1. The concept of one-dimensional linear equation;

An equation that contains only one unknown, the degree of which is 1, and the coefficient is not 0 is called a linear equation with one variable.

The standard form of a linear equation with one variable is ax+b=0 (where x is unknown, a and b are known numbers, and a≠0), and its solution is x=-.

We judge whether an equation is a linear equation by whether the simplified simplest form of the equation is the standard form ax+b=0 (a≠0). For example, the equation 3x2+5=8x+3x2 is a linear equation if it is reduced to 8x-5=0; Equation 4x-7=3x-7+x There is an unknown number x on the surface, and the degree of x is once, but it is 0x=0 after simplification, which is not a linear equation.

2. The general steps of solving linear equations:

When the (1) equation contains a denominator, the denominator should be removed first to simplify the process. The specific method is to multiply both sides of the equation by the least common multiple of each denominator. Be careful not to miss items without denominator, such as equation x+ =3. It is wrong to get 10x+3=3 without a denominator, because the right side of the equation forgets to multiply by 6, which leads to an error.

(2) bracket removal: according to the rules of bracket removal, the bracket is removed first, then the bracket is removed, and finally the braces are removed. Pay special attention to the minus sign before brackets, remove the minus sign and brackets, and change the symbols of everything in brackets. Pay attention to the distribution law when there are numerical factors before parentheses.

(3) Shift the term: move all the terms containing unknowns to one side of the equation, and other terms to the other side of the equation. Pay attention to the change of signs when moving items.

(4) Merging term: the equation is simplified to the simplest form ax=b (a≠0).

(5) Convert the coefficient of the unknown quantity into 1: divide the coefficient a of the unknown quantity on both sides of the equation to get the solution x=.

Some of the above steps may not be needed when solving the equation, or they may not be in the above order. The solution steps should be flexibly arranged according to the specific form of the equation.

(2) Example:

Example 1. Solve the equation (x-5)=3- (x-5)

Analysis: According to the law, the denominator and brackets should be removed from this equation first, but it is found that the left and right sides of the equation contain x-5 terms, so they can be regarded as a whole and shifted and merged to make the operation simple.

Solution: (x-5)+ (x-5)=3.

Total: x-5=3

∴ x=8 .

Example 2. Solve equation 2x- =-

Solution: Because the equation contains a denominator, the denominator should be removed first.

Denominator:12x-3 (x+1) = 8-2 (x+2) (note that each item should be multiplied by 6).

Bracket removal: 12x-3x-3=8-2x-4 (pay attention to the law of distribution and bracket removal).

Transfer item: 12x-3x+2x=8-4+3

Merge: 1 1x=7.

The coefficient becomes 1: x =.

Example 3. {[(+4)+6]+8} = 1

Solution 1: gradually remove the brackets from outside to inside and expand the solution:

Without braces: [(+4)+6]+8=9.

Without brackets: (+4)+6+56=63.

Finishing: (+4)= 1

Without brackets: +4=5

Remove the denominator: x+2+ 12= 15.

Move this item, the result is: x= 1.

Solution 2: gradually remove the brackets from the inside out and expand the solution:

Without brackets: {[(+6]+8} =1.

Without brackets: {++8} = 1

Without braces: ++= 1

Remove the denominator: x+2+3×4+2×45+8× 105=945.

Namely: x+2+ 12+90+840=945.

Transfer and merger: ∴x= 1.

Note: As can be seen from the above two solutions, it is not necessary to strictly follow the steps mentioned above to solve linear equations, and the methods can be used flexibly according to specific topics.

Example 4. Solve the equation [(- 1)-2]-2x=3.

Analysis: This equation contains parentheses. Because × = 1, it is easy to remove the brackets first.

Solution: Remove the brackets: (-1)-2x = 3.

Delete the brackets:-1-2x = 3 = 3.

Denominator: 5x-20-24-40x=60.

Transfer item: 5x-40x=60+44

Combined project: -35x= 104

The coefficient is converted into 1: x=-.

Example 5. Solve the equation -= 0

Analysis: Both the numerator and denominator of this equation contain decimals. If the denominator is removed directly, the operation will be complicated. However, if we use the property of fraction, that is, the denominator of the numerator is the same as the value of the fraction multiplied by a number not equal to zero, and make the first two terms on the left side of the equation have decimal Huasong integers, we can simplify the operation.

Solution: Using the properties of fractions (that is, the first numerator and denominator on the left are multiplied by 10, and the second numerator and denominator are multiplied by 100), the original equation can be transformed into:

- - =0

Denominator: 6 (4x+9)-10 (3-2x)-15 (x-5) = 0.

Unpacking: 24x+54-30+20x- 15x+75=0.

