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As the ancient philosophers said, "Wisdom is the love of knowledge, and love is the measure of all things." “Measure” in Latin is “modulus”, from which the word “module” comes. And today we will work with equations containing the modulus. I hope that everything will work out for us, and at the end of the lesson we will become wiser.

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Pirogova Tatyana Nikolaevna, Taganrog, secondary school No. 10.

Topic: "Solving Equations with Modulus and Parameter"

Grade 10, lesson of the elective course "Properties of a function".

Lesson plan.

1. Motivation.
2. Knowledge update.
3. Solving a linear equation with a modulus in different ways.
4. Solution of equations containing a module under a module.
5. Researchby determining the dependence of the number of roots of the equation

| | x| - a |= in from the values ​​a and b.

1. Reflection.

During the classes.

Motivation. As the ancient philosophers said, "Wisdom is the love of knowledge, and love is the measure of all things.""Measure" in Latin -"modulus", from which the word came"module". And today we will work with equations containing the modulus. I hope that everything will work out for us, and at the end of the lesson we will become wiser.

Knowledge update.So, let's remember what we already know about the module.

• Module definition.The modulus of a real number is the number itself, if it is non-negative, and its opposite number, if it is negative.

• The geometric meaning of the module.Real Number Modulus A is equal to the distance from the origin to the point with coordinate A on the number line.

– a 0 a

|– a | = | a | | a | x

• The geometric meaning of the magnitude difference modulus.Modulus of magnitude difference| a - in | is the distance between points with coordinates a and c on the number line

Those. segment length [ a in]

1) If a b 2) If a > b

a b b a

S = b - a S = a - b

3) If a \u003d b, then S \u003d a - b \u003d b - a \u003d 0

• Basic properties of the module

1. The modulus of a number is a non-negative number, i.e.| x | ≥ 0 for any x
2. The modules of opposite numbers are equal, i.e.| x | = |– x | for any x
3. The square of the modulus is equal to the square of the submodule expression, i.e.| x | 2 = x 2 for any x

4. The modulus of the product of two numbers is equal to the product of the modulesfactors, i.e. | a b | = | a | · | b |

5. If the denominator of a fraction is non-zero, then the modulus of the fraction is equal to the quotient of dividing the modulus of the numerator by the modulus of the denominator, i.e. for b ≠ 0

6. For equality of any numbers a and b the inequalities:

| | a | – | b | | ≤ | a+b | ≤ | a | + | b |

| | a | – | b | | ≤ | a-b | ≤ | a | + | b |

• Graph of the module y = | x | - a right angle with a vertex at the origin, whose sides are the bisectors of the 1st and 2nd quadrants.
• How to plot function graphs? y = | x –4|, y = | x +3|, y = | x –3|, y = | x | + 1 ,
• y = | x | – 3, y = | x | – 5, y = | x - 3 | + 3, y = | x - 3 | – 2, y = | x + 2 | – 5. y = || x| – a |

Example. solve the equation.

Method 1. The method of opening modules by gaps.

Method 2. Direct module expansion.

If the modulus of a number is 3, then that number is 3 or -3.

Method 3 . Using the geometric meaning of the module.

It is necessary to find on the number axis such x values ​​\u200b\u200bthat are removed from 2 by a distance equal to 3.

Method 4. Squaring both sides of the equation.

This uses the module property

And the fact that both sides of the equation are non-negative.

Method 5. Graphical solution of the equation.

Denote. Let's build graphs of functions And :

The abscissas of the intersection points of the graphs will give the roots

2 -1 0 1 2 3
2 -1 0 1 2 3 4 5
2 -1 0 1 2 3
2 -1 0 1 2 3 4 5

Independent work

solve the equations:

| x – 1| = 3 | x – 5| = 3 | x –3| = 3 | x + 3| = 3 | x + 5| = 3

(-2; 4) (2; 8) (0; 6) (-6; 0) (-8;-2)

Now add another module to the conditions and solve the equations:

| | x| – 1| = 3 | | x| -5| = 3 | | x | – 3| = 3 | | x | + 3| = 3 | | x | + 5| = 3

(no roots)

So, how many roots can an equation of the form | | x | – a |= in? What does it depend on?

Research work on the topic

«Determining the dependence of the number of roots of the equation | | x | – a |= b from a and to »

We will work in groups, using analytical, graphical and geometric methods of solution.

