We continue talking about solution of equations. In this article, we will focus on rational equations and principles of decision rational equations with one variable. First, let's figure out what kind of equations are called rational, give a definition of integer rational and fractional rational equations, and give examples. Next, we obtain algorithms for solving rational equations, and, of course, consider the solutions characteristic examples with all necessary explanations.
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Based on the sounded definitions, we give several examples of rational equations. For example, x=1 , 2 x−12 x 2 y z 3 =0 , , are all rational equations.
From the examples shown, it can be seen that rational equations, as well as equations of other types, can be either with one variable, or with two, three, etc. variables. AT the following paragraphs we will talk about solving rational equations in one variable. Solving equations with two variables and them a large number deserve special attention.
In addition to dividing rational equations by the number of unknown variables, they are also divided into integer and fractional. Let us give the corresponding definitions.
Definition.
The rational equation is called whole, if both its left and right parts are integer rational expressions.
Definition.
If at least one of the parts of a rational equation is fractional expression, then this equation is called fractionally rational(or fractional rational).
It is clear that integer equations do not contain division by a variable; on the contrary, fractional rational equations necessarily contain division by a variable (or a variable in the denominator). So 3 x+2=0 and (x+y) (3 x 2 −1)+x=−y+0.5 are entire rational equations, both of their parts are integer expressions. A and x:(5 x 3 +y 2)=3:(x−1):5 are examples of fractional rational equations.
Concluding this paragraph, let us pay attention to the fact that linear equations and quadratic equations known by this moment are entire rational equations.
Solving integer equations
One of the main approaches to solving entire equations is their reduction to equivalent algebraic equations. This can always be done by performing the following equivalent transformations of the equation:
- first, the expression from the right side of the original integer equation is transferred to the left side with opposite sign to get zero on the right side;
- after that, on the left side of the equation, the resulting standard view.
The result is algebraic equation, which is equivalent to the original whole equation. So in the most simple cases solving entire equations are reduced to solving linear or quadratic equations, and in general case– to the solution of an algebraic equation of degree n. For clarity, let's analyze the solution of the example.
Example.
Find the roots of the whole equation 3 (x+1) (x−3)=x (2 x−1)−3.
Decision.
Let us reduce the solution of this whole equation to the solution of an equivalent algebraic equation. To do this, firstly, we transfer the expression from the right side to the left, as a result we arrive at the equation 3 (x+1) (x−3)−x (2 x−1)+3=0. And, secondly, we transform the expression formed on the left side into a polynomial of the standard form by doing the necessary: 3 (x+1) (x−3)−x (2 x−1)+3= (3 x+3) (x−3)−2 x 2 +x+3= 3 x 2 −9 x+3 x−9−2 x 2 +x+3=x 2 −5 x−6. Thus, the solution of the original integer equation reduces to the solution quadratic equation x 2 −5 x−6=0 .
Calculate its discriminant D=(−5) 2 −4 1 (−6)=25+24=49, it is positive, which means that the equation has two real roots, which we find by the formula of the roots of the quadratic equation:
For complete confidence do it checking the found roots of the equation. First, we check the root 6, substitute it instead of the variable x in the original integer equation: 3 (6+1) (6−3)=6 (2 6−1)−3, which is the same, 63=63 . That's right numerical equality, therefore, x=6 is indeed the root of the equation. Now we check the root −1 , we have 3 (−1+1) (−1−3)=(−1) (2 (−1)−1)−3, whence, 0=0 . For x=−1, the original equation also turned into a true numerical equality, therefore, x=−1 is also the root of the equation.
Answer:
6 , −1 .
Here it should also be noted that the term “power of an entire equation” is associated with the representation of an entire equation in the form of an algebraic equation. We give the corresponding definition:
Definition.
The degree of the whole equation call the degree of an algebraic equation equivalent to it.
According to this definition, the whole equation from the previous example has the second degree.
