Summary of mathematics lesson "rational equations". Solving equations in two variables What does it mean to solve a rational equation
549+45y+4y=-7, 45y+4y=549-7, 49y=542, y=542:49, y≈11.
In an arbitrarily chosen (from the system) equation, insert the number 11 instead of the already found “game” and calculate the second unknown:
X=61+5*11, x=61+55, x=116.
The answer to this system of equations is x=116, y=11.
Graphic method.
It consists of practically finding the coordinates of the point at which straight lines, mathematically written in a system of equations, intersect. The graphs of both lines should be drawn separately in the same coordinate system. General form of the equation of a straight line: – у=khх+b. To construct a straight line, it is enough to find the coordinates of two points, and x is chosen arbitrarily.
Let the system be given: 2x – y=4
Y=-3x+1.
A straight line is constructed using the first equation; for convenience, you need to write it down: y = 2x-4. Come up with (easier) values for x, substituting it into the equation, solving it, and finding y. We get two points along which a straight line is constructed. (see picture)
x 0 1
y -4 -2
A straight line is constructed using the second equation: y=-3x+1.
Also construct a straight line. (see picture)
y 1 -5
Find the coordinates of the intersection point of two constructed lines on the graph (if the lines do not intersect, then the system of equations has no solution - this happens).
Video on the topic
Helpful advice
If you solve the same system of equations in three different ways, the answer will be the same (if the solution is correct).
Sources:
- 8th grade algebra
- solve an equation with two unknowns online
- Examples of solving systems of linear equations with two
Solving a system of equations is challenging and exciting. The more complex the system, the more interesting it is to solve. Most often in secondary school mathematics there are systems of equations with two unknowns, but in higher mathematics there may be more variables. Systems can be solved using several methods.
Instructions
The most common method for solving a system of equations is substitution. To do this, it is necessary to express one variable in terms of another and substitute it into the second equation of the system, thus reducing the equation to one variable. For example, given the following equations: 2x-3y-1=0;x+y-3=0.
From the second expression it is convenient to express one of the variables, moving everything else to the right side of the expression, not forgetting to change the sign of the coefficient: x = 3-y.
Open the brackets: 6-2y-3y-1=0;-5y+5=0;y=1. We substitute the resulting value y into the expression: x=3-y;x=3-1;x=2.
In the first expression all terms are 2, you can put 2 out of brackets
Math lesson notes
on the topic of:
« Rational equations with two variables.
Basic Concepts».
Prepared by:
Mathematic teacher
MBOU secondary school No. 2
Borschova E. S.
Pavlovsky Posad
Lesson type: learning new material.
Lesson topic: rational equations with two variables. Basic concepts.
Goals:
introduce basic concepts and terms of the topic;
develop students’ mathematical speech and thinking.
Equipment: board for notes, projector, screen, presentation.
Organizing time. (2 – 3 min.)
(1 slide)
Hello guys, have a seat! Today we will look at a new, quite interesting topic, which will be the key to successfully mastering future material. We open our workbooks, write down the date, today is October 16, class work and lesson topic: “Rational equations with two variables. Basic concepts". (the teacher writes the same thing on the board)
II . Updating knowledge. (5 minutes.)
(2 slide)
In order to start studying a new topic, we need to remember some material that you already know. So, let's remember the elementary functions and their graphs:
1. Graph of a Linear Function
2. Parabola. Graph of a quadratic function 
, (a ≠ 0)
Consider the canonical case:
3. Cubic parabola
A cubic parabola is given by the function
4. Hyperbola graph
Again we recall the trivial hyperbole
Very good!
III . Studying new material (accompanied by a presentation). (35 min.)
(3 slide)
In previous lessons you learned the definition of a rational equation in one variable, and now we are saying that it is very similar to the definition of a rational equation in two variables:
You don’t need to write it down, it’s in your textbooks, read it again at home and learn it!
Write down examples in your notebook:
Further, we can say that a rational equation of the form h(x; y) = g(x; y) can always be transformed to the form p(x; y) = 0, where p(x; y) = 0 is a rational expression. To do this, you need to rewrite the expression like this: h (x; y) - g (x; y) = 0, i.e. p (x; y) = 0. Write down the last two equalities in your notebook!
