Find the problem analysis of the application problem of the unary linear equation

1. Equation An equation containing unknowns is called an equation. 2. Unary Equation An equation that contains only one unknown The equation with an unknown exponent of 1 is called a unary equation. 3. The solution of the equation The value of the unknown that makes the left and right sides of the equation equal is called the solution of the equation.

4. Solving Equations The process of solving equations is called solving equations. 2. The nature of the equation The nature of the equation 1 Add or subtract the same number or subformula on both sides of the equation and the result is still equal. Properties of Equation 2 Multiplying the same number on both sides of the equation or dividing by the same number that is not 0 still results in the same.

3. General steps for solving unary equations and their basis 1. Remove the denominator --- the properties of the equation 2 2. Remove the parentheses --- the distributive property 3. Shift the properties of the --- equation 1 4. Merge --- distributive properties 5. Convert the coefficient to 1--- properties of the equation 2 6. Examine the root --- substitute the roots into the left and right sides of the equation respectively to see if the obtained values are equal 4. Precautions for solving unary equations 1. When the denominator is a decimal number Convert the denominator to an integer according to the basic properties of fractions 2. When the denominator is removed The terms on both sides of the equation are multiplied by the least common multiple of each denominator At this time, the term without the denominator should not be omitted to multiply The fractional line is equivalent to the parentheses After removing the denominator, the numerator items should be added parentheses 3. When removing the parentheses, do not omit the terms in parentheses Don't make a mistake with the symbols 4. When moving the terms, remember to change the sign Don't lose the terms Sometimes merge and then move the terms so as not to lose the terms 5. When the coefficient is 1, both sides of the equation are multiplied by the reciprocal of the coefficient or divided by the coefficient Don't make a mistake with the symbol 6. Don't rigidly copy the steps of solving the equation Specific analysis of the specific problem Find the best solution. 5. General steps for solving application problems of column equations 1, review the problem 2, set the uncounted 3, find the equality relationship 4, column equation 5, solve the equation 6, test 7, write the answer step according to the matter of note Remove the brackets Distributional law, the rule of removing brackets Do not miss the items in the brackets The parentheses are preceded by the " " sign to change the sign. Shift Shift Rule Shift Term to Change Sign Merge Similar Terms Merge Similar Terms Rule of Merging Similar Terms Coefficients Addition No Missing Terms Coefficients on both sides divided by unknown numbers Equation property 2 Multiply the reciprocal of the coefficient.

I am a sophomore science student.

Let me tell you what I think.

The main problem of a one-dimensional equation is to find a relation.

The relationship between unknowns and known numbers.

Setting the unknowns is also the key to determining the difficulty of the calculation.

You have to look at it intuitively.

It's easy to solve that unknown.

And the unknowns are not necessarily the answer.

Here's an example.

Warehouse A has 242 tons of grain and warehouse B has 142 tons of grain, how many tons can be transported from warehouse A to warehouse B to make the grain stored in two warehouses equal? How many tons can be transported from bin B to bin A to make bin A 3 times that of bin B?

Set up x tons of transportation from warehouse A to warehouse B.

242-x=142+x

Solution x=50

Set up x tons of transportation from warehouse B to warehouse A.

242+x=3(142-x)

Solution x=92

I hope you can follow up.

1. Reading questions.

2. Look for equiquantitative relationships.

3. Set an unknown quantity.

4. Use unknown quantities to express the equiquantity relationship and list the equations.

5. Solve the listed equations.

6. Write the number of answers.

Let the sales unit price be set at x yuan (x 10), and the profit obtained per day is y yuan, y = [100-10(x-10)] x-8) = -10x

2+280x-1600

10（x-14）

When the sales price is set at 14 yuan, the daily sales profit is the largest, and the maximum profit is 360 yuan.

2.Solution: (1) If the original price is 1, the jumping price is, so the percentage of the jumping price to the original price is;

2) Original price**: sales amount = 100 1 = 100, new price **: sales amount = ,; The new plan is more profitable to sell

3.Solution: Let the side length of the square be xcm

Then there is: 4x=5(x-4), and the solution is x=20, then 4x=80, so the area of the rectangle is 80cm2

Solution: After x hours, the distance between the fast train and the slow train is 600 kilometers.

120x-80x=600

40x=600

x = 1515 hours later, the distance between the fast and slow trains is 600 km.

A lot of things quack and quack.