Solve the equation in steps. Solve the equation with fractions online. Solving real examples of simple linear equations

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.

To begin with, let's define: what is a linear equation and which of them should be called the simplest?

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

The simplest equation means the construction:

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

1. Open brackets, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Bring like 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 you get something like $0\cdot x=8$, i.e. on the left is zero, and on the right is a non-zero number. 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.

And now let's see how it all works on the example of real problems.

Examples of solving equations

Today we deal 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 open the parentheses, if any (as in our last example);
2. Then bring similar
3. Finally, isolate the variable, i.e. everything that is connected with the variable - the terms in which it is contained - is transferred to one side, and everything that remains without it is transferred to the other side.

Then, as a rule, you need to bring similar on each side of the resulting equality, and after that it remains only to divide by the coefficient at "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. Usually, mistakes are made either when opening brackets, or when counting "pluses" and "minuses".

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

Scheme for solving simple linear equations

To begin with, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the parentheses, if any.
2. Seclude variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
3. We present similar terms.
4. We divide everything by the coefficient at "x".

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

Solving real examples of simple linear equations

Task #1

In the first step, we are required 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:

We give similar terms on the left and on the right, but this has already been done here. Therefore, we proceed to the fourth step: divide by a factor:

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

Here we got the answer.

Task #2

In this task, we can observe the brackets, so let's expand them:

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

Here are some like:

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

Task #3

The third linear equation is already 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 just have different signs in front of them. Let's break them down:

We perform the second step already known to us:

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

Let's calculate:

We perform the last step - we divide everything by the coefficient at "x":

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

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, then 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, zero can get in among them - there is nothing wrong with that.

Zero is the same number as the rest, you should not somehow discriminate it or assume that if you get zero, then you did something wrong.

Another feature is related to the expansion of parentheses. Please note: when there is a “minus” in front of them, we remove it, but in brackets we change the signs to opposite. And then we can open it according to 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 actions is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complicated and a quadratic function will appear when performing various transformations. However, you should not be afraid of this, because if, according to the author's intention, we solve a linear equation, then in the process of transformation all monomials containing a quadratic function will necessarily be reduced.

Example #1

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

Now let's take privacy:

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

Here are some like:

Obviously, this equation has no solutions, so in the answer we write as follows:

\[\variety \]

or no roots.

Example #2

We perform the same steps. First step:

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

Here are some like:

Obviously, this linear equation has no solution, so we write it like this:

\[\varnothing\],

or no roots.

Nuances of the solution

Both equations are completely solved. On the example of these two expressions, we once again made sure that even in the simplest linear equations, everything can be not so simple: there can be either one, or none, or infinitely many. In our case, we considered two equations, in both there are simply no roots.

But I would like to draw your attention to another fact: how to work with brackets and how to expand 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: multiply each individual term. Inside there are two terms - respectively, two terms and is multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can the bracket be opened from the point of view that there is a minus sign after it. Yes, yes: only now, when the transformations are done, we remember that there is a minus sign in front of the brackets, which means that everything below just 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 no coincidence 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 learn to solve such simple equations again.

Of course, the day will come when you will hone these skills to automatism. You no longer have to perform so many transformations each time, you will write everything in 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 #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 a retreat:

Here are some like:

Let's do 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, however, they mutually canceled out, which makes the equation exactly linear, not square.

Task #2

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

Let's do the first step carefully: multiply every element in the first bracket by every element in the second. In total, four new terms should be obtained after transformations:

And now carefully perform the multiplication in each term:

Let's move the terms with "x" to the left, and without - to the right:

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

Here are similar terms:

We have received a definitive answer.

Nuances of the solution

The most important remark about these two equations is this: as soon as we start multiplying brackets in which there is more than a term, then 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 get four terms.

