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Solving equations, inequalities, systems using function graphs. Visual Guide (2019)

Many tasks that we are used to calculating purely algebraically can be solved much easier and faster; using function graphs will help us with this. You say “how so?” draw something, and what to draw? Believe me, sometimes it is more convenient and easier. Shall we get started? Let's start with the equations!

Graphical solution of equations

Graphical solution of linear equations

As you already know, the graph of a linear equation is a straight line, hence the name of this type. Linear equations are quite easy to solve algebraically - we transfer all the unknowns to one side of the equation, everything we know to the other, and voila! We found the root. Now I'll show you how to do it graphically.

So you have the equation:

How to solve it?
Option 1, and the most common one is to move the unknowns to one side and the knowns to the other, we get:

Now let's build. What did you get?

What do you think is the root of our equation? That's right, the coordinate of the intersection point of the graphs is:

Our answer is

That's the whole wisdom of the graphic solution. As you can easily check, the root of our equation is a number!

As I said above, this is the most common option, close to an algebraic solution, but you can solve it in another way. For consideration alternative solution Let's return to our equation:

This time we will not move anything from side to side, but will construct the graphs directly, as they are now:

Built? Let's see!

What is the solution this time? That's right. The same thing - the coordinate of the intersection point of the graphs:

And, again, our answer is.

As you can see, with linear equations everything is extremely simple. It's time to look at something more complex... For example, graphical solution of quadratic equations.

Graphical solution of quadratic equations

So, now let's start solving the quadratic equation. Let's say you need to find the roots of this equation:

Of course, you can now start counting through the discriminant, or according to Vieta’s theorem, but many people, out of nerves, make mistakes when multiplying or squaring, especially if the example is with large numbers, and, as you know, you won’t have a calculator for the exam... Therefore, let’s try to relax a little and draw while solving this equation.

You can find solutions to this equation graphically different ways. Let's consider various options, and you can choose which one you like best.

Method 1. Directly

We simply build a parabola using this equation:

To do this quickly, I'll give you one little hint: It is convenient to start the construction by determining the vertex of the parabola. The following formulas will help determine the coordinates of the vertex of a parabola:

You will say “Stop! The formula for is very similar to the formula for finding the discriminant,” yes, it is, and this is a huge disadvantage of “directly” constructing a parabola to find its roots. However, let's count to the end, and then I'll show you how to do it much (much!) easier!

Did you count? What coordinates did you get for the vertex of the parabola? Let's figure it out together:

Exactly the same answer? Well done! And now we already know the coordinates of the vertex, but to construct a parabola we need more... points. How many minimum points do you think we need? Right, .

You know that a parabola is symmetrical about its vertex, for example:

Accordingly, we need two more points on the left or right branch of the parabola, and in the future we will symmetrically reflect these points on the opposite side:

Let's return to our parabola. For our case, period. We need two more points, so we can take positive ones, or we can take negative ones? Which points are more convenient for you? It’s more convenient for me to work with positive ones, so I’ll calculate at and.

Now we have three points, we can easily construct our parabola by reflecting the last two points relative to its vertex:

What do you think is the solution to the equation? That's right, points at which, that is, and. Because.

And if we say that, it means that it must also be equal, or.

Just? We have finished solving the equation with you in a complex graphical way, or there will be more!

Of course, you can check our answer algebraically - you can calculate the roots using Vieta's theorem or Discriminant. What did you get? The same? Here you see! Now let's look at a very simple graphic solution, I'm sure you'll really like it!

Method 2. Divided into several functions

Let’s take our same equation: , but we’ll write it a little differently, namely:

Can we write it like this? We can, since the transformation is equivalent. Let's look further.

Let's construct two functions separately:

1. - the graph is a simple parabola, which you can easily construct even without defining the vertex using formulas and drawing up a table to determine other points.
2. - the graph is a straight line, which you can just as easily construct by estimating the values ​​in your head without even resorting to a calculator.

Built? Let's compare with what I got:

What do you think are the roots of the equation in this case? Right! The coordinates obtained by the intersection of two graphs and, that is:

Accordingly, the solution to this equation is:

What do you say? Agree, this method of solution is much easier than the previous one and even easier than looking for roots through a discriminant! If so, try solving the following equation using this method:

What did you get? Let's compare our graphs:

The graphs show that the answers are:

Did you manage? Well done! Now let's look at the equations a little more complicated, namely, the solution of mixed equations, that is, equations containing functions of different types.