On this one could finish with the solution of entire rational equations, if not for one but .... As is known, the solution of algebraic equations of degree higher than the second is associated with significant difficulties, and for equations of degree higher than the fourth, there are no such equations at all. general formulas roots. Therefore, to solve entire equations of the third, fourth and more high degrees often have to resort to other methods of solution.
In such cases, sometimes the approach to solving entire rational equations based on factorization method. At the same time, the following algorithm is followed:
- first they seek to have zero on the right side of the equation, for this they transfer the expression from the right side of the whole equation to the left;
- then, the resulting expression on the left side is presented as a product of several factors, which allows you to go to a set of several simpler equations.
The above algorithm for solving the whole equation through factorization requires a detailed explanation using an example.
Example.
Solve the whole equation (x 2 −1) (x 2 −10 x+13)= 2 x (x 2 −10 x+13) .
Decision.
First, as usual, we transfer the expression from the right side to the left side of the equation, not forgetting to change the sign, we get (x 2 −1) (x 2 −10 x+13) − 2 x (x 2 −10 x+13)=0 . It is quite obvious here that it is not advisable to transform the left side of the resulting equation into a polynomial of the standard form, since this will give an algebraic equation of the fourth degree of the form x 4 −12 x 3 +32 x 2 −16 x−13=0, whose solution is difficult.
On the other hand, it is obvious that x 2 −10·x+13 can be found on the left side of the resulting equation, thereby representing it as a product. We have (x 2 −10 x+13) (x 2 −2 x−1)=0. The resulting equation is equivalent to the original whole equation, and it, in turn, can be replaced by a set of two quadratic equations x 2 −10·x+13=0 and x 2 −2·x−1=0 . Finding their roots known formulas roots through the discriminant is not difficult, the roots are equal. They are the desired roots of the original equation.
Answer:
It is also useful for solving entire rational equations. method for introducing a new variable. In some cases, it allows one to pass to equations whose degree is lower than the degree of the original integer equation.
Example.
Find real roots rational equation (x 2 +3 x+1) 2 +10=−2 (x 2 +3 x−4).
Decision.
Reducing this whole rational equation to an algebraic equation is, to put it mildly, not a very good idea, since in this case we will come to the need to solve a fourth-degree equation that does not have rational roots. Therefore, you will have to look for another solution.
It is easy to see here that you can introduce a new variable y and replace the expression x 2 +3 x with it. Such a replacement leads us to the whole equation (y+1) 2 +10=−2 (y−4) , which, after transferring the expression −2 (y−4) to the left side and subsequent transformation of the expression formed there, reduces to equation y 2 +4 y+3=0 . The roots of this equation y=−1 and y=−3 are easy to find, for example, they can be found based on the inverse theorem of Vieta's theorem.
Now let's move on to the second part of the method of introducing a new variable, that is, to making a reverse substitution. After performing the reverse substitution, we obtain two equations x 2 +3 x=−1 and x 2 +3 x=−3 , which can be rewritten as x 2 +3 x+1=0 and x 2 +3 x+3 =0 . According to the formula of the roots of the quadratic equation, we find the roots of the first equation. And the second quadratic equation has no real roots, since its discriminant is negative (D=3 2 −4 3=9−12=−3 ).
Answer:
In general, when we are dealing with integer equations of high degrees, we must always be ready to search non-standard method or an artificial device for their solution.
Solution of fractionally rational equations
First, it will be useful to understand how to solve fractionally rational equations of the form , where p(x) and q(x) are rational integer expressions. And then we will show how to reduce the solution of the remaining fractionally rational equations to the solution of equations of the indicated form.
One of the approaches to solving the equation is based on the following statement: the numerical fraction u/v, where v is a non-zero number (otherwise we will encounter , which is not defined), is equal to zero if and only if its numerator zero, that is, if and only if u=0 . By virtue of this statement, the solution of the equation is reduced to the fulfillment of two conditions p(x)=0 and q(x)≠0 .