On the algebraic sum

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

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

Solving equations with a fraction

To solve such tasks, one more step will have to be added to our algorithm. But first, I will remind our algorithm:

1. Open brackets.
2. Separate variables.
3. Bring similar.
4. Divide by a factor.

Alas, this wonderful algorithm, for all its efficiency, is not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on the left and on 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 performed both before the first action and after it, namely, to get rid of fractions. Thus, the algorithm will be as follows:

1. Get rid of fractions.
2. Open brackets.
3. Separate variables.
4. Bring similar.
5. Divide by a factor.

What does it mean to "get rid of fractions"? And why is it possible to do this both after and before the first standard step? In fact, in our case, all fractions are numeric in terms of the denominator, i.e. everywhere the denominator is just a number. Therefore, if we multiply both parts of the equation by this number, then we will get rid of fractions.

Example #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 brackets doesn't mean you have to multiply each of them by "four". Let's write:

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

Now let's open it:

We perform seclusion of a variable:

We carry out the reduction of similar terms:

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

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

We have received the final solution, we pass to the second equation.

Example #2

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

Here we perform all the same actions:

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

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

Problem solved.

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

Key points

The key findings are as follows:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Do not worry if you have quadratic functions somewhere, most likely, in the process of further transformations, they will be reduced.
• The roots in linear equations, even the simplest ones, are of three types: one single root, the entire number line is a root, there are 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, solve the examples presented there. Stay tuned, there are many more interesting things waiting for you!

The online fraction calculator allows you to perform simple arithmetic operations with fractions: addition of fractions, subtraction of fractions, multiplication of fractions, division of fractions. To make calculations, fill in the fields corresponding to the numerators and denominators of two fractions.

Fraction in mathematics a number representing a part of a unit or several of its parts is called.

A common fraction is written as two numbers, usually separated by a horizontal line, indicating the division sign. The number above the bar is called the numerator. The number below the bar is called the denominator. The denominator of a fraction shows the number of equal parts into which the whole is divided, and the numerator of the fraction shows the number of these parts of the whole taken.

Fractions are right and wrong.

• A correct fraction is one whose numerator is less than the denominator.
• An improper fraction is when the numerator is greater than the denominator.

A mixed fraction is a fraction written as a whole number and a proper fraction, and is understood as the sum of this number and the fractional part. Accordingly, a fraction that does not have an integer part is called a simple fraction. Any mixed fraction can be converted to an improper simple fraction.

In order to convert a mixed fraction into an ordinary one, it is necessary to add the product of the integer part and the denominator to the numerator of the fraction:

How to convert an ordinary fraction to a mixed one

In order to convert an ordinary fraction to a mixed one, you must:

1. Divide the numerator of a fraction by its denominator
2. The result of the division will be the integer part
3. The remainder of the branch will be the numerator

How to convert a common fraction to a decimal

To convert a fraction to a decimal, you need to divide its numerator by the denominator.

In order to convert a decimal to a common fraction, you must:

How to convert a fraction to a percentage

In order to convert an ordinary or mixed fraction to a percentage, you need to convert it to a decimal fraction and multiply by 100.

How to convert percentages to fractions

In order to convert percentages to fractions, it is necessary to obtain a decimal fraction from percents (dividing by 100), then convert the resulting decimal fraction to an ordinary one.

Addition of fractions

The algorithm for adding two fractions is as follows:

1. Add fractions by adding their numerators.

Subtraction of fractions

Algorithm of actions when subtracting two fractions:

1. Convert mixed fractions to common fractions (get rid of the integer part).
2. Bring fractions to a common denominator. To do this, you need to multiply the numerator and denominator of the first fraction by the denominator of the second fraction, and multiply the numerator and denominator of the second fraction by the denominator of the first fraction.
3. Subtract one fraction from another by subtracting the numerator of the second fraction from the numerator of the first.
4. Find the greatest common divisor (GCD) of the numerator and denominator and reduce the fraction by dividing the numerator and denominator by the GCD.
5. If the numerator of the final fraction is greater than the denominator, then select the whole part.

Multiplication of fractions

Algorithm of actions when multiplying two fractions:

1. Convert mixed fractions to common fractions (get rid of the integer part).
2. Find the greatest common divisor (GCD) of the numerator and denominator and reduce the fraction by dividing the numerator and denominator by the GCD.
3. If the numerator of the final fraction is greater than the denominator, then select the whole part.