Graphical solution of mixed equations

Now let's try to solve the following:

Of course, you can bring everything to a common denominator, find the roots of the resulting equation, without forgetting to take into account the ODZ, but again, we will try to solve it graphically, as we did in all previous cases.

This time let's build the following 2 graphs:

1. - the graph is a hyperbola
2. - the graph is a straight line, which you can easily construct by estimating the values ​​in your head without even resorting to a calculator.

Realized it? Now start building.

Here's what I got:

Looking at this picture, tell me what are the roots of our equation?

That's right, and. Here's the confirmation:

Try plugging our roots into the equation. Happened?

That's right! Agree, solving such equations graphically is a pleasure!

Try to solve the equation graphically yourself:

I'll give you a hint: move part of the equation to right side, so that on both sides there are the simplest functions to construct. Did you get the hint? Take action!

Now let's see what you got:

Respectively:

1. - cubic parabola.
2. - ordinary straight line.

Well, let's build:

As you wrote down long ago, the root of this equation is - .

Having decided this a large number of examples, I'm sure you realized how easily and quickly you can solve equations graphically. It's time to figure out how to solve systems in this way.

Graphic solution of systems

Graphically solving systems is essentially no different from graphically solving equations. We will also build two graphs, and their intersection points will be the roots of this system. One graph is one equation, the second graph is another equation. Everything is extremely simple!

Let's start with the simplest thing - solving systems of linear equations.

Solving systems of linear equations

Let's say we have the following system:

First, let's transform it so that on the left there is everything that is connected with, and on the right - everything that is connected with. In other words, let’s write these equations as a function in our usual form:

Now we just build two straight lines. What is the solution in our case? Right! The point of their intersection! And here you need to be very, very careful! Think about it, why? Let me give you a hint: we are dealing with a system: in the system there is both, and... Got the hint?

That's right! When solving a system, we must look at both coordinates, and not just as when solving equations! Another important point- write them down correctly and not confuse where we have the meaning and where the meaning is! Did you write it down? Now let's compare everything in order:

And the answers: and. Do a check - substitute the found roots into the system and make sure whether we solved it correctly graphically?

Solving systems of nonlinear equations

What if, instead of one straight line, we have quadratic equation? It's okay! You just build a parabola instead of a straight line! Do not believe? Try solving the following system:

What's our next step? That’s right, write it down so that it’s convenient for us to build graphs:

And now it’s all a matter of small things - build it quickly and here’s your solution! We are building:

Did the graphs turn out the same? Now mark the solutions of the system in the figure and correctly write down the identified answers!

I've done everything? Compare with my notes:

Is everything right? Well done! You are already cracking these types of tasks like nuts! If so, let’s give you a more complicated system:

What are we doing? Right! We write the system so that it is convenient to build:

I’ll give you a little hint, since the system looks very complicated! When building graphs, build them “more”, and most importantly, do not be surprised by the number of intersection points.

So, let's go! Exhaled? Now start building!

So how? Beautiful? How many intersection points did you get? I have three! Let's compare our graphs:

Also? Now carefully write down all the solutions of our system:

Now look at the system again:

Can you imagine that you solved this in just 15 minutes? Agree, mathematics is still simple, especially when looking at an expression you are not afraid to make a mistake, but just take it and solve it! You're a big lad!

Graphical solution of inequalities

Graphical solution of linear inequalities

After the last example, you can do anything! Now exhale - compared to previous sections This one will be very, very easy!

We will start, as usual, with a graphical solution linear inequality. For example, this one:

First, let's carry out the simplest transformations - open the brackets of perfect squares and present similar terms:

The inequality is not strict, therefore it is not included in the interval, and the solution will be all points that are to the right, since more, more, and so on:

Answer:

That's all! Easily? Let's solve a simple inequality with two variables:

Let's draw a function in the coordinate system.

Did you get such a schedule? Now let’s look carefully at what inequality we have there? Less? This means we paint over everything that is to the left of our straight line. What if there were more? That's right, then we would paint over everything that is to the right of our straight line. It's simple.