Let us determine under what conditions this equation has 1 root, 2 roots, 3 roots, 4 roots and has no roots.

1 group (by definition)

2 group (using the geometric sense of the module)

3 group (using function graphs)

A > 0
	1 group	2 group	3 group
no roots	V c ≥ 0 c + a	V c ≥ 0 a + b	V c ≥ 0 V A
exactly one root	b > 0 and b + a = 0	b > 0 and b + a = 0	c > 0 and c = - a
exactly two roots	b > 0 and b + a > 0 – in + a	b > 0 and b + a > 0 – in + a	in > 0 and in > | a |
exactly three roots	c > 0 and - c + a = 0	c > 0 and - c + a = 0	b > 0 and b = a
exactly four roots	c > 0 and – c + a > 0	c > 0 and – c + a > 0	in > 0 and in A

Compare the results, draw a general conclusion and draw up a general scheme.

Of course, not necessarily this scheme memorize . The main focus of our study wassee this dependency using different methods, and now it will not be difficult for us to repeat our reasoning when solving such equations.

After all, solving a task with a parameter always implies some research.

Solution of equations with two modules and a parameter.

1. Find values p, x| - R - 3| = 7 has exactly one root.

Solution: | | x| – (p + 3)| = 7

p +3= -7, p = -10. Or geometrically

p + 3 – 7 p + 3 p + 3+7 p + 3+7=0, p = -10

7 7 according to the scheme, an equation of this form has exactly one root, if c \u003d - a, where c \u003d 7, a \u003d p +3

2. Find values R, for each of which the equation | | x| - R - 6| = 11 has exactly two roots.

Solution: | | x| – (p + 6)| = 11 geometrically

P + 6 - 11 p + 6 p + 6 + 11 p + 6-11 R p + 6+11>0, p > -17

11 11

according to the scheme, an equation of this form has exactly two roots, if in + a > 0 and - in + a where a = 11, a = p +6. -17 R 5.

3. Find values R, for each of which the equation | | x| - 4 r | = 5 p -9 has exactly four roots.

Solution: according to the scheme, an equation of this kind has exactly four roots if

0p -9 p, p > and p

those. 1 R 9.

Answer: 1 R 9.

4 . . Find p values, for each of which the equation | | x| – 2 r | = 5 p +2 has no roots. Solution: 5 r +2 p +2 =0 and –2 p >0, or 5 p +2 >0 and 5 p +2 R.

R p = –0.4, or p > –0.4 and p . Answer: p

5. At what values ​​of the parameter p the equation | | x –4 | – 3| + 2 r = 0 has three roots. Find those roots.

Let's transform the equation to the form:

| | x –4 | – 3|= – 2 r.

According to the scheme, an equation of this kind has three roots,

if –2 р =3>0,

Those. p = -1.5.

|| x –4|–3| = 3,

| x –4|=0, x = 4,

|| x –4|=6, x = –2, x =10.

Answer: at r = -1.5 the equation has three roots: x 1 \u003d -2, x 2 \u003d 4, x 3 \u003d 10.

Summing up the lesson. Reflection.

Tell me, what would you highlight the main words of the lesson? (module, parameter)

What did we do today? (Definition of the module, geometric meaning of the module of the number and difference of numbers, properties of the module, different ways of solving equations)

What did we do today?

Homework.

21x 2 + 55x + 42 = 0, D = 552 - 4 21 42 = 3025 - 3582< 0.

Answer: 1; 2.

§6. Solving equations with modules and parameters

Consider several equations in which the variable x is under the modulus sign. Recall that

x , if x ≥ 0,

x = − x if x< 0.

Example 1. Solve the equation:

a) x - 2 = 3; b) x + 1 − 2x − 3 = 1;

x+2

X=1; d) x 2 −

6; e) 6x 2 −

x+1

x − 1

a) If the modulus of a number is 3, then this number is either 3 or (− 3 ) ,

i.e. x − 2 = 3, x = 5 or x − 2 = − 3, x = − 1.

b) It follows from the definition of a module that

x+1

X + 1, for x + 1 ≥ 0,

i.e., for x ≥ − 1 and

x+1

= − x − 1 for x< − 1. Выражение

2x − 3

2x − 3 if x ