Advice 1: How to solve problems for grade 7 algebra

In 7th grade algebra course is more complicated. In the program there is a lot of interesting topics. In 7th grade solve problems on different topics, for example: "on the speed (movement), move down the river", "fractions", "comparing quantities". The skill with ease to solve problems indicates a high level of mathematical and logical thinking. Of course,with pleasure can only be solved by those that can easily be obtained.
How to solve problems for grade 7 algebra
Instruction
1
Let us examine how to solve common tasks.

The solution of problems on velocity we need to know a few formulas and be able to correctly write the equation.

Formula for the solution :

S=V*t formula is the way;

V=S/t formula speed;

t =S/V time formula, where S is distance, V - velocity, t - time.

For example, let us consider how to solve tasks of this type.

Condition: the Truck on the way from city a to city B spent 1.5 hours. The second truck has spent 1.2 hours. The speed of the second car more than 15 km/h than the speed of the first. To find the distance between two cities.
Solution: For convenience, use the following table. In it, specify what is known by the condition:

1 auto 2 auto

S X X

V X/1.5 X/1,2

t 1,5 1,2

For X take what you need to find, i.e. the distance. When setting up the equation to be careful, note that all values were in the same dimension (time - in hours, speed in km/h). By the speed of the 2nd car for more speeds 1 to 15 km/h, i.e., V1 - V2=15. Knowing this, we form and solve the equation:

X/1,2 X/1,5=15

1.5 X - 1.2 X - 27=0

0.3 X=27

X=90(km) - distance between cities.

Answer: the Distance between the towns is 90 km.
2
In solving problems on motion on the water " you must know that there are several types of velocities: the velocity (V) speed current (Vпо tech.), speed against the current (V [PR]. tech.), flow rate (Vтеч.).

Remember the following formula:

Vпо tech=VC+Vтеч.

V [PR]. tech.=VC-Vтеч.

V [PR]. tech=tech Vпо. - 2Vтеч.

Vпо tech.=V [PR]. tech+2Vтеч.

VC=(Vпо tech.+V [PR] tech.)/2 or VC=Vпо tech.+Vтеч.

Vтеч.=(Vпо tech. - V [PR]. tech)/2

For example, let us consider how to solve them.

Condition: boat Speed the flow of 21.8 km/h and upstream 17.2 km/h to Find a private speed boat and speed of river flow.

Solution: According to the formula: VC=(Vпо tech.+V [PR] tech.)/2 and Vтеч.=(Vпо tech. - V [PR]. tech)/2, we find:

Vтеч = (21,8 - 17,2)/2=4,6\2=2,3 (km/h)

VC = V [PR] tech.+Vтеч=17,2+2,3=19,5 (km/h)

Answer: Vc=19,5 (km/h), Vтеч=2,3 (km/h).
3
Objectives to compare quantities

Condition: Lot of 9 bricks 20 kg more than the weight of one brick. Find the mass of one brick.

Solution: let X (K), then the mass of 9 bricks 9X (kg). From the condition it follows that:

9X - X=20

8x=20

X=2,5

Answer: the Weight of one brick is 2.5 kg.
4
Task on fractions. The main rule when dealing with such this type of tasks: to find a fraction of the number, this number multiplied by this fraction.

Condition: the Camper was on the road 3 days. The first day was it? all the way, in the second the remaining 5/9 of the way, and on the third day - the last 16 km to Find the entire path of the tourist.

Solution: Let the entire path of the tourist is equal to X (km). That first day was it? x (km) second day - 5/9(x -?) = 5/9*3/4x = 5/12x. Since the third day it was 16 km, then:

1/4x+5/12x+16=x

1/4x+5/12x-x= - 16

- 1/3x=-16

X=- 16:(-1/3)

X=48

Answer: All the way a tourist is 48 km away.
Useful advice
To easily solve problems, we must learn to translate them into the language of numbers”, using some tricks. Drawing tables and diagrams as possible helps to understand the problem statement, the relationship of the variables. Also facilitates the process of generating the equations. Of course, you have to know the necessary formulas.

Advice 2: How to solve problems on the motion

To solve the problem on the motion is relatively simple. It is enough to know only one formula: S=V*t.
How to solve problems on the motion
Instruction
1
When solving problems in the movement the main parameters are:

the path, indicated generally as S,

the speed – V and

time - t.

The dependence between these parameters is expressed by the following formulas:

S=Vt, V=S/t and t=S/V

To avoid confusion in units of measurement, these parameters must be specified in one system. For example, if the time is measured in hours and the distance traveled in kilometers, the speed, respectively, must be measured in kilometer/hour.

When solving problems of this type are usually produced by the following steps:

1. Select one of the unknown parameters and is denoted by the letter x (y, z, etc.)

2. Specify which of the three basic parameters are known.

3. A third of the remaining parameters using the above formulae is expressed through the other two.

4. Based on the conditions of the problem, make an equation that links an unknown value with known parameters.

5. Solve the resulting equation.

6. Check the found roots of the equation under tasks.

In some cases, solve the problem by means of the drawing (regardless of the quality of the picture).
2
Example 1.

