Solve equations that reduce to quadratic. Rational equations

Item

algebra

Class

Topic and lesson number in the topic

"Quadratic equations"; Lesson 14

Basic tutorial

"Algebra 8", ed. Sh.A. Alimov et al., Moscow: Education, 2009.

5. Purpose of the lesson: consolidate algorithms for solving biquadratic and fractional rational equations.

6. Tasks:

- educational: know the form of biquadratic and fractional rational equations; algorithm for solving biquadratic

and fractional rational equation; be able to solve biquadratic and fractional rational equations;

-developing : developing the ability to highlight the main thing, compare, analyze and draw conclusions;

developing the ability to formulate cognitive tasks and plan cognitive activities;

develop personality traits - hard work, accuracy, perseverance in achieving goals;

- educational: production objective assessment your achievements; formation of responsibility;

development of teamwork skills.

7. Lesson type: atrock consolidation of knowledge.

8. Forms of student work: frontal, individual;group

9. Necessary technical equipment: computer, projector, ID.

TECHNOLOGICAL LESSON MAP

Didactic sky structure lesson tour	Methodological structure of the lesson
Didactic sky structure lesson tour		Updating knowledge	Formation of ZUN	Consolidation	Control		Instructing home
	Preparing students for work in the classroom	Provide motivation, update basic knowledge and skills	Repeat the algorithm for solving biquads. ur-th and methods for solving fractional-rat. equations	Know the algorithm for solving biquads. ur-th and methods for solving fractional-rat. equations and the ability to apply them in practice	Group work	Analyze and evaluate the success of achieving the goal of the lesson and outline prospects for further work	Provide an understanding of the purpose, content and methods of performing the task
		Today we will be a dodecahedron. To fix the faces of the dodecahedron, you need to solve certain type equations What type of equations did you learn to solve in previous lessons?	1.Together with the student, formulate the purpose of the lesson 2.Repeat the algorithm for solving the biquadratic equation; conditions for equality of fraction 0; square root formula equations	1. Several options for the formula for square roots have been proposed. ur. - choose the correct Pres. L.No.2 2. Establish a correspondence between the stages of the algorithm and the points of solving the biqu. ur-I Pres. L.No.3 3. Select from three answer options the condition for the existence of the fraction Pres.	L.No.4 4.Find the error in solving equation. Pres.	L.No.5	Groups of different levels perform tasks using cards. Adj. No. 1 The slide contains the faces of the dodecahedron with the correct answers. Pres. l. No. 6. Each team is pre-assigned a color, if the team solved the equations correctly, then its selected faces will match its color in three places. At the end of the work, students check the results of constructing the net of the dodecahedron. Prez.l. No. 7
The teacher, together with the students, sums up the work done.		Correcting errors: fixing equation solutions	Correcting errors: fixing equation solutions	great job	, in which there were errors	Teaching methods	Teaching methods
Reproductive		Partial search	Partial search	Partial search	Partially search, search, self-control	Self-control	Partial search
Forms of org. cognizer-	Frontal	Group Individual, frontal	Real result	All students joined the work environment	Students prepared for active learning	cognitive activity	Students got acquainted with equations that can be reduced to quadratic equations and methods for solving them

Students know how to solve quadratic equations Students have an idea of the degree to which they have mastered the material they have studied, their achievements and gaps in the topic studied The student conducted a self-assessment of knowledge and skills on the topic, drawing a conclusion about the result of his work Conditions have been created for doing homework

In this article I will show you

seven types of solution algorithms

rational equations

1 , which can be reduced to quadratic by changing variables. In most cases, the transformations that lead to replacement are very non-trivial, and it is quite difficult to guess about them on your own.

For each type of equation, I will explain how to make a change of variable in it, and then show a detailed solution in the corresponding video tutorial.

1. You have the opportunity to continue solving the equations yourself, and then check your solution with the video lesson.

So, let's begin.

. (x-1)(x-7)(x-4)(x+2)=40

Note that on the left side of the equation there is a product of four brackets, and on the right side there is a number.

Let's group the brackets by two so that the sum of the free terms is the same.

2. Multiply them.

3. Let's introduce a change of variable.

2 .