All solutions to this inequality are “shaded out” orange. That's it, the inequality with two variables is solved. This means that the coordinates of any point from the shaded area are the solutions.

Graphical solution of quadratic inequalities

Now we will understand how to graphically solve quadratic inequalities.

But before we get down to business, let's review some material regarding the quadratic function.

What is the discriminant responsible for? That’s right, for the position of the graph relative to the axis (if you don’t remember this, then definitely read the theory about quadratic functions).

In any case, here's a little reminder for you:

Now that we have refreshed all the material in our memory, let's get down to business - solve the inequality graphically.

I’ll tell you right away that there are two options for solving it.

Option 1

We write our parabola as a function:

Using the formulas, we determine the coordinates of the vertex of the parabola (exactly the same as when solving quadratic equations):

Did you count? What did you get?

Now let's take two more different points and calculate for them:

Let's start building one branch of the parabola:

We symmetrically reflect our points onto another branch of the parabola:

Now let's return to our inequality.

We need it to be less than zero, respectively:

Since in our inequality the sign is strictly less than, we exclude the end points - “puncture out”.

Answer:

Long way, right? Now I will show you a simpler version of the graphical solution using the example of the same inequality:

Option 2

We return to our inequality and mark the intervals we need:

Agree, it's much faster.

Let us now write down the answer:

Let's consider another solution that simplifies the algebraic part, but the main thing is not to get confused.

Multiply the left and right sides by:

Try to solve the following quadratic inequality yourself in any way you like: .

Did you manage?

Look how my graph turned out:

Answer: .

Graphical solution of mixed inequalities

Now let's move on to more complex inequalities!

How do you like this:

It's creepy, isn't it? Honestly, I have no idea how to solve this algebraically... But it’s not necessary. Graphically there is nothing complicated about this! The eyes are afraid, but the hands are doing!

The first thing we will start with is by constructing two graphs:

I won’t write out a table for each one - I’m sure you can do it perfectly on your own (wow, there are so many examples to solve!).

Did you paint it? Now build two graphs.

Let's compare our drawings?

Is it the same with you? Great! Now let’s arrange the intersection points and use color to determine which graph we should have larger in theory, that is. Look what happened in the end:

Now let’s just look at where our selected graph is higher than the graph? Feel free to take a pencil and paint over this area! She will be the solution to our complex inequality!

At what intervals along the axis is we located higher than? Right, . This is the answer!

Well, now you can handle any equation, any system, and even more so any inequality!

BRIEFLY ABOUT THE MAIN THINGS

Algorithm for solving equations using function graphs:

1. Let's express it through
2. Let's define the function type
3. Let's build graphs of the resulting functions
4. Let's find the intersection points of the graphs
5. Let’s write the answer correctly (taking into account the ODZ and inequality signs)
6. Let's check the answer (substitute the roots into the equation or system)

For more information about constructing function graphs, see the topic “”.

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested this work, please download the full version.

Goals and objectives of the lesson:

• continue work on developing skills in solving systems of equations using the graphical method;
• conduct research and draw conclusions about the number of solutions to a system of two linear equations;
• develop interest in the subject through play.

DURING THE CLASSES

1. Organizing time(Planning meeting)- 2 minutes.

- Good afternoon! We are starting our traditional planning meeting. We are pleased to welcome everyone who is visiting us today in our laboratory (I represent the guests). Our laboratory is called: “WORK WITH interest and pleasure”(showing slide 2). The name serves as a motto in our work. “Create, Decide, Learn, Achieve with interest and pleasure" Dear guests, I present to you the heads of our laboratory (slide 3).
Our laboratory is engaged in the study of scientific works, research, examination, and works on the creation of creative projects.
Today the topic of our discussion is: “Graphical solution of systems of linear equations.” (I suggest writing down the topic of the lesson)

Program of the day:(slide 4)

1. Planning meeting
2. Extended academic council:

• Speeches on the topic
• Permission to work

3. Expertise
4. Research and discovery
5. Creative project
6. Report
7. Planning

2. Questioning and oral work (Extended Academic Council)- 10 min.

– Today we are holding an expanded scientific council, which is attended not only by the heads of departments, but also by all members of our team. The laboratory has just begun work on the topic: “Graphical solution of systems of linear equations.” We must try to achieve the highest achievements in this matter. Our laboratory should be renowned for the quality of its research on this topic. As a senior researcher, I wish everyone good luck!