To solve the problem:

Skier passing 5 km in the same amount of time for which the pedestrian manages to pass 2 km.

To find this time, if it is known that the speed of the skier more the walking speed of 6 km/h. Determine the speed of the pedestrian and skier.

Let us denote the desired time (in hours) using t.

Then, by the formula V=S/t, the speed of the skier equal to 5/t km/h and walking speed is equal to 2/t km/h.

Using conditions the task can write the equation:

5/t – 2/t = 6

Where is defined as: t=0.5

Therefore: speed of a pedestrian is 4 km/h, and skier - 10 km/h.

Response: 0.5 hours; 4 km/h; 10 km/h.
3
Example 2.

Solve the above problem in a different way:

Denote the walking speed using V (km/h).

Then the velocity of the skier will be (V+6) km/h.

In accordance with the formula: t=S/V, the time can be determined according to the following expression:

t=5/(V+6)=2/V

How elementary is:

V=4,

t=0.5.

Advice 3: How to solve problems with fractions

To solve the problem with fractions, you need to learn how to do arithmetic. They can be decimal, but most often uses natural fraction with the numerator and denominator. Only after that you can go to to solve math problems with fractional quantities.
How to solve problems with fractions
You will need
  • calculator;
  • - knowledge of the properties of fractions;
  • - the ability to perform operations with fractions.
Instruction
1
The roll call record of division of one number by another. Often to do this completely is impossible, therefore, leave this step "unfinished . A number that is divisible (it is above or in front of the sign of the fraction) is called the numerator and the second number (under the sign of the fraction or after it) is the denominator. If the numerator is more than denominator, the fraction is called improper, and it is possible to allocate the integer part. If the numerator less than the denominator, such a fraction is called proper, and its integer part equal to 0.
2
Tasks with fractions are divided into several types. Determine which of them is the challenge. The simplest option – finding fractions of numbers expressed in fraction. To solve this problem, it is sufficient to multiply this number by the fraction. For example, the warehouse delivered 8 tons of potatoes. In the first week, it sold 3/4 of its total. How many potatoes are left? To solve this problem, the number 8 and multiply by 3/4. Makes 8∙3/4=6 t
3
If you want to find the number of parts, multiply the known part of a number the inverse of that which shows what proportion of this part in the number. For example, 8 students constitute 1/3 of the total number of students. How many children are studying in class? Because 8 people is the part that is 1/3 of the total number, then find the inverse fraction, which is equal to 3/1 or just 3. Then, to obtain the number of students in class 8∙3=24 students.
4
When you need to find what part number is one number from another, divide the number which represents the part on that which is whole. For example, if the distance between the cities is 300 km and the car has driven 200 km, what part of this is from all the way? Share part of the journey 200 with the complete path 300, after the reduction of the fraction will get the result. 200/300=2/3.
5
To find the unknown part of the proportion of when there is known to take an integer for the standard unit, and subtract from it a certain amount. For example, if you have passed 4/7 part of the lesson, how long? Take the whole lesson as a conventional unit and subtract from it 4/7. Get 1-4/7=7/7-4/7=3/7.

Advice 4: How to solve problems with improper fractions

A fraction is a mathematical notation of simple rational numbers. It is a number that consists of one or more fractions of a unit can be either in decimal and ordinary form. Today operations convert fractions are of great importance not only in mathematics but also in other fields of knowledge.
How to solve problems with improper fractions
Instruction
1
Typically, the most common fractions are wrong, and in this case they require specific action on the part of someone who decides examples and tasks with this fraction.
2
Take the tutorial to the task. Please read condition, reading it several times and go to solution. Let's see which fractions are to be solved action. This can be incorrect, correct or decimal fraction. Put the correct fractions in the wrong, but remember that to record all response actions will have to run back, already converting improper fraction to the correct. At common fraction the number above the fraction bar (the numerator) is always greater than the number below the line is denominator. To make the translation from proper fractions to improper follow the next steps.
3
Multiply the denominator by the integer and add the result to the numerator. For example, if a fraction of 2 integers 7/9, 9 must be multiplied by 2 and then add 7 to 18 - the end result will be 25/9.
4
Perform all necessary actions according to the task (addition, subtraction, division, multiplication) using the transformed fraction.Take your answer, it will need to be present in the fractions. To do this, divide the numerator by the denominator. For example, if you have to take the number 25/9 in a proper fraction, divide 25 by 9. As 25 9 not evenly divided, the answer is 2 and as many as seven (numerator) ninth (denominator). Now obtained the proper fraction, where the numerator more than the denominator and a part of it.
5
Record the response task correct fraction. Check your action, if it requires the condition to do task or teacher.
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