Division of fractions

Algorithm of actions when dividing two fractions:

1. Convert mixed fractions to common fractions (get rid of the integer part).
2. To divide fractions, you need to convert the second fraction by swapping its numerator and denominator, and then multiply the fractions.
3. Multiply the numerator of the first fraction by the numerator of the second fraction and the denominator of the first fraction by the denominator of the second.
4. Find the greatest common divisor (GCD) of the numerator and denominator and reduce the fraction by dividing the numerator and denominator by the GCD.
5. If the numerator of the final fraction is greater than the denominator, then select the whole part.

Online calculators and converters:

An equation with one unknown, which, after opening the brackets and reducing like terms, takes the form

ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.

For example, all equations:

2x + 3 \u003d 7 - 0.5x; 0.3x = 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.

The value of the unknown that turns the equation into a true equality is called decision or the root of the equation .

For example, if in the equation 3x + 7 \u003d 13 we substitute the number 2 instead of the unknown x, then we get the correct equality 3 2 + 7 \u003d 13. Hence, the value x \u003d 2 is the solution or the root of the equation.

And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 2 + 7 ≠ 13. Therefore, the value x \u003d 3 is not a solution or a root of the equation.

The solution of any linear equations is reduced to the solution of equations of the form

ax + b = 0.

We transfer the free term from the left side of the equation to the right, while changing the sign in front of b to the opposite, we get

If a ≠ 0, then x = – b/a .

Example 1 Solve the equation 3x + 2 =11.

We transfer 2 from the left side of the equation to the right, while changing the sign in front of 2 to the opposite, we get
3x \u003d 11 - 2.

Let's do the subtraction, then
3x = 9.

To find x, you need to divide the product by a known factor, that is,
x = 9:3.

So the value x = 3 is the solution or the root of the equation.

Answer: x = 3.

If a = 0 and b = 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0, we get 0, but b is also 0. The solution to this equation is any number.

Example 2 Solve the equation 5(x - 3) + 2 = 3 (x - 4) + 2x - 1.

Let's expand the brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.

5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.

Here are similar members:
0x = 0.

Answer: x is any number.

If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when multiplying any number by 0, we get 0, but b ≠ 0.

Example 3 Solve the equation x + 8 = x + 5.

Let us group the terms containing unknowns on the left side, and the free terms on the right side:
x - x \u003d 5 - 8.

Here are similar members:
0x = - 3.

Answer: no solutions.

On the figure 1 the scheme for solving the linear equation is shown

Let us compose a general scheme for solving equations with one variable. Consider the solution of example 4.

Example 4 Let's solve the equation

1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.

2) After reduction we get
4 (x - 4) + 3 2 (x + 1) - 12 = 6 5 (x - 3) + 24x - 2 (11x + 43)

3) To separate members containing unknown and free members, open the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.

4) We group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.

5) Here are similar members:
- 22x = - 154.

6) Divide by - 22 , We get
x = 7.

As you can see, the root of the equation is seven.

In general, such equations can be solved as follows:

a) bring the equation to an integer form;

b) open brackets;

c) group the terms containing the unknown in one part of the equation, and the free terms in the other;

d) bring similar members;

e) solve an equation of the form aх = b, which was obtained after bringing like terms.

However, this scheme is not required for every equation. When solving many simpler equations, one has to start not from the first, but from the second ( Example. 2), third ( Example. thirteen) and even from the fifth stage, as in example 5.

Example 5 Solve the equation 2x = 1/4.

We find the unknown x \u003d 1/4: 2,
x = 1/8 .

Consider the solution of some linear equations encountered in the main state exam.

Example 6 Solve equation 2 (x + 3) = 5 - 6x.

2x + 6 = 5 - 6x

2x + 6x = 5 - 6

Answer: - 0.125

Example 7 Solve the equation - 6 (5 - 3x) \u003d 8x - 7.

– 30 + 18x = 8x – 7

18x - 8x = - 7 +30

Answer: 2.3

Example 8 Solve the Equation

3(3x - 4) = 4 7x + 24

9x - 12 = 28x + 24

9x - 28x = 24 + 12

Example 9 Find f(6) if f (x + 2) = 3 7's

Decision

Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.

