle brackets, and finally the braces; (Remember to change the sign if there is a minus sign outside the parentheses).

3.Shift: Move all terms with unknowns to one side of the equation and all other terms to the other side of the equation; The move item needs to be renumbered.

4.Merge similar terms: Equations are formed in the form ax=b(a≠0);

5.Coefficients are reduced to 1: Divide the coefficient a by the unknown number on both sides of the equation to get the solution of the equation x=b a

Remove the denominator and then brackets, move items and merge similar items.

The coefficient of 1 is not good, and it is not in vain to be accurate.

Unary Linear Equation Judgment Method:

By simplification, the equation that contains only one unknown and the highest order term that contains the unknowns is one is called a one-dimensional equation.

To determine whether an equation is a univariate equation, first look at whether it is an integral equation. If so, sort it out. If it can be organized in the form ax+b=0(a≠0), then the equation is a univariate equation.

There should be an equal sign in it, and there should be no unknown number in the denominator.

There are five steps to solve a one-dimensional equation, i.e., denominating, deparentheses, shifting terms, merging similar terms, and coefficients to 1, all of which are performed according to the properties of integers and equations. Below I have compiled the detailed solutions and example problems of the junior high school unary equation for your reference.

1. About moving items:

Any term in the equation can be moved from one side of the equation to the other after changing the sign, that is, the term on the right side of the equation can be moved to the left side of the equation after changing the sign You can also change the sign of the term on the left side of the equation and move it to the right side of the equation A common mistake made in shifting terms is forgetting to change signs Also note that there is a fundamental difference between shifting terms and swapping the positions of two terms on one side of the equation If the position of the terms on the same side of the equal sign changes, these terms do not change the sign because the order in which a term is arranged in the polynomial is changed, It is a variation based on the commutative law of addition and the law of conjunction, but if certain terms are moved from one side of the equal sign to the other, they are subject to a change.

2 About denominator:

Removing the denominator is based on the property of the equation 2 on both sides of the equation each term is multiplied by the least common multiple of the denominator A common mistake is to omit multiplying the term that does not contain the denominator Another easy mistake is that the understanding of the fraction line is incomplete The fraction line has two meanings, on the one hand, it is a division sign, and on the other hand, it represents the parentheses, so when removing the denominator, the numerator should be enclosed in parentheses.

3.About removing parentheses:

The common mistake of removing parentheses is that the parentheses are preceded by a negative sign, and forgetting to change the number when removing parentheses; Multiply a number by a polynomial, and omit to multiply the next terms of the polynomial when removing parentheses.

4 Ideas for solving equations:

Solving a one-dimensional equation is actually a series of deformations of an equation using the properties of the equation to finally transform it into the form of mx=n, and then solving mx=n.

Idea Analysis].

The main thing is to make use of the deformation of the equation.

Problem solving process] There are two elements of the equation, one of which is indispensable

1) The equation must be an equation;

2) The equation must contain unknowns.

Therefore, it can be said that equations are special equations, and their particularity lies in the fact that they contain unknowns. It is also because of the unknowns that the equation is an undetermined equation; When an unknown number is taken to determine a value, the values on the left and right sides of the equation may or may not be equal. For example, when x=2, the left and right values of equation 5x-7=8 are not equal; When x=3, the left and right sides of equation 5x-7=8 are equal.

If the values on the left and right sides of the equation are equal when an unknown is taken to determine a certain value, the value of the unknown is called the solution of the equation. For example, 3 is the solution of the equation 5x-y=8, which is generally expressed by x=3

1) The value of the unknowns that make the left and right sides of the equation equal can be more than one, and the solution of the equation refers to the values of all these unknowns.

2) Conversely, if the solution of the equation is known to be a certain value of an unknown number, then the value of this unknown is substituted into the left and right sides of the equation, and the values on the left and right sides of the equation are equal, that is, the equation is a definite equation.

Equations can contain one or more unknowns. For an equation that contains only one unknown, its solution is also called a root. The concept of root is a new one.

This concept will be used later, for example, in the chapter "Unary Quadratic Equations" there is a formula for finding the root, the relationship between the root and the coefficient. The concept of roots is only for unary equations, and multivariate equations do not mention roots.

There are many ways to solve equations, for example, the solution of equations 5x-7 8 can be found in the same way as in elementary school, or you can use the method learned in Chapter 1. Regardless of the method, the process of finding the solution of an equation is called solving an equation. Solving an equation requires all the solutions of the equation.