The results of the research will be reported to the head of the laboratory.

The floor for a report on solving systems of equations is... (I call the student to the board). I give the task a task (card 1).

And the laboratory assistant... (I give his last name) will remind you how to graph a function with a modulus. I give you card 2.

Card 1(solution to the task on slide 7)

Solve the system of equations:

Card 2(solution to the task on slide 9)

Graph the function: y = | 1.5x – 3 |

While the staff is preparing for the report, I will check how prepared you are to complete the research. Each of you must obtain permission to work. (We begin oral counting with writing down answers in a notebook)

Permission to work(tasks on slides 5 and 6)

1) Express at through x:

3x + y = 4 (y = 4 – 3x)
5x – y = 2 (y = 5x – 2)
1/2y – x = 7 (y = 2x + 14)
2x + 1/3y – 1 = 0 (y = – 6x + 3)

2) Solve the equation:

5x + 2 = 0 (x = – 2/5)
4x – 3 = 0 (x = 3/4)
2 – 3x = 0 (x = 2/3)
1/3x + 4 = 0 (x = – 12)

3) Given a system of equations:

Which of the pairs of numbers (– 1; 1) or (1; – 1) is the solution to this system of equations?

Answer: (1; – 1)

Immediately after each fragment of oral calculation, students exchange notebooks (with a student sitting next to them in the same section), the correct answers appear on the slides; The inspector gives a plus or minus. At the end of the work, department heads enter the results into the summary table (see below); 1 point is given for each example (it is possible to get 9 points).
Those who score 5 or more points are allowed to work. The rest receive conditional admission, i.e. will be required to work under the supervision of the department head.

Table (filled out by the boss)

(Tables are issued before the start of the lesson)

After receiving admission, we listen to the students’ answers at the blackboard. For the answer, the student receives 9 points if the answer is complete (the maximum number for admission), 4 points if the answer is not complete. Points are entered in the “admission” column.
If on the board correct solution, then slides 7 and 9 may not be shown. If the solution is correct, but not clearly executed, or the solution is incorrect, then the slides must be shown with explanations.
I always show slide 8 after the student’s answer on card 1. On this slide, conclusions are important for the lesson.

Algorithm for solving systems graphically:

• Express y in terms of x in each equation of the system.
• Graph each equation of the system.
• Find the coordinates of the intersection points of the graphs.
• Carry out a check (I draw students’ attention to the fact that the graphical method usually gives an approximate solution, but if the intersection of the graphs hits a point with whole coordinates, you can check and get an exact answer).
• Write down the answer.

3. Exercises (Examination)- 5 minutes.

Yesterday, serious mistakes were made in the work of some employees. Today you are already more competent in the matter of graphic solutions. You are invited to conduct an examination of the proposed solutions, i.e. find errors in solutions. Slide 10 is shown.
Work is going on in departments. (Photocopies of assignments with errors are given to each desk; in each department, employees must find errors and highlight them or correct them; photocopies must be handed over to the senior researcher, i.e. teacher). The boss adds 2 points to those who find and correct the mistake. Then we discuss the mistakes made and indicate them on slide 10.

Error 1

Solve the system of equations:

Answer: there are no solutions.

Students must continue the lines until they intersect and get the answer: (– 2; 1).

Error 2.

Solve the system of equations:

Answer: (1; 4).

Students must find the error in the transformation of the first equation and correct it on the finished drawing. Get another answer: (2; 5).

4. Explaining new material (Research and discovery)– 12 min.

I suggest that students solve three systems graphically. Each student solves independently in a notebook. Only those with conditional clearance can consult.

Solution

Without drawing graphs, it is clear that the straight lines will coincide.

Slide 11 shows the systems solution; It is expected that students will have difficulty writing down the answer in example 3. After working in the departments, we check the solution (the boss adds 2 points for a correct one). Now it's time to discuss how many solutions a system of two linear equations can have.
Students must draw conclusions on their own and explain them, listing the cases of the relative positions of lines on a plane (slide 12).

