Which equation is called linear with one variable. Solution of linear equations with one variable. Solving complex linear equations
When solving linear equations, we strive to find a root, that is, a value for a variable that will turn the equation into a correct equality.
To find the root of the equation you need equivalent transformations bring the equation given to us to the form
\(x=[number]\)
This number will be the root.
That is, we transform the equation, making it easier with each step, until we reduce it to a completely primitive equation “x = number”, where the root is obvious. The most commonly used in solving linear equations are the following transformations:
For example: add \(5\) to both sides of the equation \(6x-5=1\)
\(6x-5=1\) \(|+5\)
\(6x-5+5=1+5\)
\(6x=6\)
Please note that we could get the same result faster - simply by writing the five on the other side of the equation and changing its sign in the process. Actually, this is exactly how the school “transfer through equals with a change of sign to the opposite” is done.
2. Multiplying or dividing both sides of an equation by the same number or expression.
For example: Divide the equation \(-2x=8\) by minus two
\(-2x=8\) \(|:(-2)\)
\(x=-4\)
Usually this step is done at the very end, when the equation has already been reduced to \(ax=b\), and we divide by \(a\) to remove it from the left.
3. Using the properties and laws of mathematics: opening brackets, reducing like terms, reducing fractions, etc.
Add \(2x\) left and right
Subtract \(24\) from both sides of the equation
Again, we present like terms
Now we divide the equation by \ (-3 \), thereby removing before the x on the left side.
Answer : \(7\)
Answer found. However, let's check it out. If the seven is really a root, then substituting it instead of x in the original equation should result in the correct equality - the same numbers on the left and right. We try.
Examination:
\(6(4-7)+7=3-2\cdot7\)
\(6\cdot(-3)+7=3-14\)
\(-18+7=-11\)
\(-11=-11\)
Agreed. This means that the seven is indeed the root of the original linear equation.
Do not be lazy to check the answers you found by substitution, especially if you are solving an equation on a test or exam.
The question remains - how to determine what to do with the equation at the next step? How exactly to convert it? Share something? Or subtract? And what exactly to subtract? What to share?
The answer is simple:
Your goal is to bring the equation to the form \(x=[number]\), that is, on the left x without coefficients and numbers, and on the right - only a number without variables. So see what's stopping you and do the opposite of what the interfering component does.
To understand this better, let's take a step-by-step solution to the linear equation \(x+3=13-4x\).
Let's think: how does this equation differ from \(x=[number]\)? What's stopping us? What's wrong?
Well, firstly, the triple interferes, since there should be only a lone X on the left, without numbers. And what does the trio do? Added to xx. So, to remove it - subtract the same trio. But if we subtract a triple from the left, then we must subtract it from the right so that the equality is not violated.
\(x+3=13-4x\) \(|-3\)
\(x+3-3=13-4x-3\)
\(x=10-4x\)
Good. Now what's stopping you? \(4x\) on the right, because it should only contain numbers. \(4x\) subtracted- remove adding.
\(x=10-4x\) \(|+4x\)
\(x+4x=10-4x+4x\)
Now we give like terms on the left and right.
It's almost ready. It remains to remove the five on the left. What is she doing"? multiplies on x. So we remove it division.
\(5x=10\) \(|:5\)
\(\frac(5x)(5)\) \(=\)\(\frac(10)(5)\)
\(x=2\)
The solution is complete, the root of the equation is two. You can check by substitution.
notice, that most often there is only one root in linear equations. However, two special cases may occur.
Special case 1 - there are no roots in a linear equation.
Example . Solve the equation \(3x-1=2(x+3)+x\)
Solution :
Answer : no roots.
In fact, the fact that we will come to such a result was seen earlier, even when we got \(3x-1=3x+6\). Think about it: how can \(3x\) be equal, from which \(1\) was subtracted, and \(3x\) to which \(6\) was added? Obviously, no way, because they did different actions with the same thing! It is clear that the results will vary.
Special case 2 - a linear equation has an infinite number of roots.
Example . Solve the linear equation \(8(x+2)-4=12x-4(x-3)\)Solution :
Answer : any number.
By the way, this was noticeable even earlier, at the stage: \(8x+12=8x+12\). Indeed, left and right are the same expressions. Whatever x you substitute, there will be the same number both there and there.
More complex linear equations.
The original equation does not always immediately look like a linear one, sometimes it is “disguised” as other, more complex equations. However, in the process of transformation, the masking subsides.