An equation of this type is similar to the previous one with one difference: on the right side of the equation is the product of the number and . And it is solved in a completely different way:

1. We group the brackets by two so that the product of the free terms is the same.

2. Multiply each pair of brackets.

3. We take x out of each factor.

4. Divide both sides of the equation by .

5. We introduce a change of variable.

In this equation, we group the first bracket with the fourth, and the second with the third, since:

Note that in each bracket the coefficient at and the free term are the same. Let's take a factor out of each bracket:

Since x=0 is not a root of the original equation, we divide both sides of the equation by . We get:

We get the equation:

3. Let's introduce a change of variable.

3 .

Note that the denominators of both fractions contain quadratic trinomials, in which the leading coefficient and the free term are the same. Let us take x out of the bracket, as in the equation of the second type. We get:

Divide the numerator and denominator of each fraction by x:

Now we can introduce a variable replacement:

We obtain an equation for the variable t:

4 .

Note that the coefficients of the equation are symmetrical with respect to the central one. This equation is called returnable .

To solve it,

1. Divide both sides of the equation by (We can do this since x=0 is not a root of the equation.) We get:

2. Let’s group the terms in this way:

3. In each group, let’s take the common factor out of brackets:

4. Let's introduce the replacement:

5. Express through t the expression:

From here

We get the equation for t:

3. Let's introduce a change of variable.

5. Homogeneous equations.

Equations that have a homogeneous structure can be encountered when solving exponential, logarithmic and trigonometric equations, so you need to be able to recognize it.

Homogeneous equations have the following structure:

In this equality, A, B and C are numbers, and the square and circle denote identical expressions. That is, on the left side of a homogeneous equation there is a sum of monomials having the same degree (in in this case the degree of the monomials is 2), and there is no free term.

To solve homogeneous equation, divide both sides by

Attention! When dividing the right and left sides of an equation by an expression containing an unknown, you can lose roots. Therefore, it is necessary to check whether the roots of the expression by which we divide both sides of the equation are the roots of the original equation.

Let's go the first way. We get the equation:

Now we introduce variable replacement:

Let us simplify the expression and obtain a biquadratic equation for t:

3. Let's introduce a change of variable. or

7 .

This equation has the following structure:

To solve it, you need to select on the left side of the equation perfect square.

To select a full square, you need to add or subtract twice the product. Then we get the square of the sum or difference. This is crucial for successful variable replacement.

Let's start by finding twice the product. This will be the key to replacing the variable. In our equation double product equals

Now let's figure out what is more convenient for us to have - the square of the sum or the difference. Let's first consider the sum of expressions:

Great! This expression is exactly equal to twice the product. Then, in order to get the square of the sum in brackets, you need to add and subtract the double product:

A quadratic equation is the equation ax²+bx+c=0, where a, b, c are given numbers, a0, x are unknown. The coefficients a, b, c of a quadratic equation are usually called as follows: a – the first or leading coefficient, b – the second coefficient, c – the free term. For example, in the equation 3x²-x+2=0, the senior (first) coefficient is a=3, the second coefficient is b=-1, and the free term is c=2. Solving many problems in mathematics, physics, and technology comes down to solving quadratic equations: 2x²+x-1=0, x²-25=0, 4x²=0, 5t²-10t+3=0. When solving many problems, equations are obtained that, using algebraic transformations are reduced to square ones. For example, the equation 2x²+3x=x²+2x+2, after moving all its terms to the left side and bringing similar terms, is reduced to the quadratic equation x²+x-2=0.

Consider the equation general view: ax²+bx+c=0, where a0. The roots of the equation are found using the formula: The expression is called the discriminant of a quadratic equation. If D 0, then the equation has two real roots. In the case where D=0, the quadratic equation is sometimes said to have two identical roots.

Incomplete quadratic equations. If in a quadratic equation ax²+bx+c=0 the second coefficient b or the free term c are equal to zero, then the quadratic equation is called incomplete. An incomplete quadratic equation may have one of the following types: Incomplete equations are isolated because to find their roots you don’t have to use the formula for the roots of a quadratic equation - it’s easier to solve the equation by factoring its left side.

A quadratic equation of the form x 2 +px+q=0 is called reduced. In this equation the leading coefficient equal to one: a=1. The roots of the above quadratic equation are found by the formula: This formula is convenient to use when p is an even number. Example: Solve the equation x 2 -14x-15=0. Using the formula we find: Answer: x 1 =15, x 2 =-1.

