Proven by science: How to solve complex problems while half asleep. How to learn to solve simple and complex equations

52. More complex examples of equations.
Example 1.

5/(x – 1) – 3/(x + 1) = 15/(x 2 – 1)

The common denominator is x 2 – 1, since x 2 – 1 = (x + 1)(x – 1). Let's multiply both sides of this equation by x 2 – 1. We get:

or, after reduction,

5(x + 1) – 3(x – 1) = 15

5x + 5 – 3x + 3 = 15

2x = 7 and x = 3½

Let's consider another equation:

5/(x-1) – 3/(x+1) = 4(x 2 – 1)

Solving as above, we get:

5(x + 1) – 3(x – 1) = 4
5x + 5 – 3x – 3 = 4 or 2x = 2 and x = 1.

Let's see whether our equalities are justified if we replace x in each of the considered equations with the found number.

For the first example we get:

We see that there is no room for any doubts: we have found a number for x such that the required equality is justified.

For the second example we get:

5/(1-1) – 3/2 = 15/(1-1) or 5/0 – 3/2 = 15/0

Here doubts arise: we are faced with division by zero, which is impossible. If in the future we manage to give a certain, albeit indirect, meaning to this division, then we can agree that the found solution x – 1 satisfies our equation. Until then, we must admit that our equation does not have a solution that has a direct meaning.

Such cases can occur when the unknown is somehow included in the denominators of the fractions present in the equation, and some of these denominators, when the solution is found, turn to zero.

Example 2.

You can immediately see that this equation has the form of a proportion: the ratio of the number x + 3 to the number x – 1 is equal to the ratio of the number 2x + 3 to the number 2x – 2. Let someone, in view of this circumstance, decide to apply here to free the equation from fractions, the main property of proportion (the product of the extreme terms is equal to the product of the middle terms). Then he will get:

(x + 3) (2x – 2) = (2x + 3) (x – 1)

2x 2 + 6x – 2x – 6 = 2x 2 + 3x – 2x – 3.

Here, fears that we will not cope with this equation may be raised by the fact that the equation includes terms with x 2. However, we can subtract 2x 2 from both sides of the equation - this will not break the equation; then the terms with x 2 are destroyed and we get:

6x – 2x – 6 = 3x – 2x – 3

Let's move the unknown terms to the left and the known ones to the right - we get:

3x = 3 or x = 1

Remembering this equation

(x + 3)/(x – 1) = (2x + 3)/(2x – 2)

We will immediately notice that the found value for x (x = 1) makes the denominators of each fraction vanish; We must abandon such a solution until we have considered the question of division by zero.

If we also note that the application of the property of proportion has complicated the matter and that a simpler equation could be obtained by multiplying both sides of the given by a common denominator, namely 2(x – 1) - after all, 2x – 2 = 2 (x – 1) , then we get:

2(x + 3) = 2x – 3 or 2x + 6 = 2x – 3 or 6 = –3,

which is impossible.

This circumstance indicates that this equation does not have any solutions that have a direct meaning that would not turn the denominators of this equation to zero.
Let us now solve the equation:

(3x + 5)/(x – 1) = (2x + 18)/(2x – 2)

Let's multiply both sides of the equation 2(x – 1), i.e. by a common denominator, we get:

6x + 10 = 2x + 18

The found solution does not make the denominator vanish and has a direct meaning:

or 11 = 11

If someone, instead of multiplying both parts by 2(x – 1), were to use the property of proportion, they would get:

(3x + 5)(2x – 2) = (2x + 18)(x – 1) or
6x 2 + 4x – 10 = 2x 2 + 16x – 18.

Here the terms with x 2 would not be destroyed. Moving all the unknown terms to the left side, and the known ones to the right, we would get

4x 2 – 12x = –8

x 2 – 3x = –2

Now we will not be able to solve this equation. In the future, we will learn how to solve such equations and find two solutions for it: 1) you can take x = 2 and 2) you can take x = 1. It’s easy to check both solutions:

1) 2 2 – 3 2 = –2 and 2) 1 2 – 3 1 = –2

If we remember the initial equation

(3x + 5) / (x – 1) = (2x + 18) / (2x – 2),

then we will see that now we get both of its solutions: 1) x = 2 is the solution that has a direct meaning and does not turn the denominator to zero, 2) x = 1 is the solution that turns the denominator to zero and does not have a direct meaning .

