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's compose general scheme solutions of 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 more simple equations you have 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,. 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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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 most simple 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 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:
- 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:
\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]
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!
One of the most important skills in admission to 5th grade is the ability to solve simple equations. Since the 5th grade is not so far from elementary school, then there are not so many types of equations that a student can solve. We will introduce you to all the main types of equations that you need to be able to solve if you want enroll in a physics and mathematics school.
1 type: "bulbous"
These are equations that you will almost certainly encounter when admission to any school or a 5th grade circle as a separate task. They are easy to distinguish from others: they contain a variable only once. For example, or.
They are solved very simply: you just need to "get" to the unknown, gradually "removing" everything superfluous that surrounds it - as if peeling an onion - hence the name. To solve it, it is enough to remember a few rules from the second class. Let's list them all:
Addition
- term1 + term2 = sum
- term1 = sum - term2
- term2 = sum - term1
Subtraction
- minuend - subtrahend = difference
- minuend = subtrahend + difference
- subtrahend = minuend - difference
Multiplication
- multiplier1 * multiplier2 = product
- multiplier1 = product: multiplier2
- multiplier2 = product: multiplier1
Division
- dividend: divisor = quotient
- dividend = divisor * quotient
- divisor = dividend: quotient
Let's look at an example of how to apply these rules.
Note that we share 
on and we get . In this situation, we know the divisor and the quotient. To find the dividend, you need to multiply the divisor by the quotient:
We got a little closer to ourselves. Now we see that to 
added and obtained. So, to find one of the terms, you need to subtract the known term from the sum:
And one more "layer" is removed from the unknown! Now we see a situation with a known value of the product () and one known multiplier ().
Now the situation is "reduced - subtracted = difference"
And the last step is the known product () and one of the factors () 
2 type: equations with brackets
Equations of this type are most often found in problems - 90% of all problems for admission to grade 5. Unlike "onion equations" the variable here can occur several times, so it is impossible to solve it using the methods from the previous paragraph. Typical equations: or
The main difficulty is to correctly open the brackets. After we managed to do this correctly, we should bring like terms (numbers to numbers, variables to variables), and after that we get the simplest "onion equation" which we can solve. But first things first.
Bracket expansion. We will give a few rules that should be used in this case. But, as practice shows, the student begins to correctly open the brackets only after 70-80 solved problems. The basic rule is this: any factor outside the brackets must be multiplied by each term inside the brackets. And the minus before the bracket changes the sign of all expressions that are inside. So, the basic rules of disclosure: 
Bringing similar. Everything is much easier here: by transferring the terms through the equal sign, you need to ensure that on the one hand there are only terms with the unknown, and on the other - only numbers. The basic rule is this: each term carried through changes its sign - if it was with, then it will become with, and vice versa. After a successful transfer, it is necessary to count the total number of unknowns, the final number on the other side of equality than the variables, and solve a simple "onion equation".
Parentheses are used to indicate the order in which operations are performed in numerical and literal expressions, as well as in expressions with variables. It is convenient to pass from an expression with brackets to an identically equal expression without brackets. This technique is called parenthesis opening.
To expand brackets means to rid the expression of these brackets.
Another point deserves special attention, which concerns the peculiarities of writing solutions when opening brackets. We can write the initial expression with brackets and the result obtained after opening the brackets as equality. For example, after opening the parentheses, instead of the expression
3−(5−7) we get the expression 3−5+7. We can write both of these expressions as the equality 3−(5−7)=3−5+7.
And one more important point. In mathematics, to reduce entries, it is customary not to write a plus sign if it is the first in an expression or in brackets. For example, if we add two positive numbers, for example, seven and three, then we write not +7 + 3, but simply 7 + 3, despite the fact that seven is also a positive number. Similarly, if you see, for example, the expression (5 + x) - know that there is a plus in front of the bracket, which is not written, and there is a plus + (+5 + x) in front of the five.
Bracket expansion rule for addition
When opening brackets, if there is a plus before the brackets, then this plus is omitted along with the brackets.
Example. Open the brackets in the expression 2 + (7 + 3) Before the brackets plus, then the characters in front of the numbers in the brackets do not change.
