How to solve an equation with double brackets. Expansion of brackets: rules and examples (grade 7). Solving real-life examples of simple linear equations
An equation with one unknown, which after opening the brackets and reducing similar terms takes the form
ax + b \u003d 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 \u003d 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 instead of the unknown x to substitute the number 2, 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. Hence, the value x \u003d 3 is not a solution or a root of the equation.
Solving any linear equations is reduced to solving equations of the form
ax + b \u003d 0.
Moving the free term from the left side of the equation to the right, changing the sign in front of b to the opposite, we get
If a ≠ 0, then x \u003d - b / a .
Example 1. Solve the 3x + 2 \u003d 11 equation.
Move 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.
Subtract, then
3x \u003d 9.
To find x, you need to divide the product by a known factor, that is
x \u003d 9: 3.
Hence, the value x \u003d 3 is the solution or root of the equation.
Answer: x \u003d 3.
If a \u003d 0 and b \u003d 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. Any number is a solution to this equation.
Example 2.Solve the equation 5 (x - 3) + 2 \u003d 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 terms:
0x \u003d 0.
Answer: x is any number.
If a \u003d 0 and b ≠ 0, then we get the equation 0x \u003d - b. This equation has no solutions, since multiplying any number by 0 we get 0, but b ≠ 0.
Example 3.Solve the equation x + 8 \u003d x + 5.
Let us group the members containing unknowns on the left, and free members on the right:
x - x \u003d 5 - 8.
Here are similar terms:
0x \u003d - 3.
Answer: there are no solutions.
On the picture 1 shows the scheme for solving the linear equation
Let's draw up a general scheme for solving equations with one variable. Consider the solution to Example 4.
Example 4. Let the equation be solved
1) Multiply all the terms of the equation by the least common multiple of the denominators, equal to 12.
2) After reduction, we get
4 (x - 4) + 32 (x + 1) - 12 \u003d 6 5 (x - 3) + 24x - 2 (11x + 43)
3) To separate the members containing unknown and free members, we expand the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) Let us group in one part the members containing unknowns, and in the other - free members:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) Here are similar terms:
- 22x \u003d - 154.
6) Divide by - 22, We get
x \u003d 7.
As you can see, the root of the equation is seven.
Generally such equations can be solved according to the following scheme:
a) bring the equation to its whole form;
b) open the brackets;
c) group the terms containing the unknown in one part of the equation, and free terms in the other;
d) bring similar members;
e) solve an equation of the form ax \u003d b, which was obtained after bringing similar terms.
However, this scheme is not required for every equation. When solving many simpler equations, one has to start not with the first, but with the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.
Example 5.Solve the equation 2x \u003d 1/4.
Find the unknown x \u003d 1/4: 2,
x \u003d 1/8 .
Consider the solution of some linear equations found in the main state exam.
Example 6.Solve the equation 2 (x + 3) \u003d 5 - 6x.
2x + 6 \u003d 5 - 6x
2x + 6x \u003d 5 - 6
Answer: - 0, 125
Example 7.Solve the equation - 6 (5 - 3x) \u003d 8x - 7.
- 30 + 18x \u003d 8x - 7
18x - 8x \u003d - 7 +30
Answer: 2.3
Example 8. Solve the equation
3 (3x - 4) \u003d 4.7x + 24
9x - 12 \u003d 28x + 24
9x - 28x \u003d 24 + 12
Example 9.Find f (6) if f (x + 2) \u003d 3 7th
Decision
Since we need to find f (6), and we know f (x + 2),
then x + 2 \u003d 6.
Solve the linear equation x + 2 \u003d 6,
we get x \u003d 6 - 2, x \u003d 4.
If x \u003d 4, then
f (6) \u003d 3 7-4 \u003d 3 3 \u003d 27
Answer: 27.
If you have any questions, if you want to understand the solution of equations more thoroughly, sign up for my lessons in the SCHEDULE. I will be glad to help you!
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Linear equations. Solution, examples.
Attention!
There are additional
materials in Special section 555.
For those who are "not very ..."
And for those who are "very even ...")
Linear equations.
Linear equations are not the most difficult topic in school mathematics. But there are tricks there that can puzzle even a trained student. Shall we figure it out?)
Usually, a linear equation is defined as an equation of the form:
ax + b = 0 Where a and b - any numbers.