(4 slide)
We listen carefully and remember the following definition; there is no need to write it down!
And in your notebook write down only examples:
(5 slide)
Let’s solve the following equation (students write the solution in their notebooks, the teacher comments on each step of the solution, while simultaneously answering the children’s questions):
(6 slide)
The next definition is the definition of equivalence of two equations, you also already know this from the previous paragraphs, so just watch and listen:
Now let's remember what equivalent transformations you know:
Transferring terms of an equation from one part to another with opposite signs (examples on the board, you don’t have to write them down, if you want, write them down);
Multiplying or dividing both sides of an equation by the same number different from zero or (we also know) by an expression that is everywhere different from zero (pay attention to this!); (Write down examples for anyone who needs them).
What unequal transformations do you know?
1) exemption from denominators containing variables;
2) squaring both sides of the equation.
Wonderful!
(7 slide)
The next concept that we will consider today is the formula for the distance between two points.
Write:
(students write both theorems in their notebooks)
We redraw this drawing in a notebook, label the coordinate axes, the center of the circle, and mark the radius.
Do you have any questions? (if there are no questions, we continue working)
(8 slide)
Let's look at examples, write down:
(Fig. to P1) 
(Fig. to P2)
Children gradually, based on the above written theorem, answering the teacher’s questions, decide independently, write down the solution in a notebook, and redraw the drawings.
Well done! Now, redraw such a table for yourself, it will become a good assistant in the future when solving problems.
(9 slide)
Students carefully draw this table in their notebooks and enter the data into it.
V. Homework (2 – 3 min.).
(10 slide)
There are 2 minutes left until the end of the lesson, open the diaries, write down your homework:
1) Chapter 2, §5;
2) p. 71 self-test questions;
3) No. 5.1; No. 5.3 (a, b); No. 5.7.
Introspection.
The beginning of the lesson was quite friendly, sincere, open and organized. The class was prepared for the lesson. The children showed good performance throughout the lesson.
I immediately announced the goals of the lesson. The goals proposed for the children for the lesson corresponded to the program requirements and the content of the material.
At the beginning of the lesson, as a way to intensify cognitive activity, the children were asked to recall some material from previously studied material, which they coped with without any particular difficulties.
The content of the lesson met the requirements of the educational standard.
The structure of the lesson is suggested above. In my opinion, it corresponds to the goals and type of lesson. The stages of the lesson were logically connected and smoothly transitioned into one another. At each stage the results were summed up. Time was allocated to individual stages differently depending on which of them was the main one. In my opinion, it was distributed rationally. The beginning and end of the lesson were organized. The pace of the lesson was optimal.
After the first stage of updating knowledge, the main stage of the lesson came - an explanation of the new material. This stage was the main one, so most of the time was devoted to it.
The presentation of the new material was logical, competent, at a high theoretical and at the same time accessible to children level. I always highlighted the main thoughts on the topic and wrote them down in their workbooks.
The study of new material was carried out in the form of a short lecture with the completion of basic practical tasks, for the fastest and most correct assimilation of the material.
I made a presentation in PowerPoint. The presentation had a mainly auxiliary function.
In order to control the assimilation of knowledge, throughout the lesson, students solved problems, based on the results of which I could judge the degree of assimilation of theoretical material by each of the children. After monitoring the knowledge, the teacher carried out correction work. Those questions that caused the most difficulty for students were considered again.
After this, the lesson was summed up and the students were given homework. The homework was of a reinforcing, developmental nature. In my opinion, it was feasible for all children.
The content of the lesson was optimal, the teaching methods were oral, visual and practical. The form of work is conversation. I used techniques for activating cognitive activity - posing problematic questions, generalizing according to plans of a general nature.
The students were active in the lesson. They showed the ability to work productively, draw conclusions from what they saw, and the ability to analyze and generalize their knowledge. The children also showed the presence of self-control skills, but only a few were restless, and they received the most attention from me.
The class was prepared for the lesson.
I believe that the goals set at the beginning of the lesson have been achieved.