This conclusion is consistent with the following algorithm for solving a fractionally rational equation. To solve a fractional rational equation of the form
- solve the whole rational equation p(x)=0 ;
- and check whether the condition q(x)≠0 is satisfied for each found root, while
- if true, then this root is the root of the original equation;
- if not, then this root is extraneous, that is, it is not the root of the original equation.
Let's analyze an example of using the voiced algorithm when solving a fractional rational equation.
Example.
Find the roots of the equation.
Decision.
This is a fractionally rational equation of the form , where p(x)=3 x−2 , q(x)=5 x 2 −2=0 .
According to the algorithm for solving fractionally rational equations of this kind, we first need to solve the equation 3·x−2=0 . This is linear equation, whose root is x=2/3 .
It remains to check for this root, that is, to check whether it satisfies the condition 5·x 2 −2≠0 . We substitute the number 2/3 instead of x into the expression 5 x 2 −2, we get . The condition is met, so x=2/3 is the root of the original equation.
Answer:
2/3 .
The solution of a fractional rational equation can be approached from a slightly different position. This equation is equivalent to the whole equation p(x)=0 on the variable x of the original equation. That is, you can follow this algorithm for solving a fractionally rational equation :
- solve the equation p(x)=0 ;
- find ODZ variable x ;
- take the roots belonging to the area allowed values, - they are the desired roots of the original fractional rational equation.
For example, let's solve a fractional rational equation using this algorithm.
Example.
Solve the equation.
Decision.
First, we solve the quadratic equation x 2 −2·x−11=0 . Its roots can be calculated using the root formula for an even second coefficient, we have D 1 =(−1) 2 −1 (−11)=12, and .
Secondly, we find the ODZ of the variable x for the original equation. It consists of all numbers for which x 2 +3 x≠0 , which is the same x (x+3)≠0 , whence x≠0 , x≠−3 .
It remains to check whether the roots found at the first step are included in the ODZ. Obviously yes. Therefore, the original fractionally rational equation has two roots.
Answer:
Note that this approach is more profitable than the first one if the ODZ is easily found, and it is especially beneficial if the roots of the equation p(x)=0 are irrational, for example, , or rational, but with a rather large numerator and/or denominator, for example, 127/1101 and -31/59 . This is due to the fact that in such cases, checking the condition q(x)≠0 will require significant computational efforts, and it is easier to exclude extraneous roots from the ODZ.
In other cases, when solving the equation, especially when the roots of the equation p(x)=0 are integers, it is more advantageous to use the first of the above algorithms. That is, it is advisable to immediately find the roots of the whole equation p(x)=0 , and then check whether the condition q(x)≠0 is satisfied for them, and not find the ODZ, and then solve the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to make a check than to find the ODZ.
Consider the solution of two examples to illustrate the stipulated nuances.
Example.
Find the roots of the equation.
Decision.
First we find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, compiled using the numerator of the fraction. The left side of this equation is a product, and the right side is zero, therefore, according to the method of solving equations through factorization, this equation is equivalent to the set of four equations 2 x−1=0 , x−6=0 , x 2 −5 x+ 14=0 , x+1=0 . Three of these equations are linear and one is quadratic, we can solve them. From the first equation we find x=1/2, from the second - x=6, from the third - x=7, x=−2, from the fourth - x=−1.
With the roots found, it is quite easy to check them to see if the denominator of the fraction on the left side of the original equation does not vanish, and it is not so easy to determine the ODZ, since this will have to solve an algebraic equation of the fifth degree. Therefore, let's give up finding ODZ in favor of checking the roots. To do this, we substitute them in turn instead of the variable x in the expression x 5 −15 x 4 +57 x 3 −13 x 2 +26 x+112, obtained after substitution, and compare them with zero: (1/2) 5 −15 (1/2) 4 + 57 (1/2) 3 −13 (1/2) 2 +26 (1/2)+112= 1/32−15/16+57/8−13/4+13+112=
122+1/32≠0
;
6 5 −15 6 4 +57 6 3 −13 6 2 +26 6+112= 448≠0
;
7 5 −15 7 4 +57 7 3 −13 7 2 +26 7+112=0;
(−2) 5 −15 (−2) 4 +57 (−2) 3 −13 (−2) 2 + 26 (−2)+112=−720≠0 ;
(−1) 5 −15 (−1) 4 +57 (−1) 3 −13 (−1) 2 + 26·(−1)+112=0 .