2 + (7 + 3) = 2 + 7 + 3
The rule for expanding brackets when subtracting
If there is a minus before the brackets, then this minus is omitted along with the brackets, but the terms that were in the brackets change their sign to the opposite. The absence of a sign before the first term in parentheses implies a + sign.
Example. Open brackets in expression 2 − (7 + 3)
There is a minus before the brackets, so you need to change the signs before the numbers from the brackets. There is no sign in brackets before the number 7, which means that the seven is positive, it is considered that the + sign is in front of it.
2 − (7 + 3) = 2 − (+ 7 + 3)
When opening the brackets, we remove the minus from the example, which was before the brackets, and the brackets themselves 2 − (+ 7 + 3), and change the signs that were in the brackets to the opposite ones.
2 − (+ 7 + 3) = 2 − 7 − 3
Expanding parentheses when multiplying
If there is a multiplication sign in front of the brackets, then each number inside the brackets is multiplied by the factor in front of the brackets. At the same time, multiplying a minus by a minus gives a plus, and multiplying a minus by a plus, like multiplying a plus by a minus, gives a minus.
Thus, parentheses in products are expanded in accordance with the distributive property of multiplication.
Example. 2 (9 - 7) = 2 9 - 2 7
When multiplying parenthesis by parenthesis, each term of the first parenthesis is multiplied with every term of the second parenthesis.
(2 + 3) (4 + 5) = 2 4 + 2 5 + 3 4 + 3 5
In fact, there is no need to remember all the rules, it is enough to remember only one, this one: c(a−b)=ca−cb. Why? Because if we substitute one instead of c, we get the rule (a−b)=a−b. And if we substitute minus one, we get the rule −(a−b)=−a+b. Well, if you substitute another bracket instead of c, you can get the last rule.
Expand parentheses when dividing
If there is a division sign after the brackets, then each number inside the brackets is divisible by the divisor after the brackets, and vice versa.
Example. (9 + 6) : 3=9: 3 + 6: 3
How to expand nested parentheses
If the expression contains nested brackets, then they are expanded in order, starting with external or internal.
At the same time, when opening one of the brackets, it is important not to touch the other brackets, just rewriting them as they are.
Example. 12 - (a + (6 - b) - 3) = 12 - a - (6 - b) + 3 = 12 - a - 6 + b + 3 = 9 - a + b
Not all equations containing parentheses are solved the same way. Of course, most often they need to open the brackets and give like terms (however, the ways of opening the brackets differ). But sometimes you don't need to open the brackets. Let's consider all these cases with specific examples:
- 5x - (3x - 7) = 9 + (-4x + 16).
- 2x - 3(x + 5) = -12.
- (x + 1)(7x - 21) = 0.
Solving Equations Through Bracket Opening
This method of solving equations is the most common, but even with all its apparent universality, it is divided into subspecies depending on the way the brackets are opened.
1) Solution of the equation 5x - (3x - 7) = 9 + (-4x + 16).
In this equation, there are minus and plus signs in front of the brackets. To open the brackets in the first case, where they are preceded by a minus sign, all the signs inside the brackets should be reversed. The second pair of brackets is preceded by a plus sign, which will not affect the signs in brackets, so they can simply be omitted. We get:
5x - 3x + 7 = 9 - 4x + 16.
The terms with x will be transferred to the left side of the equation, and the rest to the right (the signs of the transferred terms will change to the opposite):
5x - 3x + 4x = 9 + 16 - 7.
Here are similar terms:
To find the unknown factor x, divide the product 18 by the known factor 6:
x \u003d 18 / 6 \u003d 3.
2) Solution of the equation 2x - 3(x + 5) = -12.
In this equation, you also first need to open the brackets, but applying the distributive property: in order to multiply -3 by the sum (x + 5), you should multiply -3 by each term in brackets and add the resulting products:
2x - 3x - 15 = -12
x = 3 / (-1) = 3.
Solving equations without opening parentheses
The third equation (x + 1) (7x - 21) \u003d 0 can also be solved by opening the brackets, but it is much easier in such cases to use the multiplication property: the product is zero when one of the factors is zero. Means:
x + 1 = 0 or 7x - 21 = 0.