We solve the linear equation x + 2 = 6,
we get x \u003d 6 - 2, x \u003d 4.

If x = 4 then
f(6) = 3 7-4 = 3 3 = 27

Answer: 27.

If you still have questions, there is a desire to deal with the solution of equations more thoroughly, sign up for my lessons in the SCHEDULE. I will be glad to help you!

TutorOnline also recommends watching a new video tutorial from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.

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What are irrational equations and how to solve them

Equations in which the variable is contained under the sign of the radical or under the sign of raising to a fractional power are called irrational. When we deal with a fractional power, we deprive ourselves of many mathematical operations for solving the equation, so irrational equations are solved in a special way.

Irrational equations are usually solved by raising both sides of the equation to the same power. At the same time, raising both parts of the equation to the same odd power is an equivalent transformation of the equation, and to an even one is not equivalent. Such a difference is obtained due to such features of exponentiation, such as if raised to an even power, then negative values ​​\u200b\u200bare “lost”.

The point of raising both sides of an irrational equation to a power is to get rid of "irrationality". Thus, we need to raise both parts of the irrational equation to such a degree that all fractional powers of both parts of the equation turn into integers. After that, you can look for a solution to this equation, which will coincide with the solutions of the irrational equation, with the difference that in the case of raising to an even power, the sign is lost and the final solutions will require verification and not all will work.

Thus, the main difficulty is associated with raising both parts of the equation to the same even power - due to the non-equivalence of the transformation, extraneous roots may appear. Therefore, it is obligatory to check all found roots. Checking the found roots is most often forgotten by those who solve an irrational equation. It is also not always clear to what degree it is necessary to raise an irrational equation in order to get rid of irrationality and solve it. Our smart calculator is just designed to solve an irrational equation and automatically check all the roots, which will eliminate forgetfulness.

Free online calculator of irrational equations

Our free solver will allow you to solve an irrational equation online of any complexity in a matter of seconds. All you need to do is just enter your data into the calculator. You can also learn how to solve the equation on our website. And if you have any questions, you can ask them in our VKontakte group.

Service assignment. The matrix calculator is designed to solve systems of linear equations in a matrix way (see an example of solving similar problems).

Instruction. For an online solution, you must select the type of equation and set the dimension of the corresponding matrices. where A, B, C are given matrices, X is the desired matrix. Matrix equations of the form (1), (2) and (3) are solved through the inverse matrix A -1 . If the expression A X - B = C is given, then it is necessary to first add the matrices C + B and find a solution for the expression A X = D , where D = C + B . If the expression A*X = B 2 is given, then the matrix B must first be squared.

It is also recommended to familiarize yourself with the basic operations on matrices.

Example #1. Exercise. Find a solution to a matrix equation
Decision. Denote:
Then the matrix equation will be written in the form: A·X·B = C.
The determinant of matrix A is detA=-1
Since A is a nonsingular matrix, there is an inverse matrix A -1 . Multiply both sides of the equation on the left by A -1: Multiply both sides of this equation on the left by A -1 and on the right by B -1: A -1 A X B B -1 = A -1 C B -1 . Since A A -1 = B B -1 = E and E X = X E = X, then X = A -1 C B -1

Inverse matrix A -1:
Find the inverse matrix B -1 .
Transpose matrix B T:
Inverse matrix B -1:
We are looking for the matrix X by the formula: X = A -1 C B -1

Answer:

Example #2. Exercise. Solve matrix equation
Decision. Denote:
Then the matrix equation will be written in the form: A X = B.
The determinant of matrix A is detA=0
Since A is a degenerate matrix (the determinant is 0), therefore, the equation has no solution.

Example #3. Exercise. Find a solution to a matrix equation
Decision. Denote:
Then the matrix equation will be written in the form: X·A = B.
The determinant of matrix A is detA=-60
Since A is a nonsingular matrix, there is an inverse matrix A -1 . Multiply on the right both sides of the equation by A -1: X A A -1 = B A -1 , from which we find that X = B A -1
Find the inverse matrix A -1 .
Transposed matrix A T:
Inverse matrix A -1:
We are looking for the matrix X by the formula: X = B A -1


Answer: >