Thus, 1/2, 6 and −2 are the desired roots of the original fractionally rational equation, and 7 and −1 are extraneous roots.
Answer:
1/2 , 6 , −2 .
Example.
Find the roots of a fractional rational equation.
Decision.
First we find the roots of the equation (5x2 −7x−1)(x−2)=0. This equation is equivalent to a set of two equations: the square 5·x 2 −7·x−1=0 and the linear x−2=0 . According to the formula of the roots of the quadratic equation, we find two roots, and from the second equation we have x=2.
Checking if the denominator does not vanish at the found values of x is rather unpleasant. And to determine the range of acceptable values of the variable x in the original equation is quite simple. Therefore, we will act through the ODZ.
In our case, the ODZ of the variable x of the original fractional rational equation is made up of all numbers, except for those for which the condition x 2 +5·x−14=0 is satisfied. The roots of this quadratic equation are x=−7 and x=2, from which we conclude about the ODZ: it is made up of all x such that .
It remains to check whether the found roots and x=2 belong to the region of admissible values. The roots - belong, therefore, they are the roots of the original equation, and x=2 does not belong, therefore, it is an extraneous root.
Answer:
It will also be useful to dwell separately on cases where a number is in the numerator in a fractional rational equation of the form, that is, when p (x) is represented by some number. Wherein
- if this number is different from zero, then the equation has no roots, since the fraction is zero if and only if its numerator is zero;
- if this number is zero, then the root of the equation is any number from the ODZ.
Example.
Decision.
Since there is a non-zero number in the numerator of the fraction on the left side of the equation, for no x can the value of this fraction be equal to zero. Hence, given equation has no roots.
Answer:
no roots.
Example.
Solve the equation.
Decision.
The numerator of the fraction on the left side of this fractional rational equation is zero, so the value of this fraction is zero for any x for which it makes sense. In other words, the solution to this equation is any value of x from the DPV of this variable.
It remains to determine this range of acceptable values. It includes all such values x for which x 4 +5 x 3 ≠0. The solutions of the equation x 4 +5 x 3 \u003d 0 are 0 and −5, since this equation is equivalent to the equation x 3 (x + 5) \u003d 0, and it, in turn, is equivalent to the combination of two equations x 3 \u003d 0 and x +5=0 , from where these roots are visible. Therefore, the desired range of acceptable values are any x , except for x=0 and x=−5 .
Thus, a fractionally rational equation has infinitely many solutions, which are any numbers except zero and minus five.
Answer:
Finally, it's time to talk about solving fractional rational equations arbitrary type. They can be written as r(x)=s(x) , where r(x) and s(x) are rational expressions, and at least one of them is fractional. Looking ahead, we say that their solution is reduced to solving equations of the form already familiar to us.
It is known that the transfer of a term from one part of the equation to another with the opposite sign leads to equivalent to the equation, so the equation r(x)=s(x) is equivalent to the equation r(x)−s(x)=0 .
We also know that any can be identically equal to this expression. Thus, we can always transform the rational expression on the left side of the equation r(x)−s(x)=0 into an identically equal rational fraction of the form .
So we go from the original fractional rational equation r(x)=s(x) to the equation , and its solution, as we found out above, reduces to solving the equation p(x)=0 .
But here it is necessary to take into account the fact that when replacing r(x)−s(x)=0 with , and then with p(x)=0 , the range of allowable values of the variable x may expand.