5. Creative project (Exercises)– 12 min.

The task is given for the department. The boss gives each laboratory assistant, according to his abilities, a fragment of his performance.

Solve systems of equations graphically:

After opening the brackets, students should receive the system:

After opening the parentheses, the first equation looks like: y = 2/3x + 4.

6. Report (checking the completion of the task)- 2 minutes.

After completing a creative project, students turn in their notebooks. On slide 13 I show what should have happened. The bosses hand over the table. The last column is filled out by the teacher and marked (the marks can be communicated to the students at the next lesson). In the project, the solution to the first system is assessed with three points, and the second – with four.

7. Planning (summarizing and homework)- 2 minutes.

Let's summarize our work. We did a good job. We’ll talk specifically about the results tomorrow at the planning meeting. Of course, all laboratory assistants, without exception, mastered the graphical method of solving systems of equations and learned how many solutions a system can have. Tomorrow each of you will have a personal project. For additional preparation: paragraph 36; 647-649(2); repeat analytical methods for solving systems. 649(2) and solve analytically.

Our work was supervised throughout the day by the director of the laboratory, Nouman Nou Manovich. He has the floor. (Showing the final slide).

Approximate grading scale

Mark	Tolerance	Expertise	Study	Project	Total
3	5	2	2	2	11
4	7	2	4	3	16
5	9	3	5	4	21

Video lesson “Graphical method for solving systems of equations” presents educational material to master this topic. Material contains general concept about solving a system of equations, as well as a detailed explanation using an example of how a system of equations is solved graphically.

The visual aid uses animation to make it more user-friendly and clear implementation constructions, as well as different ways highlighting important concepts and details for an in-depth understanding of the material and better memorization.

The video lesson begins by introducing the topic. Students are reminded what a system of equations is and what systems of equations they were already familiar with in 7th grade. Previously, students had to solve systems of equations of the form ax+by=c. Deepening the concept of solving systems of equations and in order to develop the ability to solve them, this video lesson examines the solution of a system consisting of two equations of the second degree, as well as one equation of the second degree, and the second of the first degree. We are reminded of what solving a system of equations is. The definition of a solution to a system as a pair of values ​​of variables that reverse its equations when substituted into a correct equality is displayed on the screen. In accordance with the definition of the system solution, the task is specified. It is displayed on the screen to remember that solving a system means finding suitable solutions or prove their absence.

It is proposed to master a graphical method for solving a certain system of equations. Application this method is considered using the example of solving a system consisting of the equations x 2 +y 2 =16 and y=-x 2 +2x+4. The graphical solution of the system begins with plotting each of these equations. Obviously, the graph of the equation x 2 + y 2 = 16 will be a circle. The points belonging to a given circle are the solution to the equation. Next to the equation, a circle of radius 4 with center O at the origin is constructed on the coordinate plane. The graph of the second equation is a parabola, the branches of which are lowered down. This parabola corresponding to the graph of the equation is constructed on the coordinate plane. Any point belonging to a parabola represents a solution to the equation y = -x 2 + 2x + 4. It is explained that the solution to a system of equations is points on the graphs that simultaneously belong to the graphs of both equations. This means that the intersection points of the constructed graphs will be solutions to the system of equations.

It is noted that the graphical method consists of finding the approximate value of the coordinates of points located at the intersection of two graphs, which reflect the set of solutions to each equation of the system. The figure shows the coordinates of the found intersection points of the two graphs: A, B, C, D[-2;-3.5]. These points are solutions to a system of equations found graphically. You can check their correctness by substituting them into the equation and obtaining a fair equality. After substituting the points into the equation, it is clear that some of the points give the exact value of the solution, and some represent the approximate value of the solution to the equation: x 1 = 0, y 1 = 4; x 2 =2, y 2 ≈3.5; x 3 ≈3.5, y 3 = -2; x 4 = -2, y 4 ≈-3.5.

The video tutorial explains in detail the essence and application of the graphical method of solving a system of equations. This makes it possible to use it as a video tutorial in an algebra lesson at school when studying this topic. The material will also be useful for students to study independently and can help explain the topic during distance learning.