Teaching objectives

1. Make students understand and master the linear equation of letter coefficient and its solution;

2. Understand the meaning of formula deformation and master the method of formula deformation;

3. Improve students' ability of calculation and reasoning.

Emphasis and difficulty of education

Emphasis: the sum and solution of a linear equation with letter coefficient.

Difficulties: the application of letter coefficient condition and the deformation of formula.

Teaching process design

First, the introduction of new courses.

Q: What is an equation? What is a linear equation with one variable?

Answer: An equation with an unknown number is called an equation, and an equation with an unknown number 1 is called a linear equation.

Example solution equation 2x-13-10x+16 = 2x+14-1.

Solve the denominator, and multiply both sides of the equation by 12 to get

4(2x- 1)-2( 10x+ 1)= 3(2x+ 1)- 12，

Remove the bracket and get

8x-4-20x-2=6x+3- 12

Move items, get

8x-20x-6x=3- 12+4+2，

Merge similar projects to obtain

- 18x=-3，

Divide both sides of the equation by-18, and you get

X=3 18, which means x= 1 6.

Second, the new lesson

The letter coefficient solution of 1.

We express the one-dimensional linear equation as a general form.

ax=b (a≠0)，

Where x represents an unknown number and a and b are known numbers in letters. For the unknown X, the letter A is the coefficient of X, which is called the letter coefficient, and the letter B is a constant term.

If the coefficients in a unitary linear equation are expressed by letters, then this equation is called a unitary letter coefficient equation.

Quadratic equation.

Unless otherwise specified, in equations with letter coefficients, known numbers are generally represented by A, B and C, and unknowns are represented by X, Y and Z, etc.

The solution of linear equation with letter coefficient is the same as that of linear equation with number coefficient. According to this solution

The steps of one-dimensional linear equation are finally transformed into the form of ax=b(a≠0). It should be noted here that when multiplying or dividing two sides of the equation with a formula containing letters, the value of this formula cannot be equal to zero. For example, (m-2) x = 3, it must be when m-2 ≠ 0, that is, when m-2 ≠ 0 2.

Example 1 Solve equation ax+b2=bx+a2(a≠b).

Analysis: The letters A and B in this equation are known numbers, and X is unknown, which is a linear equation with letter coefficients. The condition a≠b given here is the key to solving the equation, and it should be used in the process of solving the equation.

To delete this term, you must

ax-bx=a2-b2，

Merge similar projects to obtain

(a-b)x=a2-b2。

Because a≠b, A-B ≠ 0. Divide both sides of the equation by A-B, and you get

x=a2-b2 a-b=(a+b)(a-b) a-b，

So x = a+b.

Point out:

A≠b is given in (1). In the process of solving the equation, it is guaranteed that the solution of the equation obtained by removing both sides of the equation with formula A-B which is not equal to zero is the solution of the original equation.

(2) If the solution of the equation is in fractional form, it should generally be reduced to the simplest fractional or algebraic expression.

Example 2 x-b a = 2-x-a b (a+b ≠ 0).

Observe the characteristics of the equation structure, please tell the method of solving the equation.

Answer: This equation contains fractions. You can first remove the denominator and convert the equation into a linear equation with letter coefficients.

In the transformation of the equation, the known condition a+b≠0 should be applied.

Solve the denominator and multiply both sides of the equation by ab.

b(x-b)=2ab-a(x-a)，

Remove the bracket and get

bx-b2=2ab-ax+a2，

Move items, get

ax+bx=a2+2ab+b2

Merge similar projects to obtain

(a+b)x=(a+b)2。

X=a+b because a+b≠0.

It is pointed out that ab≠0 is an implicit condition, because letters A and B are denominators of two fractions in the equation, so A ≠ 0 and b≠0, so ab≠0.

Example 3 solving the equation about x

a2+(x- 1)ax+3a=6x+2(a≠2，a≠-3)。

This solution transforms this equation into, get

a2x-a2+ax+3a=6x+2，

Move projects, merge similar projects, and get

a2x+ax-6x=a2-3a+2，

(a2+a-6)x=a2-3a+2，

(a+3)(a-2)x=(a- 1)(a-2)。

Because a-2≠0 2, a =-3, a+3≠0, a-2≠0. Divide both sides of the equation by (a+3) (a-2), and you get

x=a- 1 a+3。

2. Formula deformation.

In physics class, we learned many physical formulas. If q represents the combustion value and m represents the mass of fuel, then the heat generated by the complete combustion of these fuels is W=qm. For another example, q represents the electric quantity passing through the cross section of the foreign body, t represents the time, and I represents the current passing through the conductor. The relationship between them is I = Q = It. In this formula, if I and t represent q, I and t are known. If I and q are used to represent t, that is, I and q are known, and t is obtained.

As mentioned above, transforming a formula from one form to another is called formula deformation.

Taking one letter in the formula as an unknown quantity and other letters as known quantities, finding the unknown quantity is to solve the problem of containing letters.

Coefficient number equation. In other words, the formula deformation is actually to solve the equation with letter coefficient. Formula deformation is very important not only in mathematics, but also in physical chemistry. We should master the skills of formula deformation skillfully.