Instructions
Substitution MethodExpress one variable and substitute it into another equation. You can express any variable at your discretion. For example, express y from the second equation:
x-y=2 => y=x-2Then substitute everything into the first equation:
2x+(x-2)=10 Move everything without “x” to the right side and calculate:
2x+x=10+2
3x=12 Next, to get x, divide both sides of the equation by 3:
x=4. So, you found “x. Find "y. To do this, substitute “x” into the equation from which you expressed “y”:
y=x-2=4-2=2
y=2.
Do a check. To do this, substitute the resulting values into the equations:
2*4+2=10
4-2=2
The unknowns have been found correctly!
A way to add or subtract equations Get rid of any variable right away. In our case, this is easier to do with “y.
Since in the equation “y” has a “+” sign, and in the second one “-”, then you can perform the addition operation, i.e. fold the left side with the left, and the right with the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4Substitute “x” into any equation and find “y”:
2*4+y=10
8+y=10
y=10-8
y=2 Using the 1st method, you can check that the roots are found correctly.
If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have “2x”, and in the second we simply have “x”. To reduce x when adding or subtracting, multiply the second equation by 2:
x-y=2
2x-2y=4Then subtract the second from the first equation:
2x+y-(2x-2y)=10-4 Note that if there is a minus in front of the bracket, then after opening, change the signs to the opposite ones:
2x+y-2x+2y=6
3у=6
find y=2x by expressing from any equation, i.e.
x=4
Video on the topic
When solving differential equations, the argument x (or time t in physical problems) is not always explicitly available. Nevertheless, this is a simplified special case of specifying a differential equation, which often helps to simplify the search for its integral.
Instructions
Consider a physics problem that results in a differential equation in which the argument t is missing. This is a problem about oscillations of a mass m suspended on a thread of length r located in a vertical plane. The equation of motion of the pendulum is required if it was initially motionless and tilted from the equilibrium state by an angle α. The forces should be neglected (see Fig. 1a).
Solution. A mathematical pendulum is a material point suspended on a weightless and inextensible thread at point O. Two forces act on the point: the force of gravity G=mg and the tension force of the thread N. Both of these forces lie in the vertical plane. Therefore, to solve the problem, you can apply the equation of rotational motion of a point around a horizontal axis passing through point O. The equation of rotational motion of a body has the form shown in Fig. 1b. In this case, I is the moment of inertia of the material point; j is the angle of rotation of the thread together with the point, measured from the vertical axis counterclockwise; M is the moment of forces applied to a material point.
Calculate these values. I=mr^2, M=M(G)+M(N). But M(N)=0, since the line of action of the force passes through point O. M(G)=-mgrsinj. The “-” sign means that the moment of force is directed in the direction opposite to the movement. Substitute the moment of inertia and the moment of force into the equation of motion and get the equation shown in Fig. 1s. By reducing the mass, a relationship emerges (see Fig. 1d). There is no t argument here.
§ 1 Integer and fractional rational equations
In this lesson we will look at concepts such as rational equation, rational expression, whole expression, fractional expression. Let's consider solving rational equations.
A rational equation is an equation in which the left and right sides are rational expressions.
Rational expressions are:
Fractional.
An integer expression is made up of numbers, variables, integer powers using the operations of addition, subtraction, multiplication, and division by a number other than zero.
For example:
Fractional expressions involve division by a variable or an expression with a variable. For example:
A fractional expression does not make sense for all values of the variables included in it. For example, the expression
at x = -9 it does not make sense, since at x = -9 the denominator goes to zero.
This means that a rational equation can be integer or fractional.
A whole rational equation is a rational equation in which the left and right sides are whole expressions.
For example:
A fractional rational equation is a rational equation in which either the left or right sides are fractional expressions.
For example:
§ 2 Solution of an entire rational equation
Let's consider the solution of an entire rational equation.
For example:
Let's multiply both sides of the equation by the least common denominator of the denominators of the fractions included in it.
For this:
1. find the common denominator for denominators 2, 3, 6. It is equal to 6;
2. find an additional factor for each fraction. To do this, divide the common denominator 6 by each denominator
additional factor for fraction
additional factor for fraction
3. multiply the numerators of the fractions by their corresponding additional factors. Thus, we obtain the equation
which is equivalent to the given equation
Let's open the brackets on the left, move the right part to the left, changing the sign of the term when transferred to the opposite one.