2x + 7 \u003d 0. Here a \u003d 2, b \u003d 7
0.1x - 2.3 \u003d 0 Here a \u003d 0.1, b \u003d -2.3
12x + 1/2 \u003d 0 Here a \u003d 12, b \u003d 1/2
Nothing complicated, right? Especially if you don't notice the words: "where a and b are any numbers"... And if you notice, but carelessly think?) After all, if a \u003d 0, b \u003d 0 (any numbers are possible?), then you get a funny expression:
But that's not all! If, say, a \u003d 0, and b \u003d 5, it turns out quite something out of the ordinary:
Which strains and undermines confidence in mathematics, yes ...) Especially in exams. But from these strange expressions it is also necessary to find the X! Which is not there at all. And, surprisingly, this X is very easy to find. We will learn how to do this. In this tutorial.
How do you know a linear equation by its appearance? It depends on what appearance.) The trick is that linear equations are not only equations of the form ax + b = 0 , but also any equations that are reduced to this form by transformations and simplifications. And who knows whether it can be reduced or not?)
A linear equation can be clearly recognized in some cases. Say, if we have an equation in which there are only unknowns in the first degree, and numbers. And in the equation there is no fractions divided by unknown , it is important! And division by number, or a numeric fraction - please! For example:
This is a linear equation. There are fractions here, but there are no x's in the square, in the cube, etc., and there are no x's in the denominators, i.e. not division by x... And here is the equation
cannot be called linear. Here the x's are all in the first degree, but there is division by expression with x... After simplifications and transformations, you can get a linear equation, and a quadratic, and anything you like.
It turns out that it is impossible to find out a linear equation in some tricky example until you almost solve it. This is upsetting. But assignments usually don't ask about the type of equation, right? The tasks are given equations solve. This makes me happy.)
Solving linear equations. Examples.
The entire solution to linear equations consists of identical transformations of equations. By the way, these transformations (as many as two!) Underlie the solutions all equations of mathematics. In other words, the solution any the equation begins with these very transformations. In the case of linear equations, it (the solution) is based on these transformations and ends with a full-fledged answer. It makes sense to go to the link, right?) Moreover, there are also examples of solving linear equations.
Let's start with the simplest example. No pitfalls. Suppose we need to solve this equation.
x - 3 \u003d 2 - 4x
This is a linear equation. X is all in the first degree, there is no division by X. But, in fact, we do not care what equation it is. We need to solve it. The scheme is simple. Collect everything with x on the left side of the equation, everything without x (number) on the right.
To do this, you need to transfer - 4x to the left, with a change of sign, of course, but - 3 - to the right. By the way, this is first identical transformation of equations. Are you surprised? So, we didn't follow the link, but in vain ...) We get:
x + 4x \u003d 2 + 3
We give similar ones, we believe:
What do we lack for complete happiness? Yes, so that there was a clean X on the left! The five is in the way. Getting rid of the top five with second identical transformation of equations. Namely, we divide both sides of the equation by 5. We get a ready answer:
An elementary example, of course. This is for warm-up.) It is not very clear why I was recalling identical transformations here? Okay. We take the bull by the horns.) Let's decide something more impressive.
For example, here's the equation:
Where do we start? With x - to the left, without x - to the right? Could be so. Little steps along the long road. Or you can immediately, in a universal and powerful way. If, of course, you have in your arsenal identical transformations of equations.
I ask you a key question: what do you dislike most about this equation?
95 people out of 100 will answer: fractions ! The answer is correct. So let's get rid of them. Therefore, we start right away with second identity transformation... What do you need to multiply the fraction on the left so that the denominator can be reduced completely? Right, 3. And on the right? By 4. But mathematics allows us to multiply both sides by the same number... How do we get out? And let's multiply both sides by 12! Those. by a common denominator. Then both the three and the four will decrease. Do not forget that you need to multiply each part wholly... This is what the first step looks like:
Expand the brackets:
Note! Numerator (x + 2) I bracketed! This is because when you multiply fractions, the numerator is multiplied entirely, entirely! And now the fractions can be reduced:
Expand the remaining brackets:
Not an example, but sheer pleasure!) Now we recall the spell from the elementary grades: with an x \u200b\u200b- to the left, without an x \u200b\u200b- to the right! And apply this transformation:
Here are similar ones:
And we divide both parts by 25, i.e. apply the second transform again:
That's all. Answer: x=0,16
Note: to bring the original tricky equation to a pleasant form, we used two (only two!) identical transformations - transfer left-right with a change of sign and multiplication-division of the equation by the same number. This is a universal way! We will work in this way with any equations! Absolutely any. That is why I am repeating these identical transformations all the time.)
As you can see, the principle of solving linear equations is simple. Take the equation and simplify it with identical transformations until a reply is received. The main problems here are in computation, not in principle of solution.
But ... There are such surprises in the process of solving the most elementary linear equations that they can drive you into a strong stupor ...) Fortunately, there can be only two such surprises. Let's call them special cases.
Special cases when solving linear equations.
First surprise.