Francois Viet? Vieta's theorem. If the reduced quadratic equation x 2 +px+q=0 has real roots, then their sum is equal to -p, and the product is equal to q, that is, x 1 +x 2 = -p, x 1 x 2 = q (the sum of the roots of the reduced square equation is equal to the second coefficient taken from opposite sign, and the product of the roots is equal to the free term). Study of the relationship between the roots and coefficients of a quadratic equation.

Statement 1: Let x 1 and x 2 be the roots of the equation x 2 +pх+q=0. Then the numbers x 1, x 2, p, q are related by the equalities: x 1 + x 2 = - p, x 1 x 2 =q Statement 2: Let the numbers x 1, x 2, p, q be related by the equalities x 1 + x 2 = - p, x 1 x 2 =q. Then x 1 and x 2 are the roots of the equation x 2 +pх+q=0 Corollary: x 2 +pх+q=(x-x 1)(x-x 2). Situations in which Vieta's theorem can be used. Checking the correctness of the found roots. Determining the signs of the roots of a quadratic equation. Orally finding the integer roots of a given quadratic equation. Drawing up quadratic equations with given roots. Decomposition quadratic trinomial by multipliers.

Biquadratic equations A biquadratic equation is an equation of the form where a 0. Biquadratic equation is solved by introducing a new variable: putting, we get a quadratic equation Example: Solve the equation x 4 +4x 2 -21=0 Putting x 2 =t, we get a quadratic equation t 2 +4t -21=0, from where we find t 1 = -7, t 2 =3. Now the problem comes down to solving the equations x 2 = -7, x 2 =3. The first equation has no real roots; from the second we find: which are the roots of the given biquadratic equation.

Solving problems using quadratic equations Problem 1: The bus departed from the bus station to the airport, located 40 km away. 10 minutes later, a passenger left behind the bus in a taxi. The speed of a taxi is 20 km/h faster than the speed of a bus. Find the speed of the taxi and bus if they arrived at the airport at the same time. Speed V (km/h) Time t (h) Path S (km) Busx40 TaxiX+2040 For 10 min 10 min = h Let’s create and solve the equation:

Multiplying both sides of the equation by 6x(x+20), we get: The roots of this equation: For these values of x, the denominators of the fractions included in the equation are not equal to 0, therefore they are the roots of the equation. Since the speed of the bus is positive, only one root satisfies the problem conditions: x=60. Therefore, the taxi speed is 80 km/h. Answer: The speed of a bus is 60 km/h, the speed of a taxi is 80 km/h.

Problem 2: The first typist spends 3 hours less on retyping the manuscript than the second. Working simultaneously, they completed retyping the entire manuscript in 6 hours and 40 minutes. How long would it take each of them to retype the entire manuscript? Amount of work per hour Time t (h) Amount of work First typist x1 Second typist x+31 Together in 6 hours 40 minutes 6 hours 40 minutes = 6 hours Let’s create and solve the equation:

This equation can be written as follows: Multiplying both sides of the equation by 20x(x+3), we get: The roots of this equation: For these values of x, the denominators of the fractions included in the equation are not equal to 0, therefore - the roots of the equation. Since time is positive, then x=12h. Therefore, the first typist spends 12 hours on work, the second - 12 hours + 3 hours = 15 hours Answer: 12 hours and 15 hours 15

Francois Viet Francois Viet was born in 1540 in France. Viet's father was a prosecutor. The son chose his father's profession and became a lawyer, graduating from the university in Poitou. In 1563, he left jurisprudence and became a teacher in a noble family. It was teaching that aroused the young lawyer’s interest in mathematics. Viet moves to Paris, where it is easier to learn about the achievements of Europe's leading mathematicians. Since 1571, Viet has occupied important government posts, but in 1584 he was suspended and expelled from Paris. Now he had the opportunity to take mathematics seriously. In 1591, he published the treatise “Introduction to the Analytical Art,” where he showed that by operating with symbols, one can obtain a result applicable to any corresponding quantities. The famous theorem was published that same year. He gained great fame under Henry III during the Franco-Spanish War. Within two weeks, working days and nights, he found the key to the Spanish cipher. Died in Paris in 1603, there are suspicions that he was murdered.