Therefore, the original equation r(x)=s(x) and the equation p(x)=0 , which we came to, may not be equivalent, and by solving the equation p(x)=0 , we can get roots that will be extraneous roots of the original equation r(x)=s(x) . It is possible to identify and not include extraneous roots in the answer, either by checking, or by checking their belonging to the ODZ of the original equation.
We summarize this information in algorithm for solving a fractional rational equation r(x)=s(x). To solve the fractional rational equation r(x)=s(x) , one must
- Get zero on the right by moving the expression from the right side with the opposite sign.
- Perform actions with fractions and polynomials on the left side of the equation, thereby converting it into a rational fraction of the form.
- Solve the equation p(x)=0 .
- Identify and exclude extraneous roots, which is done by substituting them into the original equation or by checking their belonging to the ODZ of the original equation.
For greater clarity, we will show the entire chain of solving fractional rational equations:
.
Let's go through the solutions of several examples with a detailed explanation of the solution in order to clarify the given block of information.
Example.
Solve a fractional rational equation.
Decision.
We will act in accordance with the just obtained solution algorithm. And first we transfer the terms from the right side of the equation to the left side, as a result we pass to the equation .
In the second step, we need to convert the fractional rational expression on the left side of the resulting equation to the form of a fraction. To do this, we perform a cast rational fractions to common denominator and simplify the resulting expression: . So we come to the equation.
In the next step, we need to solve the equation −2·x−1=0 . Find x=−1/2 .
It remains to check whether the found number −1/2 is an extraneous root of the original equation. To do this, you can check or find the ODZ variable x of the original equation. Let's demonstrate both approaches.
Let's start with a check. We substitute the number −1/2 instead of the variable x into the original equation, we get , which is the same, −1=−1. The substitution gives the correct numerical equality, therefore, x=−1/2 is the root of the original equation.
Now we will show how the last step of the algorithm is performed through the ODZ. The range of admissible values of the original equation is the set of all numbers except −1 and 0 (when x=−1 and x=0, the denominators of fractions vanish). The root x=−1/2 found at the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.
Answer:
−1/2 .
Let's consider another example.
Example.
Find the roots of the equation.
Decision.
We need to solve a fractionally rational equation, let's go through all the steps of the algorithm.
First, we transfer the term from the right side to the left, we get .
Secondly, we transform the expression formed on the left side: . As a result, we arrive at the equation x=0 .
Its root is obvious - it is zero.
At the fourth step, it remains to find out if the root found is not an outside one for the original fractionally rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it does not make sense, since it contains division by zero. Whence we conclude that 0 is an extraneous root. Therefore, the original equation has no roots.
7 , which leads to the equation . From this we can conclude that the expression in the denominator of the left side must be equal to from the right side, that is, . Now we subtract from both parts of the triple: . By analogy, from where, and further.
The check shows that both found roots are the roots of the original fractional rational equation.
Answer:
Bibliography.
- Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
- Mordkovich A. G. Algebra. 8th grade. At 2 p.m. Part 1. Student's textbook educational institutions/ A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
- Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
So far, we have only solved integer equations with respect to the unknown, that is, equations in which the denominators (if any) did not contain the unknown.
Often you have to solve equations that contain the unknown in the denominators: such equations are called fractional.
To solve this equation, we multiply both sides of it by that is, by a polynomial containing the unknown. Will the new equation be equivalent to the given one? To answer the question, let's solve this equation.
Multiplying both sides of it by , we get:
Solving this equation of the first degree, we find:
So, equation (2) has a single root
Substituting it into equation (1), we get:
Hence, is also the root of equation (1).
Equation (1) has no other roots. In our example, this can be seen, for example, from the fact that in equation (1)
how unknown divisor must be equal to the dividend 1 divided by the quotient 2, i.e.
So, equations (1) and (2) have a single root. Hence, they are equivalent.
2. We now solve the following equation:
The simplest common denominator: ; multiply all the terms of the equation by it:
After reduction we get:
Let's expand the brackets:
Bringing like terms, we have:
Solving this equation, we find:
Substituting into equation (1), we get:
On the left side, we received expressions that do not make sense.