In the formula of example 4, υ=υo+at, υ, υo, a are known, and a≠0, find t.

Analysis: If you know υ, υo and A, you can find T, that is, if υ, υo and A are known quantities, you can solve the equation about the letter coefficient of the unknown quantity T. 。

To delete this term, you must

υ-υ0=at。

Because a≠0, divide both sides of the equation by a, and you get

t=υ-υo a

Example 5 In the trapezoid area formula s= 12(a+b)h, it is known that A, B and H are positive numbers.

(1) H is represented by s, a and b; (2) use s, b, h b and h to represent a.

Q: Which of (1) and (2) is a known quantity? What are the unknown;

A: In (1), S, A and B are known quantities, and H is unknown; (2) where s, b and h are known quantities and a is unknown quantity.

Multiply both sides of the equation (1) by 2 to obtain

2s=(a+b)h。

Because both A and B are positive numbers, a≠0, b≠0, that is, a+b≠0, divide both sides of the equation by a+b, and you get

h=2sa+b。

(2) multiply both sides of the equation by 2 to obtain

2s=(a+b)h，

Pack up, take it

ah=2s-bh。

Because H is a positive number, so h≠0, divide both sides of the equation by H, and you get

a=2s-bh h。

It is pointed out that the problem is to solve the equation about h, (a+b) can be regarded as the coefficient of unknown h, and (a+b)h should not be expanded in operation.

Third, classroom exercises.

1. Solve the following equation about x:

( 1)3a+4x = 7x-5b； (2)xa-b = x b-a(a≠b)；

(3)m2(x-n)= N2(x-m)(m2≠N2)；

(4)a b+ xa = x B- ba(a≠b)；

(5)a2x+2=a(x+2)(a≠0，a≠ 1)。

2. Fill in the blanks:

(1) If y=rx+b r≠0 is known, then x = _ _ _ _ _ _ _

(2) if f = ma and a ≠ 0 are known, then m = _ _ _ _ _ _ _

(3) given ax+by=c, a≠0, then x = _ _ _ _ _ _

3. None of the letters in the following formula is equal to zero.

(1) find n in the formula m=pn+2;

(2) given xa+ 1b= 1m, find x;

(3) in the formula S=a+b2h, find a;

(4) In the formula S=υot+ 12t2x, find x. 。

Answer:

1.( 1)x = 3a+5b 3； (2)x = ab； (3)x = Mn m+n； (4)x=a2+b2 a-b (5)x=2a。

2.( 1)x = y-b r； (2)m = Fa； (3)x = c- times a.

3.( 1)n = p-2m m； (2)x = a b-am BM； (3)a = 2s-BH h；

(4)x=2s-2υott2。

Four. abstract

1. The solution of one-dimensional linear equation with letter coefficient is the same as that of one-dimensional linear equation with only digital coefficient, but it should be paid special attention to that when multiplying or dividing two sides of the equation with letter formula, the value of this formula cannot be zero. Our examples and the conditions given in the classroom exercises ensure this.

2. For formula deformation, we must first find out which quantities in the formula are known and which are unknown. Take the known quantity as the word.

The process of finding the unknown quantity is the process of solving the equation about the letter coefficient.

Verb (short for verb) homework

1. Solve the following equation about x.

( 1)(m2+N2)x = m2-N2+2 mnx(m-n≠0)；

(2)(x-a)2-(x-b)2 = 2 a2-2 B2(a-b≠0)；

(3)x+XM = m(m≦- 1)；

(4)XB+b = xa+a(a≠b)；

(5)m+nx m+n=a+bx a+b(mb≠na)。

2. In the formula M = D-D2L, all the letters are not equal to zero.

(1) It is known that m, l, d l and d find d; (2) Given m, l, d, find d. 。

3. In the formula S = 1 2n [a1+(n-1) d], all letters are positive numbers, and n is an integer greater than1. Find D.

Answer:

1.( 1)x = m+n m-n； (2)x =-a+B2； (3)x = m2 m+ 1； (4)x = ab； (5)x= 1。

2.( 1)D = 2lM+D； (2)d=D-2lM。

3.d=2S-na 1 n(n- 1)。

Description of classroom mathematical design

1. It is difficult for students to understand and solve the letter coefficient equation and formula deformation.

When the equation contains alphabetic coefficients, the relationship between the generated coefficient equation and the equation containing only numerical coefficients is both general and special.

When the letters in it are given a specific number, it is an equation with only numerical coefficients. Therefore, in the instructional design, we should review the solution, which only includes

Starting from the one-dimensional linear equation of digital coefficient, the transition is made to discuss the solution and formula deformation of the one-dimensional linear equation of letter coefficient.

It embodies the law of following students' way of thinking and understanding things from concrete to abstract and from special to general.

2. Infiltration reasoning factors should be paid attention to in algebra teaching. In the process of solving the letter coefficient and formula deformation of the one-dimensional linear equation, students should be guided to pay attention to what the known conditions in the given problem are and correctly use the known conditions in the problem for equation deformation. For example, when solving an equation, the two sides of the equation are often multiplied (or divided) by a formula with letters, and how to ensure that the value of this formula is not equal to zero according to known conditions is discussed, so as to consciously train and improve students' ability.