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Standard types of equations and methods for solving them

1. Equation of the form
= b↔ f(x) = b 2, for b ≥ 0; has no solutions for b

Golden Rule. To solve, the root needs to be isolated.

Examples.

3)
. There are no solutions, because...

2. Equation of the form

Examples.

Answer: x = - 1

2) In examples that boil down to this species equations, when using equivalent transitions it is necessary to find the range of permissible values.

Example.

Answer

3. Equation of the form

or

Choose the inequality that is simpler.

Examples.

, sinх = t, |t| ≤ 1, t ≥ 0, 0 ≤ t ≤ 1

2t 2 + t – 1 = 0

t = -1 , t = ½ Subject to restrictions t = ½

3. Let's introduce a change of variable.

4. Equations reducing to quadratic

Such equations contain roots with identical radical expressions, the degrees of which differ by a factor of two (
). Solved by replacing the root
, subject to restrictions.

Examples.

= t, where t ≥ 0

t 2 – 2 t – 3 = 0, t = - 1, t = 3, taking into account that t ≥ 0, t = 3

= 3

Answer: x = ± 7

= t, then

= 2 or = ½

= 32 = 1/32

16z =32 16·32z – z = - 1

z = 2 z = - 1/511
5. Equations containing more than one root in the form of terms

In equations of this type it is necessary to get rid of the roots. Most often this is done by squaring both parts. It should be noted that when squaring the ODZ of the unknown, it expands, which can lead to foreign roots equations Squaring does not provide an equivalent transition, so the resulting values of the unknown must be checked.

When making a decision, the following rules must be observed:

Spread the roots across different sides, since the transformations in this case are simpler;
Find the set of values for which the roots exist;
Square both parts;
Bring the equation to standard form;
Solve according to types 1 – 3;
Eliminate extraneous roots;
Check remaining roots.

Examples.

solve by performing step 5 (equation of the form)

Checking x = 3

Equality is true.

Answer: x = 3.
2)

3x – 4 - 2
= x – 2

2x – 2 = (1) x – 1 =

Note that, based on equivalence, we solve only equation (1), and not the original one, so we need to check.

You can solve without taking into account the ODZ and not use equivalence, but in this case all obtained values of x must be checked. In some equations this is quite complicated.

Examination. x = 3

Equality is true.

Answer: x = 3
6. Equations solved by changing variables.

6.1 Obvious substitutions.

If the example contains terms with repeated expressions, then it is advisable to replace variables, which in essence is not a direct solution, but greatly simplifies the transformation of expressions and bringing the equation to a standard form.

Golden Rule . Made a replacement - determine the area of change of the new variable. (put restrictions on the new variable)

Examples.

Let = t, where t ≥ 0, since the root is an arithmetic one.

We get: t 2 – 2t – 3 = 0

t = - 1, t = 3

Because t ≥ 0, t = 3

Let's move on to x

= 3 x 2 + 32 = 81, x = ± 7.

Answer: x = ± 7.

Because
And
mutually inverse expressions, then if
= t,

= , where t > 0.

We get t + = , 2t 2 – 5t + 2 = 0,

t = ½, t = 2,

= or = 2

8x = 1+2x, 2x = 4 + 8x

x = 1/6. x = - 2/3

The largest root x = 1/6.

= t, t ≥ 0 Replace the root and express the right-hand side through t.

= t 2,
t 2 – 20

t = - (t 2 – 20), t 2 + t – 20 = 0. t = - 5 or t = 4.

Because t ≥ 0, then t = 4

= 4,

x 2 + 2x + 8 = 16,

x 2 + 2x - 8 = 0, x = - 4 or x = 2.

Answer: x = - 4, x = 2.

4)
. We will produce double replacement:

t =
, where t ≥ 0, d =
, where d ≥ 0.

Let's express x from each: x = 5 - t 2 or x = d 2 + 3. Let's get the system:

. t = 0 or d = 0

= 0 or = 0

x = 5 or x = 3

Answer: x = 5; x = 3.

6.2 Non-obvious replacement

The replacement of a variable may not occur immediately, but after the transformations have been carried out.

Examples.

ODZ: - 1 ≤ x ≤ 3

Let's reschedule
right for more complex expression
there's only one left.

Let's square both sides, expecting to get the same expressions:

The expectations were justified.