Hence, the root of equation (1) is not. This implies that equations (1) and are not equivalent.
In this case, we say that equation (1) has acquired an extraneous root.
Let us compare the solution of equation (1) with the solution of the equations we considered earlier (see § 51). In solving this equation, we had to perform two such operations that had not been seen before: first, we multiplied both sides of the equation by an expression containing the unknown (common denominator), and, second, we reduced algebraic fractions by factors containing the unknown .
Comparing Equation (1) with Equation (2), we see that not all x values valid for Equation (2) are valid for Equation (1).
It is the numbers 1 and 3 that are not admissible values of the unknown for equation (1), and as a result of the transformation they became admissible for equation (2). One of these numbers turned out to be a solution to equation (2), but, of course, it cannot be a solution to equation (1). Equation (1) has no solutions.
This example shows that when both sides of the equation are multiplied by a factor containing the unknown and when the algebraic fractions an equation may be obtained that is not equivalent to the given one, namely: extraneous roots may appear.
Hence we draw the following conclusion. When solving an equation containing an unknown in the denominator, the resulting roots must be checked by substitution into the original equation. extraneous roots must be discarded.
Equations with fractions themselves are not difficult and very interesting. Consider the types of fractional equations and ways to solve them.
How to solve equations with fractions - x in the numerator
In case given fractional equation, where the unknown is in the numerator, the solution does not require additional conditions and is solved without unnecessary trouble. General form such an equation is x/a + b = c, where x is an unknown, a, b and c are ordinary numbers.
Find x: x/5 + 10 = 70.
In order to solve the equation, you need to get rid of the fractions. Multiply each term of the equation by 5: 5x/5 + 5x10 = 70x5. 5x and 5 is reduced, 10 and 70 are multiplied by 5 and we get: x + 50 = 350 => x = 350 - 50 = 300.
Find x: x/5 + x/10 = 90.
This example is a slightly more complicated version of the first. There are two solutions here.
- Option 1: Get rid of fractions by multiplying all terms of the equation by a larger denominator, that is, by 10: 10x/5 + 10x/10 = 90x10 => 2x + x = 900 => 3x = 900 => x=300.
- Option 2: Add the left side of the equation. x/5 + x/10 = 90. The common denominator is 10. Divide 10 by 5, multiply by x, we get 2x. 10 divided by 10, multiplied by x, we get x: 2x+x/10 = 90. Hence 2x+x = 90×10 = 900 => 3x = 900 => x = 300.
Often there are fractional equations in which the x's are in different sides equal sign. In such a situation, it is necessary to transfer all fractions with x in one direction, and the numbers in another.
- Find x: 3x/5 = 130 - 2x/5.
- Move 2x/5 to the right with the opposite sign: 3x/5 + 2x/5 = 130 => 5x/5 = 130.
- We reduce 5x/5 and get: x = 130.
How to solve an equation with fractions - x in the denominator
This type of fractional equations requires writing additional conditions. The indication of these conditions is a mandatory and integral part right decision. By not attributing them, you run the risk, since the answer (even if it is correct) may simply not be counted.
The general form of fractional equations, where x is in the denominator, is: a/x + b = c, where x is an unknown, a, b, c are ordinary numbers. Note that x may not be any number. For example, x cannot be zero, since you cannot divide by 0. This is what is additional condition, which we must specify. This is called the range of acceptable values, abbreviated - ODZ.
Find x: 15/x + 18 = 21.
We immediately write the ODZ for x: x ≠ 0. Now that the ODZ is indicated, we solve the equation according to the standard scheme, getting rid of the fractions. We multiply all terms of the equation by x. 15x/x+18x = 21x => 15+18x = 21x => 15 = 3x => x = 15/3 = 5.
Often there are equations where the denominator contains not only x, but also some other operation with it, such as addition or subtraction.
Find x: 15/(x-3) + 18 = 21.