Transfer items: 24x+20x- 15x=-54+30-75.

Total: 29x=-99

The coefficient becomes 1: x =-.

Example 6. In the formula S= (a+b)h, it is known that a=5, S=44 and h=8, and the value of b is found.

Analysis: This is a trapezoidal area formula. If we know the values of any three of the four quantities S, A, B and H, we can find the value of the fourth quantity.

Solution 1: Substitute A = 5, S = 44 and H = 8 into the formula.

44= (5+b)×8 This is a linear equation about B.

Simplification: b+5= 1 1

Move this item, and the combined result is: b=6.

Scheme 2: First, consider B as an unknown number, all other quantities as known numbers, deform the formula, use the other three quantities to represent B, and then substitute the values of known numbers to get B. ..

S= (a+b)h

Denominator: 2S=(a+b)h

Not wearing a seat belt: 2S=ah+bh

Transposition: 2S-ah=bh, that is, bh=2S-ah.

The coefficient is 1:∵ h≠0: ∵ h≠0, ∴ b= -a (don't forget the condition h≠0).

When a=5, S=44 and h=8,

b= -5= 1 1-5=6

∴ b=6 .

Example 7. When x=2, the value of the formula x2+bx+4 is 0. Find the value of x2+bx+4 when x=3.

Analysis: This is also an example of the application of linear equations with one variable. It needs the value of x2+bx+4, first find the value of b, and finally find the value of x2+bx+4 when x=3.

Solution: ∵ When x=2, the value of x2+bx+4 is 0.

∴ 4+2b+4=0 (get a linear equation about b)

Solving this equation leads to 2b=-8, ∴ b=-4,

∴ x2+bx+4 is x2-4x+4,

When x=3, x2-4x+4 = 32-4x3+4 = 9-12+4 =1,

When x=3, the value of this formula is 1.

Example 8. Understanding absolute value equations:

( 1)| 2x- 1 | = 8(2)= 4(3)= 4

(4)| 3x- 1 |+9 = 5(5)| 1-| x | | = 2

Explanation: Solving absolute value equations is also the application of one-dimensional linear equations. The main solutions are as follows: ① First, take |ax+b| as a whole, treat the absolute value equation as a linear equation with |ax+b| as an unknown, and change it into the form of |ax+b|=c; ② Discuss |ax+b|=c, when c>0, correctly remove the absolute value, and get two linear equations of ax+b=c or ax+b=-c, thus finding the value of X; When c=0, the linear equation of ax+b=0 is obtained, and then x is obtained; When c<0, this equation has no solution because the absolute value is non-negative.

(1) solution: ∫| 2x- 1 | = 8

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

* 2x = 9 or 2x=-7

∴ x= or x=-

The solution of the original equation is x= or x=-.

(2) Solution: ∫= 4

Denominator deleted: |3x+2|= 12.

∴ 3x+2= 12 or 3x+2=- 12

∴ 3x= 10 or 3x=- 14

∴ x= or x=-

The solution of the original equation is x= or x=-.

(3) Solution: ∫= 4

Denominator: 2|x|+5= 12.

Move items and merge similar items: 2|x|=7

The coefficient is 1: | x | =

∴ x=

The solution of the original equation is x= or x=-.

(4) Solution: ∵ |3x- 1|+9=5

∴ |3x- 1|=-4

The absolute value of any rational number is non-negative,

There is no solution to this equation.

(5) Solution: ∵ | 1-|x||=2,

∴1-x | = 2 or1-x | =-2,

∴ |x|=- 1 or |x|=3, ∴ x= 3,

From the concept of absolute value, this equation has no solution;

∴ x= 3 is the solution of the original equation.

In equation (5), the absolute value must be processed twice, and the value of x can be obtained smoothly only by strictly following the law.

(3) Practice:

I. Fill in the blanks:

1. The solution of equation 3 (x-2)-5 (2x-1) = 4 (1-2x) is _ _ _ _ _ _ _.

2. If |3x-2|=2, then X is _ _ _ _ _ _.

3. When x = _ _ _ _ _ _ _ the values of the algebraic expressions 3x-2 and 3- x are opposite to each other.

4. Equation 2 is a linear equation about X (x3m-2+3x) = 3x3m-2+6x-2, so m = _ _ _ _ _ _

5. If the algebraic value +5 is the reciprocal of the algebraic value, then x = _ _ _ _ _ _ _ _

6. If |2x+3|+(x-3y+4)2=0, then x = _ _ _ _ _ _ and y = _ _ _ _ _ _

Second, solve the equation:

1. 1- + =

2.{[(+ 1)- 1]+x } = 1

3.- =

Practice reference answer:

