In previous lessons, we got acquainted with expressions, and also learned how to simplify and calculate them. Now we pass to more difficult and interesting, namely to the equations.
Equation and its roots
Equality containing variable(s) are called equations. solve the equation , means to find the value of the variable for which the equality will be true. The value of the variable is called the root of the equation .
Equations can have one root, several or none at all.
When solving equations, the following properties are used:
- if in the equation we transfer the term from one part of the equation to another, while changing the sign to the opposite, then we get an equation equivalent to the given one.
- If both sides of the equation are multiplied or divided by the same number, then an equation equivalent to the given one will be obtained.
Example #1Which of the numbers: -2, -1, 0, 2, 3 are the roots of the equation:
To solve this task, you just need to alternately substitute each of the numbers instead of the variable x and select those numbers for which equality is considered true.
With "x \u003d -2":
\((-2)^2=10-3 \cdot (-2) \)
\ (4 \u003d 4 \) - the equality is true, which means (-2) is the root of our equation
With "x \u003d -1"
\((-1)^2=10-3 \cdot (-1) \)
\(1=7 \) - the equality is wrong, therefore (-1) is not the root of the equation
\(0^2=10-3 \cdot 0 \)
\(0=10 \) - the equality is wrong, so 0 is not the root of the equation
\(2^2=10-3 \cdot 2\)
\ (4 \u003d 4 \) - the equality is true, which means 2 is the root of our equation
\(3^2=10-3 \cdot 3 \)
\(9=1 \) - the equality is wrong, so 3 is not the root of the equation
Answer: from the presented numbers, the roots of the equation \(x^2=10-3x \) are the numbers -2 and 2.
Linear equation with one variable are equations of the form ax = b, where x is a variable and a and b are some numbers.
There are a large number of types of equations, but the solution of many of them comes down to solving linear equations, so knowledge of this topic is mandatory for further learning!
Example #2 Solve the equation: 4(x+7) = 3-x
To solve this equation, first of all, you need to get rid of the bracket, and for this we multiply each of the terms in the bracket by 4, we get:
4x + 28 = 3 - x
Now you need to transfer all the values from "x" to one side, and everything else to the other side (remembering to change the sign to the opposite), we get:
4x + x = 3 - 28
Now subtract the value on the left and right:
To find the unknown factor (x), you need to divide the product (25) by the known factor (5):
Answer x = -5
If in doubt about the answer, you can check by substituting the resulting value in our equation instead of x:
4(-5+7) = 3-(-5)
8 = 8 - the equation is solved correctly!
Now to solve something more difficult:
Example #3 Find the roots of the equation: \((y+4)-(y-4)=6y \)
First of all, also get rid of the parentheses:
We immediately see y and -y on the left side, which means that they can simply be crossed out, and the resulting numbers can simply be added and the expression written down:
Now you can move the values with "y" to the left, and the values with numbers to the right. But this is not necessary, because it does not matter which side the variables are on, the main thing is that they be without numbers, which means that we will not transfer anything. But for those who do not understand, we will do as the rule says and divide both parts by (-1), as the property says:
To find the unknown factor, you need to divide the product by the known factor:
\(y=\frac(8)(6) = \frac(4)(3) = 1\frac(1)(3) \)
Answer: y = \(1\frac(1)(3) \)
You can also check the answer, but do it yourself.
Example #4\((0.5x+1.2)-(3.6-4.5x)=(4.8-0.3x)+(10.5x+0.6) \)
Now I will just solve, without explanation, and you look at the progress of the solution and the correct notation of the solution of equations:
\((0.5x+1.2)-(3.6-4.5x)=(4.8-0.3x)+(10.5x+0.6) \)
\(0.5x+1.2-3.6+4.5x=4.8-0.3x+10.5x+0.6 \)
\(0.5x+4.5x+0.3x-10.5x=4.8+0.6-1.2+3.6 \)
\(x=\frac(7,8)(-5,2)=\frac(3)(-2) =-1,5 \)
Answer: x = -1.5
If something is not clear along the way, write in the comments
Solving Problems with Equations
Knowing what equations are and learning how to calculate them, you also open up access to solving many problems where equations are used for solving.