We already know that the denominator cannot be zero, which means x-3 ≠ 0. We transfer -3 to right side, while changing the “-” sign to “+” and we get that x ≠ 3. The ODZ is indicated.
Solve the equation, multiply everything by x-3: 15 + 18x(x - 3) = 21x(x - 3) => 15 + 18x - 54 = 21x - 63.
Move the x's to the right, the numbers to the left: 24 = 3x => x = 8.
Instruction
Perhaps the most obvious point here is, of course, . Numerical fractions do not pose any danger (fractional equations, where only numbers are in all denominators, will generally be linear), but if there is a variable in the denominator, then this must be taken into account and prescribed. Firstly, it is that x, which turns the denominator to 0, cannot be, and in general it is necessary to separately register the fact that x cannot be equal to this number. Even if you succeed that when substituting into the numerator, everything converges perfectly and satisfies the conditions. Secondly, we cannot multiply either or both sides of the equation by equal to zero.
After this, such an equation is reduced to transferring all its terms to the left side so that 0 remains on the right side.
It is necessary to bring all the terms to a common denominator, multiplying, where necessary, the numerators by the missing expressions.
Next, we solve the usual equation written in the numerator. We can endure common factors out of brackets, apply abbreviated multiplication, give similar ones, calculate the roots of a quadratic equation through the discriminant, etc.
The result should be a factorization in the form of a product of brackets (x-(i-th root)). It may also include polynomials that do not have roots, for example, square trinomial with a discriminant less than zero (unless, of course, there are only real roots in the problem, as most often happens).
Be sure to factorize and denominator from the location of the brackets there, already contained in the numerator. If the denominator contains expressions like (x-(number)), then it is better, when reducing to a common denominator, not to multiply the parentheses in it “head-on”, but to leave them in the form of a product of the original simple expressions.
The same brackets in the numerator and denominator can be reduced by pre-writing, as mentioned above, the conditions on x.
The answer is written in curly braces, as a set of x values, or simply by enumeration: x1=..., x2=..., etc.
Sources:
- Fractional rational equations
Something that cannot be dispensed with in physics, mathematics, chemistry. Least. We learn the basics of their solution.
Instruction
In the most general and simplest classification, it can be divided according to the number of variables they contain, and according to the degrees in which these variables stand.
Solve the equation all its roots or prove that they do not exist.
Any equation has at most P roots, where P is the maximum of the given equation.
But some of these roots may coincide. So, for example, the equation x ^ 2 + 2 * x + 1 = 0, where ^ is the exponentiation icon, folds into the square of the expression (x + 1), that is, into the product of two identical brackets, each of which gives x = - 1 as a solution.
If there is only one unknown in the equation, this means that you will be able to explicitly find its roots (real or complex).
To do this, you will most likely need various transformations: abbreviated multiplication, calculating the discriminant and roots of a quadratic equation, transferring terms from one part to another, reducing to a common denominator, multiplying both parts of the equation by the same expression, squaring, and so on.
Transformations that do not affect the roots of the equation are identical. They are used to simplify the process of solving an equation.
You can also use instead of the traditional analytical graphic method and write this equation in the form , after conducting its study.
If there is more than one unknown in the equation, then you will only be able to express one of them in terms of the other, thereby showing a set of solutions. Such, for example, are equations with parameters in which there is an unknown x and a parameter a. Decide parametric equation- means for all a to express x through a, that is, to consider all possible cases.
If the equation contains derivatives or differentials of unknowns (see picture), congratulations, this is differential equation, and here you can not do without higher mathematics).
Sources:
To solve the problem with fractions gotta learn to do with them arithmetic operations. They can be decimal, but are most commonly used natural fractions with numerator and denominator. Only then can you move on to solutions. math problems with fractional values.
You will need
- - calculator;
- - knowledge of the properties of fractions;
- - Ability to work with fractions.