Let us bring similar terms of the polynomial and get
We see that the equation is linear.
Having solved it, we find that x = 0.5.
§ 3 Solution of a fractional rational equation
Let's consider solving a fractional rational equation.
For example:
1.Multiply both sides of the equation by the least common denominator of the denominators of the rational fractions included in it.
Let's find the common denominator for the denominators x + 7 and x - 1.
It is equal to their product (x + 7)(x - 1).
2. Let's find an additional factor for each rational fraction.
To do this, divide the common denominator (x + 7)(x - 1) by each denominator. Additional factor for fractions
equal to x - 1,
additional factor for fraction
equals x+7.
3.Multiply the numerators of the fractions by their corresponding additional factors.
We obtain the equation (2x - 1)(x - 1) = (3x + 4)(x + 7), which is equivalent to this equation
4.Multiply the binomial by the binomial on the left and right and get the following equation
5. We move the right side to the left, changing the sign of each term when transferring to the opposite:
6. Let us present similar terms of the polynomial:
7. Both sides can be divided by -1. We get a quadratic equation:
8. Having solved it, we will find the roots
Since in Eq.
the left and right sides are fractional expressions, and in fractional expressions, for some values of the variables, the denominator can become zero, then it is necessary to check whether the common denominator does not go to zero when x1 and x2 are found.
At x = -27, the common denominator (x + 7)(x - 1) does not vanish; at x = -1, the common denominator is also not zero.
Therefore, both roots -27 and -1 are roots of the equation.
When solving a fractional rational equation, it is better to immediately indicate the range of acceptable values. Eliminate those values at which the common denominator goes to zero.
Let's consider another example of solving a fractional rational equation.
For example, let's solve the equation
We factor the denominator of the fraction on the right side of the equation
We get the equation
Let's find the common denominator for the denominators (x - 5), x, x(x - 5).
It will be the expression x(x - 5).
Now let's find the range of acceptable values of the equation
To do this, we equate the common denominator to zero x(x - 5) = 0.
We obtain an equation, solving which we find that at x = 0 or at x = 5 the common denominator goes to zero.
This means that x = 0 or x = 5 cannot be the roots of our equation.
Additional multipliers can now be found.
Additional factor for rational fractions
additional factor for the fraction
will be (x - 5),
and the additional factor of the fraction
We multiply the numerators by the corresponding additional factors.
We get the equation x(x - 3) + 1(x - 5) = 1(x + 5).
Let's open the brackets on the left and right, x2 - 3x + x - 5 = x + 5.
Let's move the terms from right to left, changing the sign of the transferred terms:
X2 - 3x + x - 5 - x - 5 = 0
And after bringing similar terms, we obtain a quadratic equation x2 - 3x - 10 = 0. Having solved it, we find the roots x1 = -2; x2 = 5.
But we have already found out that at x = 5 the common denominator x(x - 5) goes to zero. Therefore, the root of our equation
will be x = -2.
§ 4 Brief summary of the lesson
Important to remember:
When solving fractional rational equations, proceed as follows:
1. Find the common denominator of the fractions included in the equation. Moreover, if the denominators of fractions can be factored, then factor them and then find the common denominator.
2.Multiply both sides of the equation by a common denominator: find additional factors, multiply the numerators by additional factors.
3.Solve the resulting whole equation.
4. Eliminate from its roots those that make the common denominator vanish.
List of used literature:
- Makarychev Yu.N., N.G. Mindyuk, Neshkov K.I., Suvorova S.B. / Edited by Telyakovsky S.A. Algebra: textbook. for 8th grade. general education institutions. - M.: Education, 2013.
- Mordkovich A.G. Algebra. 8th grade: In two parts. Part 1: Textbook. for general education institutions. - M.: Mnemosyne.
- Rurukin A.N. Lesson developments in algebra: 8th grade. - M.: VAKO, 2010.
- Algebra 8th grade: lesson plans based on the textbook by Yu.N. Makarycheva, N.G. Mindyuk, K.I. Neshkova, S.B. Suvorova / Auth.-comp. T.L. Afanasyeva, L.A. Tapilina. -Volgograd: Teacher, 2005.
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