Example 3.

Let’s find the common denominator of the fractions included in this equation by factoring each of the denominators:

1) x 2 – 5x + 6 = x 2 – 3x – 2x + 6 = x(x – 3) – 2(x – 3) = (x – 3)(x – 2),

2) x 2 – x – 2 = x 2 – 2x + x – 2 = x (x – 2) + (x – 2) = (x – 2)(x + 1),

3) x 2 – 2x – 3 = x 2 – 3x + x – 3 = x (x – 3) + (x – 3) = (x – 3) (x + 1).

The common denominator is (x – 3)(x – 2)(x + 1).

Let's multiply both sides of this equation (and we can now rewrite it as:

by a common denominator (x – 3) (x – 2) (x + 1). Then, after reducing each fraction we get:

3(x + 1) – 2(x – 3) = 2(x – 2) or
3x + 3 – 2x + 6 = 2x – 4.

From here we get:

–x = –13 and x = 13.

This solution has a direct meaning: it does not make any of the denominators vanish.

If we took the equation:

then, doing exactly the same as above, we would get

3(x + 1) – 2(x – 3) = x – 2

3x + 3 – 2x + 6 = x – 2

3x – 2x – x = –3 – 6 – 2,

where would you get it from?

which is impossible. This circumstance shows that it is impossible to find a solution for the last equation that has a direct meaning.

How to learn to solve simple and complex equations

Dear parents!

Without basic mathematical training, the education of a modern person is impossible. At school, mathematics serves as a supporting subject for many related disciplines. In post-school life, continuous education becomes a real necessity, which requires basic school-wide training, including mathematics.

In elementary school, not only knowledge on basic topics is laid, but also logical thinking, imagination and spatial concepts are developed, and interest in this subject is formed.

Observing the principle of continuity, we will focus on the most important topic, namely, “The relationship between the components of actions in solving compound equations.”

With this lesson you can easily learn how to solve complex equations. During the lesson you will learn in detail step-by-step instructions for solving complex equations.

Many parents are perplexed by the question of how to get their children to learn to solve simple and complex equations. If the equations are simple, that’s half the problem, but there are also complex ones - for example, integral ones. By the way, for information, there are also equations that the best minds of our planet are struggling to solve, and for the solution of which very significant monetary bonuses are given. For example, if you rememberPerelmanand an unclaimed cash bonus of several million.

However, let us first return to simple mathematical equations and repeat the types of equations and the names of the components. A little warm-up:

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WARM-UP

Find the extra number in each column:

2) What word is missing in each column?

3) Connect the words from the first column with the words from the 2nd column.

"Equation" "Equality"

4) How do you explain what “equality” is?

5) What about the “equation”? Is this equality? What's special about it?

sum term

minuend difference

subtractive product

factorequality

dividend

the equation

Conclusion: An equation is an equality with a variable whose value must be found.

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I invite each group to write equations on a piece of paper with a felt-tip pen: (on the board)

One of the group must read their equation in mathematical language and comment on their solution, i.e., speak out the operation being performed with the known components of the actions (algorithm).

Conclusion: We can solve simple equations of all types using an algorithm, read and write literal expressions.

I propose to solve a problem in which a new type of equation appears.

Conclusion: We got acquainted with the solution of equations, one of the parts of which contains a numerical expression, the value of which must be found and a simple equation must be obtained.

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Let's consider another version of the equation, the solution of which is reduced to solving a chain of simple equations. Here's one introduction to compound equations.

I suggest everyone write down a sentence in mathematical language:

The product of the difference between the numbers x and 4 and the number 3 is 15.