Example . Find the root of the equation \(2x^(2)-(x-4)^(2)=(3+x)^(2)-15\)
Solution :
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\(2x^(2)-(x-4)^(2)=(3+x)^(2)-15\) |
It would seem that there is an x squared here - this is not a linear equation! But don't rush. Let's Apply |
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\(2x^(2)-(x^(2)-8x+16)=9+6x+x^(2)-15\) |
Why is the result of expansion \((x-4)^(2)\) in parentheses, but the result of \((3+x)^(2)\) is not? Because there is a minus before the first square, which will change all the signs. And in order not to forget about it, we take the result in brackets, which we now open. |
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\(2x^(2)-x^(2)+8x-16=9+6x+x^(2)-15\) |
We give like terms |
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\(x^(2)+8x-16=x^(2)+6x-6\) |
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\(x^(2)-x^(2)+8x-6x=-6+16\) |
Again, here are similar ones. |
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Like this. It turns out that the original equation is quite linear, and x squared is nothing more than a screen to confuse us. :) We complete the solution by dividing the equation by \(2\), and we get the answer. |
Answer : \(x=5\)
Example . Solve the linear equation \(\frac(x+2)(2)\) \(-\) \(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6 )\)
Solution :
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\(\frac(x+2)(2)\) \(-\) \(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6)\) |
The equation does not look like a linear one, some fractions ... However, let's get rid of the denominators by multiplying both parts of the equation by the common denominator of all - six |
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\(6\cdot\)\((\frac(x+2)(2)\) \(-\) \(\frac(1)(3))\) \(=\) \(\frac( 9+7x)(6)\)\(\cdot 6\) |
Open bracket on the left |
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\(6\cdot\)\(\frac(x+2)(2)\) \(-\) \(6\cdot\)\(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6)\) \(\cdot 6\) |
Now we reduce the denominators |
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\(3(x+2)-2=9+7x\) |
Now it looks like a regular linear one! Let's solve it. |
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By transferring through equals, we collect x's on the right, and numbers on the left |
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Well, dividing by \ (-4 \) the right and left parts, we get the answer |
Answer : \(x=-1.25\)
Equality with variable f(x) = g(x) is called an equation with one variable x. Any value of the variable at which f(x) and g(x) take equal numerical values is called the root of such an equation. Therefore, to solve an equation means to find all the roots of the equation or to prove that there are none.
The equation x 2 + 1 \u003d 0 does not have real roots, but has imaginary roots: in this case, these are the roots x 1 \u003d i, x 2 \u003d -i. In what follows, we will only be interested in the real roots of the equation.
If the equations have the same roots, then they are called equivalent. Those equations that have no roots are equivalent.
Let's determine if the equations are equivalent:
a) x + 2 = 5 and x + 5 = 8
1. Solve the first equation
2. Solve the second equation
The roots of the equations are the same, so x + 2 = 5 and x + 5 = 8 are equivalent.
b) x 2 + 1 = 0 and 2x 2 + 5 = 0
Both of these equations do not have real roots, therefore they are equivalent.
c) x - 5 \u003d 1 and x 2 \u003d 36
1. Find the roots of the first equation
2. Find the roots of the second equation
x 1 = 6, x 2 = -6
The roots of the equations do not match, so x - 5 \u003d 1 and x 2 \u003d 36 are not equivalent.
When solving an equation, they try to replace it with an equivalent, but simpler equation. Therefore, it is important to know, as a result of what transformations, this equation turns into an equation equivalent to it.
Theorem 1. If any term in an equation is transferred from one part to another, changing the sign, then an equation equivalent to the given one will be obtained.
For example, the equation x 2 + 2 = 3x is equivalent to the equation x 2 + 2 - 3x = 0.
Theorem 2. If both parts of the equation are multiplied or divided by the same number (not equal to zero), then an equation is obtained that is equivalent to the given one.
For example, the equation (x 2 - 1) / 3 \u003d 2x is equivalent to the equation x 2 - 1 \u003d 6x. We multiply both sides of the first equation by 3.
A linear equation with one variable is an equation of the form ax \u003d b, where a and b are real numbers, and a is called the coefficient of the variable, and b is the free term.
Consider three cases for the linear equation ax = b.
1. a ≠ 0. In this case, x \u003d b / a (because a is non-zero).
2. a \u003d 0, b \u003d 0. The equation will take the form: 0 ∙ x \u003d 0. This equation is true for any x, i.e. the root of the equation is any real number.
3. a \u003d 0, b ≠ 0. In this case, the equation will not have roots, because division by zero is prohibited (0 ∙ x = b).
As a result of transformations, many equations are reduced to linear ones.
Solving Equations
a) (1/5) x + 2/15 = 0
1. Move the component 2/15 from the left side of the equation to the right side with the opposite sign. Such a transformation is governed by Theorem 1. So, the equation will take the form: (1/5)x = -2/15.