= t, t ≥0
= t 2 + 4

4t = t 2 + 4, t 2 – 4t + 4 = 0, (t – 2) 2 = 0, t = 2

= 2,
= 4,

x = 1 is the root of the equation, since the sum of the coefficients and the free term is zero.

let's divide
on x – 1. We get x 2 – 2x + 1 = 0. x = 1 ±
.

All three roots are solutions because they satisfy the condition - 1 ≤ x ≤ 3.

Answer: x = 1, x = 1 ±
7. Equations of the form product equals zero.

The product is equal to zero when at least one of the factors is equal to zero, while the other does not lose its meaning.

f(x) g(x) = 0

Examples.

= 0

no solutions x = - 1, x = 2.

Answer: x = - 1, x = 2.

The inequalities included in the system can not be solved immediately, but the resulting root can be substituted into the inequality.

2) It is necessary to factorize.

= 4

no solutions x = 0, x = 5.

Answer: x = 0, x = 5.

Equations containing square and cube roots.

These equations should be solved by replacing each root, expressing the unknown in terms of the replaced variables, and constructing a system of equations.

Examples.

= t,
= d, where d ≥ 0

x = 2 - t 3 , x = d 2 + 1. Let's create a system:

Because for all found values t d ≥ 0, then d can not be found from the system, but x can be found from the condition x = 2 - t 3 .

x = 2, x = 10, x = 1

Answer: x = 2, x = 10, x = 1

2)
.

1 way. Solve as the previous equation.

Method 2. Note that the left side of the equation represents an increasing function, since it consists of the sum of two increasing functions on the domain of definition: x ≥ - 1. Right part– constant. The graphs of these functions intersect at one point, the abscissa of which will be a solution to this equation, i.e. the equation has one solution. Let's try to pick it up.

Obviously, the selection must be carried out in the ODZ equations. It must be assumed that the roots must be removed, because... the sum is 3.

We make sure that x = 3 is the root of the equation.

Answer: x = 3.

3)
.

Because
Let's reduce the roots to the same degree.

, x = - 1

(x + 1)(x 2 – 4x + 4)

x 2 – 4x + 4 =0 x = 2.

Both roots satisfy the ODZ.

Answer: x = - 1, x = 2

An equation containing the sum (difference) of two third roots.

To solve such equations it is convenient to use the formula:

(a + b) 3 = a 3 + b 3 + 3ab(a + b),

(a - b) 3 = a 3 - b 3 - 3ab(a - b) .

Note that the bracket (a ± b) =

Examples.

1)
. Let's cube both parts:

But
= 2, so replace the last parenthesis with 2.

We get

x = 0

answer: x = 0.

Note that the expressions 2 – x and 7 + x are repeated. Let's make a replacement:

t =
, d =
. Where does x = 2 - t 3 or x = d 3 – 7

You don’t have to find t and d, but use the fact that td = 2

= 2

- x 2 – 5x + 14 = 8, x 2 + 5x - 6 = 0, x = - 6, x = 1.

Answer: x = - 6, x = 1.

Equations containing complex radicals.

In the presence of complex radicals, for example, a root under a root, use the following program of action:

Determine whether the radical expression is a perfect square;
Select a complete square;
In the absence of paragraph 1, apply the formulas of complex radicals;
In the absence of paragraphs 1–3, apply standard transformations (replacement, factorization, exponentiation, etc.)

Examples.

Let's try to find a perfect square. (a ± b) 2 = a 2 ± 2ab + b 2. In this case, one should reason as follows:

Let
- double product, 2ab.

Let
- first number a.

Then the second number b = 1. Therefore, the sum of the squares of the first and second numbers is x – 3. The radical expression is a complete square.

Let
- double product.

Let be the first number a.

Then the second number b = 2. Therefore, the sum of the squares of the first and second numbers is equal to x. The radical expression is a complete square.

+ = 1

Because
│a│, then we get the equation:
│
│ + │
│ = 1

Now let’s make the replacement = t , = t – 1

│t │ + │t – 1 │ = 1

Let's find the zeros of the modules: t = 0, t = 1

│t │

- │ + │ +

│t – 1 │- 0 - 1 + x

no solutions
no solutions

0 ≤ ≤ 1

1 ≤ ≤ 2 Because Let's square all parts of the inequality.

1 ≤ x – 4 ≤ 4, 5 ≤ x ≤ 8.