I will not go into theory, it is better to show everything at once with examples
Example #5 There were 2 times fewer apples in the basket than in the box. After 10 apples were transferred from the basket to the box, there were 5 times more apples in the box than in the basket. How many apples were in the basket and how many were in the box?
First of all, we need to determine what we will take for "x", in this problem we can accept both boxes and baskets, but I will take apples in a basket.
So, let there be x apples in the basket, since there were twice as many apples in the box, then we take this for 2x. After the apples were transferred from the basket to the box, the basket of apples became: x - 10, which means that there were - (2x + 10) apples in the box.
Now you can make an equation:
5(x-10) - there are 5 times more apples in the box than in the basket.
Equate the first value and the second:
2x+10 = 5(x-10) and solve:
2x + 10 = 5x - 50
2x - 5x = -50 - 10
x \u003d -60 / -3 \u003d 20 (apples) - in the basket
Now, knowing how many apples were in the basket, we will find how many apples were in the box - since there were twice as many, we simply multiply the result by 2:
2*20 = 40 (apples) - in a box
Answer: There are 40 apples in a box and 20 apples in a basket.
I understand that many of you may not have fully understood how to solve problems, but I assure you that we will return to this topic more than once in our lessons, but for now, if you still have questions, ask them in the comments.
At the end, a few more examples for solving equations
Example #6\(2x - 0.7x = 0 \)
Example #7\(3p - 1 -(p+3) = 1 \)
Example #8\(6y-(y-1) = 4+5y \)
\(6y-y+1=4+5y\)
\(6y-y-5y=4-1\)
\(0y=3 \) - no roots, because You can't divide by zero!
Thank you all for your attention. If something is unclear, ask in the comments.
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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.
A linear equation with one variable has the general form
ax + b = 0.
Here x is a variable, a and b are coefficients. In another way, a is called the “coefficient of the unknown”, b is the “free term”.
The coefficients are some numbers, and solving the equation means finding the value x for which the expression ax + b = 0 is true. For example, we have a linear equation 3x - 6 \u003d 0. Solving it means finding what x must be equal to so that 3x - 6 is equal to 0. Performing transformations, we get:
3x=6
x=2
Thus the expression 3x - 6 = 0 is true for x = 2:
3 * 2 – 6 = 0
2 is the root of this equation. When you solve an equation, you find its roots.
The coefficients a and b can be any numbers, however, there are such values when there is more than one root of a linear equation with one variable.
If a = 0, then ax + b = 0 turns into b = 0. Here x is "destroyed". The expression b = 0 itself can be true only if the knowledge of b is 0. That is, the equation 0*x + 3 = 0 is false, because 3 = 0 is a false statement. However, 0*x + 0 = 0 is the correct expression. From here it is concluded that if a \u003d 0 and b ≠ 0, a linear equation with one variable has no roots at all, but if a \u003d 0 and b \u003d 0, then the equation has an infinite number of roots.
If b \u003d 0, and a ≠ 0, then the equation will take the form ax \u003d 0. It is clear that if a ≠ 0, but the result of multiplication is 0, then x \u003d 0. That is, the root of this equation is 0.
If neither a nor b are equal to zero, then the equation ax + b = 0 is transformed to the form
x \u003d -b / a.
The value of x in this case will depend on the values of a and b . However, it will be the only one. That is, it is impossible to obtain two or more different x values for the same coefficients. For example,
-8.5x - 17 = 0
x = 17 / -8.5
x = -2
No number other than -2 can be obtained by dividing 17 by -8.5.
There are equations that at first glance do not look like the general form of a linear equation with one variable, but are easily converted to it. For example,
-4.8 + 1.3x = 1.5x + 12
If we move everything to the left side, then 0 will remain on the right:
–4.8 + 1.3x – 1.5x – 12 = 0
Now the equation is reduced to the standard form and you can solve it:
x = 16.8 / 0.2
x=84
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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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
Here again 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 when substituting it instead of x in the original equation, the correct equality should be obtained - same numbers 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 for 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"? multiplied 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 differ.
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 : \(x=-1.25\)