CONCLUSION: The problematic situation that has arisen motivates the setting of the lesson goal: to learn to solve equations in which the unknown component is an expression. Such equations are compound equations.

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Or maybe the types of equations we have already studied will help us? (algorithms)

Which of the famous equations is our equation similar to? X * a = b

VERY IMPORTANT QUESTION: What is the expression on the left side - sum, difference, product or quotient?

(x - 4) * 3 = 15 (Product)

Why? (since the last action is multiplication)

Conclusion:Such equations have not yet been considered. But we can solve it if the expressionx - 4put a card (y - igrek), and you get an equation that can be easily solved using a simple algorithm for finding the unknown component.

When solving compound equations, it is necessary at each step to select an action at an automated level, commenting and naming the components of the action.

(y - 5) * 4 = 28
y - 5 = 28: 4
y - 5 = 7
y = 5 +7
y = 12
(12 - 5) * 4 = 28
28 = 28 (i)

Conclusion:In classes with different backgrounds, this work can be organized differently. In more prepared classes, even for primary consolidation, expressions can be used in which not two, but three or more actions, but their solution requires a greater number of steps, with each step simplifying the equation until a simple equation is obtained. And each time you can observe how the unknown component of actions changes.

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CONCLUSION:

When we are talking about something very simple and understandable, we often say: “The matter is as clear as two and two are four!”

But before they figured out that two and two equal four, people had to study for many, many thousands of years.

Many rules from school textbooks on arithmetic and geometry were known to the ancient Greeks more than two thousand years ago.

Wherever you need to count, measure, compare something, you cannot do without mathematics.

It is difficult to imagine how people would live if they did not know how to count, measure, and compare. Mathematics teaches this.

Today you plunged into school life, played the role of students, and I invite you, dear parents, to rate your skills on a scale.

In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.

First, let's define: what is a linear equation and which one is called the simplest?

A linear equation is one in which there is only one variable, and only to the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest using the algorithm:

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Give similar terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$.

Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to expand the parentheses, if there are any (as in our last example);
2. Then combine similar
3. Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.

Then, as a rule, you need to bring similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.

In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.

Scheme for solving simple linear equations

First, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the brackets, if any.
2. We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
3. We present similar terms.
4. We divide everything by the coefficient of “x”.

Of course, this scheme does not always work; there are certain subtleties and tricks in it, and now we will get to know them.

Solving real examples of simple linear equations

Task No. 1

The first step requires us to open the brackets. But they are not in this example, so we skip this step. In the second step we need to isolate the variables. Please note: we are talking only about individual terms. Let's write it down:

We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:

\[\frac(6x)(6)=-\frac(72)(6)\]

So we got the answer.

Task No. 2

We can see the parentheses in this problem, so let's expand them:

Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:

Here are some similar ones:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task No. 3

The third linear equation is more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's do the math:

We carry out the last step - divide everything by the coefficient of “x”:

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, there may be zero among them - there is nothing wrong with that.

Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.

Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it using standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such things is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.

Example No. 1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take a look at privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some similar ones:

Obviously, this equation has no solutions, so we’ll write this in the answer:

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some similar ones:

Obviously, this linear equation has no solution, so we’ll write it this way:

\[\varnothing\],

or there are no roots.

Nuances of the solution

Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.

But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.

Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task No. 1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do some privacy:

Here are some similar ones:

Let's complete the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.

Task No. 2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:

Now let’s carefully perform the multiplication in each term:

Let’s move the terms with “X” to the left, and those without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.

About the algebraic sum

With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.

Solving equations with fractions

To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:

1. Open the brackets.
2. Separate variables.
3. Bring similar ones.
4. Divide by the ratio.

Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:

1. Get rid of fractions.
2. Open the brackets.
3. Separate variables.
4. Bring similar ones.
5. Divide by the ratio.

What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.

Example No. 1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, let's move on to the second equation.

Example No. 2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

The problem is solved.

That, in fact, is all I wanted to tell you today.

Key points

Key findings are:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Don't worry if you have quadratic functions somewhere; most likely, they will be reduced in the process of further transformations.
• There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!