2. To get rid of the denominator, we multiply both sides of the equation by 15. Theorem 2 allows us to do this. So, the equation will take the form:
(1/5)x ∙ 15= - 2/15 ∙ 15
Thus, the root of the equation is -2/3.
b) 2/3 + x / 4 + (1 - x) / 6 \u003d 5x / 12 - 1
1. To get rid of the denominator, we multiply both parts of the equation 
on 12 (by Theorem 2). The equation will take the form:
12(2/3 + x/4 + (1 - x)/6) = 12(5x/12 - 1)
8 + 3x + 2 - 2x \u003d 5x - 12
10 + x = 5x - 12
2. Using Theorem 1, we “collect” all the numbers on the right, and the components with x on the left. The equation will take the form:
10 +12 \u003d 5x - x
Thus, the root of the equation is 5.5.
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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. 13) 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
Solution
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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And so on, it is logical to get acquainted with equations of other types. Next in line are linear equations, the purposeful study of which begins in algebra lessons in grade 7.
It is clear that first you need to explain what a linear equation is, give a definition of a linear equation, its coefficients, show it general form. Then you can figure out how many solutions a linear equation has depending on the values of the coefficients, and how the roots are found. This will allow you to move on to solving examples, and thereby consolidate the studied theory. In this article we will do this: we will dwell in detail on all theoretical and practical points regarding linear equations and their solution.
Let's say right away that here we will consider only linear equations with one variable, and in a separate article we will study the principles of solving linear equations in two variables.
Page navigation.
What is a linear equation?
The definition of a linear equation is given by the form of its notation. Moreover, in different textbooks of mathematics and algebra, the formulations of the definitions of linear equations have some differences that do not affect the essence of the issue.
For example, in an algebra textbook for grade 7 by Yu. N. Makarycheva and others, a linear equation is defined as follows:
Definition.
Type equation ax=b, where x is a variable, a and b are some numbers, is called linear equation with one variable.
Let us give examples of linear equations corresponding to the voiced definition. For example, 5 x=10 is a linear equation with one variable x , here the coefficient a is 5 , and the number b is 10 . Another example: −2.3 y=0 is also a linear equation, but with the variable y , where a=−2.3 and b=0 . And in the linear equations x=−2 and −x=3.33 a are not explicitly present and are equal to 1 and −1, respectively, while in the first equation b=−2 and in the second - b=3.33 .
A year earlier, in the textbook of mathematics by N. Ya. Vilenkin, linear equations with one unknown, in addition to equations of the form a x = b, were also considered equations that can be reduced to this form by transferring terms from one part of the equation to another with the opposite sign, as well as by reducing like terms. According to this definition, equations of the form 5 x=2 x+6 , etc. are also linear.
In turn, the following definition is given in the algebra textbook for 7 classes by A. G. Mordkovich:
Definition.
Linear equation with one variable x is an equation of the form a x+b=0 , where a and b are some numbers, called the coefficients of the linear equation.
For example, linear equations of this kind are 2 x−12=0, here the coefficient a is equal to 2, and b is equal to −12, and 0.2 y+4.6=0 with coefficients a=0.2 and b =4.6. But at the same time, there are examples of linear equations that have the form not a x+b=0 , but a x=b , for example, 3 x=12 .
Let's, so that we do not have any discrepancies in the future, under a linear equation with one variable x and coefficients a and b we will understand an equation of the form a x+b=0 . This type of linear equation seems to be the most justified, since linear equations are algebraic equations first degree. And all the other equations indicated above, as well as equations that are reduced to the form a x+b=0 with the help of equivalent transformations, will be called equations reducing to linear equations. With this approach, the equation 2 x+6=0 is a linear equation, and 2 x=−6 , 4+25 y=6+24 y , 4 (x+5)=12, etc. are linear equations.
How to solve linear equations?
Now it's time to figure out how the linear equations a x+b=0 are solved. In other words, it's time to find out if the linear equation has roots, and if so, how many and how to find them.
The presence of roots of a linear equation depends on the values of the coefficients a and b. In this case, the linear equation a x+b=0 has
- the only root at a≠0 ,
- has no roots for a=0 and b≠0 ,
- has infinitely many roots for a=0 and b=0 , in which case any number is a root of a linear equation.
Let us explain how these results were obtained.
We know that in order to solve equations, it is possible to pass from the original equation to equivalent equations, that is, to equations with the same roots or, like the original one, without roots. To do this, you can use the following equivalent transformations:
- transfer of a term from one part of the equation to another with the opposite sign,
- and also multiplying or dividing both sides of the equation by the same non-zero number.