Solving an equation is actually purposefully and progressively deforming the original equation to obtain the form x=a. These deformations should ensure that the equations obtained after deformation are the same as the original equation solutions, so that the final solution is the solution of the original equation. The deformations mentioned by the properties of the equation are guaranteed except that both sides of the equation are multiplied by 0, and these deformations are suitable for solving more complex equations, so that the solution of the unary equation can take advantage of the properties of the equation.

There are several ways to deform an equation. Negative negative becomes positive, positive positive becomes negative, and the number that does not contain unknowns is put aside as much as possible.

The first step is to simplify.

For example, solution: ax+dx+b+c=0 = (a+d)x =-b-c

The second step is to discuss the categories.

When a+d=0.

1) When -b-c=0, i.e. b=c.

The original equation has an infinite number of solutions.

2) When -b-c = 0, i.e. b = c ( = is not equal to).

There is no solution to the original equation.

When a+d=0.

The original equation is solved as x=(-c-b) (a+d).

The shift is b=2 = 0=2-b

The first step is to simplify.

For example, solution: ax+dx+b+c=0 = (a+d)x =-b-c

The second step is to discuss the categories.

When a+d=0.

1) When -b-c=0, i.e. b=c.

The original equation has an infinite number of solutions.

2) When -b-c = 0, i.e. b = c ( = is not equal to) the original equation has no solution.

When a+d=0.

The original equation is solved as x=(-c-b) (a+d).

The shift is b=2 = 0=2-b

Try to put aside numbers that don't contain unknowns.

1.Wheat x kg x - 28% x = 1802Set the air ticket** to be x yuan x+(25-20) 3If the number of junior high school students in Guangzhou is x0,000, the number of primary school students is (2x+14) million.

x+（2x+14）=128

1) Set x for white and 1 2 x+2 for black

x+1/2 x+2=32

3/2 x =30

x = 20

1/2 x+2 =12

20 white.

Black 12 pcs.

2) Set up a total of x potted flowers.

The first class is 5+1 10 (x-5).

The second shift is 10+1 10 [x-5-1 10(x-5)] It seems that this can be calculated... I don't have a pen and paper. I'm sorry for the number of flower pots, but I'll ask for the rest.

3) Let the velocity of A be x km h and the velocity of B be -y km h, 8y=18 x+y=10

13 6 x+3 2 y =18 13 x+9 y=108 gives x=y=

(1) If black x is set to white, then white is 32-x

x=(32-x) 2+2, move the term to get 3 2 x=16+2 x=12，32-x=20.

2) 1) Set up a total of x pots of flowers, and take away 5+(x-5) 10=

The second class takes away 15+(x-(5+(x-5) 10)) 10 = 15+(the two classes are equal to know, and the solution is obtained.)

x=1005。

2) Take 5+ pots in one shift, 3) Set the speed of A x meters, and the speed of B y meters has x (60+48)+y (60+48)=18000, x+y=1800 108=

A departs 40 minutes earlier than B, x (40+60+30)+y (60+30)=18000 40x+90×（x+y）=1800.

40x+1500=1800. x=30/4=y=16+2/3-7-1/2=9+1/6

(1) Equipped with x white leather blocks and (32-x) black leather blocks.

32-x=（1/2）x+2

x=20 32-20=12

A: 20 pieces of white leather and 12 pieces of black leather.

2) The school purchased x potted flowers.

x-5)×1/10+5=[(x-5)×9/10-15]×1/10+15

x=855 Number of pots of flowers distributed in each shift: (855-5) 1 10+5=90

Number of classes at this school: 855 90=

3) Set A xkm h B ykm h

1 hour 48 minutes = hours 1 hour 30 minutes = hours 40 minutes = 2 3 hours.

18/(x+y)= [18-(2/3)x]/(x+y)=

(1) Solution: Let the number of black skins be x, and the number of white skins be (32-x), according to the title

32-x 2+2=x solution: x=12 black leather digit 12The number of white leather pieces is 32-12=20 pieces (2), the meaning of which is not clear!

5 pots were taken away in one shift; The third class took 15 pots, and each class took the same number of flower pots. What a thing.

The total number of working hours required for the work is 80h

Assuming that the number of "some" originally arranged is x, then there is:

2x+(5+x)*8=80*

Solution: x=2

That is, 2 people do 2 hours first, and then 5 people do 8 hours, and 3 4 complete this work.