Instruction
A fraction is a record of dividing one number by another. Often this cannot be done completely, and therefore this action is left “unfinished. The number that is divisible (it is above or before the fraction sign) is called the numerator, and the second number (under or after the fraction sign) is called the denominator. If the numerator is greater than the denominator, the fraction is called an improper fraction, and an integer part can be extracted from it. If the numerator less than the denominator, then such a fraction is called proper, and its whole part equals 0.
Tasks are divided into several types. Determine which one is the task. The simplest option- finding the fraction of a number expressed as a fraction. To solve this problem, it is enough to multiply this number by a fraction. For example, 8 tons of potatoes were brought in. In the first week, 3/4 of her total. How many potatoes are left? To solve this problem, multiply the number 8 by 3/4. It will turn out 8 ∙ 3/4 \u003d 6 t.
If you need to find a number by its part, multiply the known part of the number by the reciprocal of the fraction that shows what proportion of this part is in the number. For example, 8 out of 1/3 of the total number of students. How many in ? Since 8 people is the part that represents 1/3 of the total, then find reciprocal, which is equal to 3/1 or just 3. Then to get the number of students in the class 8∙3=24 students.
When you need to find what part of a number is one number from another, divide the number that represents the part by the one that is the whole number. For example, if the distance is 300 km and the car has traveled 200 km, how much of this will be from the total journey? Divide the part of the path 200 by full path 300, after reducing the fraction you will get the result. 200/300=2/3.
To find the part of the unknown fraction of a number, when there is a known one, take the integer as a conventional unit, and subtract the known fraction from it. For example, if 4/7 of the lesson has already passed, is there still left? Take the whole lesson as a conventional unit and subtract 4/7 from it. Get 1-4/7=7/7-4/7=3/7.
Solving equations with fractions let's look at examples. The examples are simple and illustrative. With their help, you can understand in the most understandable way,.
For example, you need to solve a simple equation x/b + c = d.
An equation of this type is called linear, because the denominator contains only numbers.
The solution is performed by multiplying both sides of the equation by b, then the equation takes the form x = b*(d – c), i.e. the denominator of the fraction on the left side is reduced.
For example, how to solve a fractional equation:
x/5+4=9
We multiply both parts by 5. We get:
x+20=45
x=45-20=25
Another example where the unknown is in the denominator:
Equations of this type are called fractional rational or simply fractional.
We would solve a fractional equation by getting rid of fractions, after which this equation, most often, turns into a linear or quadratic one, which is solved in the usual way. You should only take into account the following points:
- the value of a variable that turns the denominator to 0 cannot be a root;
- you cannot divide or multiply the equation by the expression =0.
Here comes into force such a concept as the area of permissible values (ODZ) - these are the values \u200b\u200bof the roots of the equation for which the equation makes sense.
Thus, solving the equation, it is necessary to find the roots, and then check them for compliance with the ODZ. Those roots that do not correspond to our DHS are excluded from the answer.
For example, you need to solve a fractional equation:
Based on the above rule, x cannot be = 0, i.e. ODZ in this case: x - any value other than zero.
We get rid of the denominator by multiplying all terms of the equation by x
And solve the usual equation
5x - 2x = 1
3x=1
x = 1/3
Answer: x = 1/3
Let's solve the equation more complicated:
ODZ is also present here: x -2.
Solving this equation, we will not transfer everything in one direction and bring fractions to a common denominator. We immediately multiply both sides of the equation by an expression that will reduce all the denominators at once.
To reduce the denominators, you need to multiply the left side by x + 2, and the right side by 2. So, both sides of the equation must be multiplied by 2 (x + 2):
Exactly this ordinary multiplication fractions, which we have already discussed above
We write the same equation, but in a slightly different way.
The left side is reduced by (x + 2), and the right side by 2. After the reduction, we get the usual linear equation:
x \u003d 4 - 2 \u003d 2, which corresponds to our ODZ
Answer: x = 2.
Solving equations with fractions not as difficult as it might seem. In this article, we have shown this with examples. If you are having any difficulty with how to solve equations with fractions, then unsubscribe in the comments.