So, in a linear equation with one variable of the form a x+b=0, we can move the term b from the left side to the right side with the opposite sign. In this case, the equation will take the form a x=−b.
And then the division of both parts of the equation by the number a suggests itself. But there is one thing: the number a can be equal to zero, in which case such a division is impossible. To deal with this problem, we will first assume that the number a is different from zero, and consider the case of zero a separately a bit later.
So, when a is not equal to zero, then we can divide both parts of the equation a x=−b by a , after that it is converted to the form x=(−b):a , this result can be written using a solid line as .
Thus, for a≠0, the linear equation a·x+b=0 is equivalent to the equation , from which its root is visible.
It is easy to show that this root is unique, that is, the linear equation has no other roots. This allows you to do the opposite method.
Let's denote the root as x 1 . Suppose that there is another root of the linear equation, which we denote x 2, and x 2 ≠ x 1, which, due to definitions of equal numbers through the difference is equivalent to the condition x 1 − x 2 ≠0 . Since x 1 and x 2 are the roots of the linear equation a x+b=0, then the numerical equalities a x 1 +b=0 and a x 2 +b=0 take place. We can subtract the corresponding parts of these equalities, which the properties of numerical equalities allow us to do, we have a x 1 +b−(a x 2 +b)=0−0 , whence a (x 1 −x 2)+( b−b)=0 and then a (x 1 − x 2)=0 . And this equality is impossible, since both a≠0 and x 1 − x 2 ≠0. So we have come to a contradiction, which proves the uniqueness of the root of the linear equation a·x+b=0 for a≠0 .
So we have solved the linear equation a x+b=0 with a≠0 . The first result given at the beginning of this subsection is justified. There are two more that meet the condition a=0 .
For a=0 the linear equation a·x+b=0 becomes 0·x+b=0 . From this equation and the property of multiplying numbers by zero, it follows that no matter what number we take as x, when we substitute it into the equation 0 x+b=0, we get the numerical equality b=0. This equality is true when b=0 , and in other cases when b≠0 this equality is false.
Therefore, for a=0 and b=0, any number is the root of the linear equation a x+b=0, since under these conditions, substituting any number instead of x gives the correct numerical equality 0=0. And for a=0 and b≠0, the linear equation a x+b=0 has no roots, since under these conditions, substituting any number instead of x leads to an incorrect numerical equality b=0.
The above justifications make it possible to form a sequence of actions that allows solving any linear equation. So, algorithm for solving a linear equation is:
- First, by writing a linear equation, we find the values of the coefficients a and b.
- If a=0 and b=0 , then this equation has infinitely many roots, namely, any number is a root of this linear equation.
- If a is different from zero, then
- the coefficient b is transferred to the right side with the opposite sign, while the linear equation is transformed to the form a x=−b ,
- after which both parts of the resulting equation are divided by a non-zero number a, which gives the desired root of the original linear equation.
The written algorithm is an exhaustive answer to the question of how to solve linear equations.
In conclusion of this paragraph, it is worth saying that a similar algorithm is used to solve equations of the form a x=b. Its difference lies in the fact that when a≠0, both parts of the equation are immediately divided by this number, here b is already in the desired part of the equation and it does not need to be transferred.
To solve equations of the form a x=b, the following algorithm is used:
- If a=0 and b=0 , then the equation has infinitely many roots, which are any numbers.
- If a=0 and b≠0 , then the original equation has no roots.
- If a is non-zero, then both sides of the equation are divided by a non-zero number a, from which the only root of the equation equal to b / a is found.
Examples of solving linear equations
Let's move on to practice. Let us analyze how the algorithm for solving linear equations is applied. Let us present solutions of typical examples corresponding to different values of the coefficients of linear equations.
Example.
Solve the linear equation 0 x−0=0 .
Solution.
In this linear equation, a=0 and b=−0 , which is the same as b=0 . Therefore, this equation has infinitely many roots, any number is the root of this equation.
Answer:
x is any number.
Example.
Does the linear equation 0 x+2.7=0 have solutions?
Solution.
In this case, the coefficient a is equal to zero, and the coefficient b of this linear equation is equal to 2.7, that is, it is different from zero. Therefore, the linear equation has no roots.
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:
- Open brackets, if any;
- Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
- Bring like terms to the left and right of the equal sign;
- 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:
- 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.
- 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:
- First of all, you need to open the parentheses, if any (as in our last example);
- Then bring similar
- 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:
- Expand the parentheses, if any.
- Seclude variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
- We present similar terms.
- 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 carry out last step- 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:
- Open brackets.
- Separate variables.
- Bring similar.
- 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:
- Get rid of fractions.
- Open brackets.
- Separate variables.
- Bring similar.
- 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:
\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot four\]
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!