Answer:

Methods for solving irrational equations

Using the properties of monotonicity of functions.

11.1 If f(x) = g(x), and f(x) is increasing (decreasing) and g(x) is decreasing (increasing) or one of the functions is a constant, then the graphs of these functions intersect at one point. The solution to the equation is the abscissa of the intersection point. The equation has one solution, which can be determined by selection.

The following must be kept in mind:

The sum of two increasing (decreasing) functions is an increasing (decreasing) function.
The increase and decrease of a function can be determined by its derivative.

Examples.

1)
.

Let f(x) =
. f(x) – decreases by D(f) = (-∞; 3]

g(x) = 6 – constant. The graphs of functions intersect at one point. The equation has one solution.

We select from D(f) = (-∞; 3], taking into account that the roots must be extracted.

x = - 1.

Examination.

, 4 + 2 = 6, the equality is true.

Answer: x = - 1.

Let f(x) =
. The function is decreasing.

Let's prove it. D(f) =

f ′(x) =

f ′(x) = 0, = 0, x = 2 D(f)

f(1) =
, f(2) = 3, f(3) =

E(f) = [; 3]

g(x) =
, D(g) =

g ′(x) =

g ′(x) = 0 = 0, x = 1 D(g)

g(0) = 3, g(1) = 4, g(2) = 3

E(g) =

Note that we obtain the same function values only for x = 2

You can also reason as follows: highest value one function is equal to lowest value another function for the same values of x. Therefore, the solution to the equation f(x) = g(x) is these values of x.

max f = 3, min g = 3, max f = min g = 3 at x = 2

Answer: x = 2

1 way.

Let f(x) =
, D(f) = R.

f ′(x) = 4x 3 + 12x 2 + 12x + 4

f ′(x) = 0 4х 3 + 12х 2 + 12х + 4 = 0,

x 3 + 3x 2 + 3x + 1 = 0, (x + 1) 3 = 0

x = - 1
f ′(x) - │ +

f(x) - 1
f min = f(-1) = - 1 E(f) = [ - 1; ∞)
g(x) =
D(g) = R.

g ′(x) =
, g ′(x) = 0 x = - 1

g ′(x) + -

g(x) │- 1

g max = g(-1) = - 1 E(g) =(- ∞; - 1]
min f = max g = - 1 at x = -1.

Answer: x = - 1.

Method 2.

Select the perfect square of a polynomial:

(x 2 + 2x) 2 + 2x 2 + 4x. We get:

(X 2 + 2x) 2 + 2(x 2 + 2x) +
.

Now you can make a replacement:

x 2 + 2x = t

t 2 + 2 t +
= 2

It is possible that in given equation Method 2 is preferable. But it is necessary to master the estimation method well, since many equations, systems, and inequalities are solved precisely in this way.

Use of DZ

Sometimes it is useful to find the ODZ of an unknown, which can lead to narrowing the search for a solution and solving the equation itself.
The analysis shows that the use of any methods is difficult. Let's try to find the ODZ.
Thus x = 4 is the only possible value.
Examination.
, 0 = 0 the equality is true.
Answer: x = 4.
14. Using obvious inequalities
It is known that
(the arithmetic mean is greater than or equal to the geometric mean). In this case, equality is observed if a = b.
If there is a product under the root in the equation, it is advisable to apply this property.
Examples.
1)

Let's factorize the radical expression.
Let a = x + 1, b = 2x + 3, then a + b = 3x + 4.
On the left side is the geometric mean, on the right side is the arithmetic mean.
There will be equality if a = b.
x + 1 = 2x + 3, x = - 2.
Answer: x = - 2.
15. Using the dot product
Let the vector has coordinates (a 1 ; a 2), vector (b 1; b 2).
Then scalar product· = a 1 b 1 + a 2 b 2 . Because a 1 b 1 + a 2 b 2 = ││∙ ││· cosα, therefore, a 1 b 1 + a 2 b 2 ≤ ││∙ ││
││ =
││=
│ =

Consider using ODZ;

Consider using function monotonicity;

Consider using the properties of a function (range of values, greatest, least), i.e. apply assessments;

Consider using conjugate expressions;

Consider the possibility of using obvious inequalities and the scalar product.
Note that the same equation can be solved different ways. You need to choose the method that is better learned, which is more rational for the given equation.