I. Fill in the blanks:

1.x = 5 ^ 2。 X= or x = 0 3. x=-

4.m= 1 5。 x = 92 ^ 6。 x=-，y=

Second, solve the equation:

1.x= 2。 x= 3。 y=

Multiple choice

1. In the equation, if x= 1, then the value of a is equal to ().

a 、- 1 B、0 C、 1 D、2

2. In the following equation, the solution of 2 is ().

a、4y+2=6 B、

c、y- 1=3+ y D、x=0.25x+ 100

3. Equation 2x-3=3 and equation =0 are the same solution, so the value of a is equal to ().

a、B、2 C、 1 D、0

4. If x= 1 is the solution of the equation, then the solution of the equation m(y-3)-2=m(2y-5) about y is ().

A,-10 B, 0 C, d, all of the above are wrong.

5. When solving the equation, after removing the denominator, the correct result is ()

A, B,

C, D,

Answer and analysis

Answer: 1, C 2, B 3, B 4, B 5, c.

Analysis:

1, Analysis: Because x= 1 satisfies the equation, a linear equation about A is obtained by substituting it into the equation. When solving this linear equation about A, we can adopt the order of removing braces first, then brackets, and finally parentheses.

2. Analysis: According to the definition of the solution of a linear equation, substitute 2 into each equation to test whether the left and right sides of the equation are equal.

3. Analysis: According to the definition of homosolution equation, the solution of the first equation should satisfy the second equation, and the solution of the first equation is 3. Substitute it into the second equation, and the value of a is 2.

4. Analysis: Because x= 1 is the solution of the equation, you can get m= 1 by substituting x= 1 into the equation, and then you can get y=0 by substituting it into the following equation.

5. Analysis: Multiply both sides of the equation by 6, and get rid of the denominator.

Remove the brackets and get, so choose C.

One-dimensional linear equation and its solution

Test site scanning: understand the concept of linear equation of one variable; Using the basic properties of the equation flexibly to solve the linear equation of one variable will test the solution of the equation.

The famous teacher said:

1. One-dimensional linear equation: An equation with only one unknown number, degree 1 and coefficient not equal to 0 is called a one-dimensional linear equation. The characteristic of a linear equation with one variable is that it either has no denominator or has a denominator but does not contain unknowns, which can be simplified to the simplest form AX = remove the denominator, remove the brackets, shift the term, and merge.

2. Shift term: after changing the sign of an item in the equation, it moves from one side of the equation to the other, which is called shift term. This law comes from the property of equation 1 and is the basis of solving the equation. Understanding the shift term is to move a term from left to right or from right to left of the equation according to the need of solving the deformation of the equation, and the moved term will inevitably change its property sign, which is often overlooked.

3. Steps to solve the linear equation:

(1) denominator: the simplest common denominator of each denominator multiplied by both sides of the equation.

(2) Remove the brackets: When there is a "-"sign in front of the brackets, remember to change all the items in the brackets. There are multiple brackets, usually first remove the brackets, then remove the brackets, and finally remove the braces.

(3) Move items: move items containing unknowns to the same side, and move other items to the other side. Note that the moved item must change its symbol.

(4) Merge: transform the equation into the form of ax=b(a≠0).

(5) The coefficient is 1: the two sides of the equation are divided by the coefficient of the unknown quantity to get the solution of the equation.

Note: (1) Don't leave out items when removing the denominator, and put parentheses when the numerator is polynomial;

(2) When the conversion coefficient is 1, make clear the numerator and denominator, and don't make the mistake of x=.

Typical examples of senior high school entrance examination:

1. (Nanjing, Jiangsu) If X is the root of the equation 3x+2a=0, then A is equal to _ _ _ _ _.

Test site: the solution of a linear equation with one variable

Comments: Because 2 is the root of the equation 3x+2a=0, according to the meaning of the root of the equation, if 2 is substituted into the equation, both sides must be equal, that is, 6+2a=0. This is an equation about A, and a=-3 can be obtained by solving this equation.

2. (Kunming, Yunnan) It is known that A is an integer, 0.

Test site: the solution of a linear equation with one variable

Comments: This question examines students' ability to flexibly use the knowledge of solving linear equations with one variable. Because the equation contains two letters, A is an integer and 0.

Special training for real questions:

1. (Jingzhou) If x=0 is the root of the equation 3x-2m=4, then the value of m is ().

a、B、C、2 D、2

2. (Wuxi) If x=2 is the solution of the equation 2x+3k- 1=0 about x, then the value of k is _ _ _ _.

Answer: 1, d (hint: when x=0, the equation 3x-2m=4 becomes -2m=4, and the equation about m is solved by m=-2) 2,-1.