Suppose you come across an elementary equation, something like:
2x + 3 \u003d 5x + 5 - 3x - 2
Slightly bored, we transfer it with an x \u200b\u200bto the left, without an x \u200b\u200bto the right ... With a change of sign, everything is a chin-chinar ... We get:
2x-5x + 3x \u003d 5-2-3
We consider, and ... oh shit !!! We get:
This equality in itself is not objectionable. Zero is indeed zero. But X is gone! And we must write in the answer what is x. Otherwise, the decision does not count, yes ...) Dead end?
Calm! In such doubtful cases, the most general rules save. How to solve equations? What does it mean to solve an equation? It means, find all the x values \u200b\u200bthat, when substituted into the original equation, will give us the correct equality.
But we have true equality already happened! 0 \u003d 0, how much more accurate ?! It remains to figure out at what xx it turns out. What values \u200b\u200bof x can be substituted in initial equation if these x's will shrink to zero anyway? Come on?)
Yes!!! X can be substituted any! What you want. At least 5, at least 0.05, at least -220. They will shrink anyway. If you don’t believe it, you can check it.) Substitute any x values \u200b\u200bin initial equation and count. All the time, the pure truth will be obtained: 0 \u003d 0, 2 \u003d 2, -7.1 \u003d -7.1 and so on.
Here's the answer: x - any number.
The answer can be written in different mathematical symbols, the essence does not change. This is an absolutely correct and complete answer.
Second surprise.
Let's take the same elementary linear equation and change only one number in it. This is what we will solve:
2x + 1 \u003d 5x + 5 - 3x - 2
After the same identical transformations, we get something intriguing:
Like this. Solved a linear equation, got a strange equality. Mathematically speaking, we got wrong equality. And speaking simple language, it is not true. Rave. But nevertheless, this nonsense is a very good reason for solving the equation correctly.)
Again, we think based on the general rules. What x, when substituted into the original equation, will give us true equality? Yes, none! There are no such x's. Whatever you substitute, everything will be reduced, delirium will remain.)
Here's the answer: no solutions.
This is also a complete answer. In mathematics, such answers are common.
Like this. Now, I hope, the loss of x in the process of solving any (not only linear) equation will not confuse you at all. The matter is already familiar.)
Now that we have figured out all the pitfalls in linear equations, it makes sense to solve them.
If you like this site ...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Instant validation testing. Learning - with interest!)
you can get acquainted with functions and derivatives.
Parentheses are used to indicate the order in which actions are performed in numeric, literal, and variable expressions. It is convenient to switch from an expression with parentheses to an identically equal expression without parentheses. This technique is called parenthesis expansion.
Expand parentheses means to get rid of the expression from those parentheses.
One more point deserves special attention, which concerns the peculiarities of recording decisions when opening brackets. We can write the initial expression with parentheses and the result obtained after expanding the parentheses as equality. For example, after expanding 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) \u003d 3−5 + 7.
And one more important point. In mathematics, to shorten records, it is customary not to write a plus sign if it appears first in an expression or in parentheses. 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 parenthesis, which is not written, and in front of the five there is plus + (+ 5 + x).
The rule for expanding parentheses in addition
When expanding parentheses, if there is a plus in front of the brackets, this plus is omitted together with the parentheses.
Example. Expand parentheses in the expression 2 + (7 + 3) Before the parentheses, plus, so the signs in front of the numbers in parentheses do not change.
2 + (7 + 3) = 2 + 7 + 3
Parenthesis expansion rule for subtraction
If there is a minus in front of the brackets, then this minus is omitted together with the brackets, but the terms that were in the brackets change their sign to the opposite. The absence of a sign in front of the first term in parentheses implies a + sign.
Example. Expand parentheses in expression 2 - (7 + 3)
There is a minus in front of the brackets, which means you need to change the signs before the numbers from the brackets. There is no sign in parentheses before the number 7, this means that the seven is positive, it is considered that there is a + sign in front of it.
2 − (7 + 3) = 2 − (+ 7 + 3)
When expanding the brackets, we remove from the example the minus that was in front of the brackets, and the brackets themselves 2 - (+ 7 + 3), and the signs that were in the brackets are reversed.
2 − (+ 7 + 3) = 2 − 7 − 3
Expanding parentheses during multiplication
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. In this case, multiplying minus by minus gives plus, and multiplying minus by plus, as well as multiplying plus by minus gives minus.
Thus, parentheses in works are expanded in accordance with the distributional property of multiplication.
Example. 2 (9 - 7) \u003d 2 9 - 2 7
When you multiply a parenthesis by a parenthesis, each member of the first parenthesis is multiplied with each member of the second parenthesis.
(2 + 3) (4 + 5) \u003d 2 4 + 2 5 + 3 4 + 3 5
In fact, there is no need to memorize all the rules, it is enough to remember only one, this is: c (a-b) \u003d ca-cb. Why? Because if you substitute one in it instead of c, you get the rule (a - b) \u003d a - b. And if we substitute minus one, we get the rule - (a - b) \u003d - a + b. Well, if instead of c you substitute another parenthesis, you can get the last rule.
Expanding parentheses in division
If there is a division sign after the brackets, then each number inside the brackets is divided by the divisor after the brackets, and vice versa.
Example. (9 + 6): 3 \u003d 9: 3 + 6: 3
How to expand nested parentheses
If there are nested parentheses in the expression, then they are expanded in order, starting with the outer or inner ones.
At the same time, when opening one of the brackets, it is important not to touch the other brackets, simply rewriting them as they are.
Example. 12 - (a + (6 - b) - 3) \u003d 12 - a - (6 - b) + 3 \u003d 12 - a - 6 + b + 3 \u003d 9 - a + b
An equation with one unknown, which after opening the brackets and reducing similar terms takes the form
ax + b \u003d 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 \u003d 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 instead of the unknown x to substitute the number 2, 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. Hence, the value x \u003d 3 is not a solution or a root of the equation.
Solving any linear equations is reduced to solving equations of the form
ax + b \u003d 0.
Moving the free term from the left side of the equation to the right, changing the sign in front of b to the opposite, we get
If a ≠ 0, then x \u003d - b / a .
Example 1. Solve the 3x + 2 \u003d 11 equation.
Move 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.
Subtract, then
3x \u003d 9.
To find x, you need to divide the product by a known factor, that is
x \u003d 9: 3.
Hence, the value x \u003d 3 is the solution or root of the equation.
Answer: x \u003d 3.
If a \u003d 0 and b \u003d 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. Any number is a solution to this equation.
Example 2.Solve the equation 5 (x - 3) + 2 \u003d 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 terms:
0x \u003d 0.
Answer: x is any number.
If a \u003d 0 and b ≠ 0, then we get the equation 0x \u003d - b. This equation has no solutions, since multiplying any number by 0 we get 0, but b ≠ 0.
Example 3.Solve the equation x + 8 \u003d x + 5.
Let us group the members containing unknowns on the left, and free members on the right:
x - x \u003d 5 - 8.
Here are similar terms:
0x \u003d - 3.
Answer: there are no solutions.
On the picture 1 shows the scheme for solving the linear equation
Let's draw up a general scheme for solving equations with one variable. Consider the solution to Example 4.
Example 4. Let the equation be solved
1) Multiply all the terms of the equation by the least common multiple of the denominators, equal to 12.
2) After reduction, we get
4 (x - 4) + 32 (x + 1) - 12 \u003d 6 5 (x - 3) + 24x - 2 (11x + 43)
3) To separate the members containing unknown and free members, we expand the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) Let us group in one part the members containing unknowns, and in the other - free members:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) Here are similar terms:
- 22x \u003d - 154.
6) Divide by - 22, We get
x \u003d 7.
As you can see, the root of the equation is seven.
Generally such equations can be solved according to the following scheme:
a) bring the equation to its whole form;
b) open the brackets;
c) group the terms containing the unknown in one part of the equation, and free terms in the other;
d) bring similar members;
e) solve an equation of the form ax \u003d b, which was obtained after bringing similar terms.
However, this scheme is not required for every equation. When solving many simpler equations, one has to start not with the first, but with the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.
Example 5.Solve the equation 2x \u003d 1/4.
Find the unknown x \u003d 1/4: 2,
x \u003d 1/8 .
Consider the solution of some linear equations found in the main state exam.
Example 6.Solve the equation 2 (x + 3) \u003d 5 - 6x.
2x + 6 \u003d 5 - 6x
2x + 6x \u003d 5 - 6
Answer: - 0, 125
Example 7.Solve the equation - 6 (5 - 3x) \u003d 8x - 7.
- 30 + 18x \u003d 8x - 7
18x - 8x \u003d - 7 +30
Answer: 2.3
Example 8. Solve the equation
3 (3x - 4) \u003d 4.7x + 24
9x - 12 \u003d 28x + 24
9x - 28x \u003d 24 + 12
Example 9.Find f (6) if f (x + 2) \u003d 3 7th
Decision
Since we need to find f (6), and we know f (x + 2),
then x + 2 \u003d 6.
Solve the linear equation x + 2 \u003d 6,
we get x \u003d 6 - 2, x \u003d 4.
If x \u003d 4, then
f (6) \u003d 3 7-4 \u003d 3 3 \u003d 27
Answer: 27.
If you still have questions, there is a desire to understand the solution of equations more thoroughly,. I will be glad to help you!
TutorOnline also advises you to watch a new video tutorial from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.
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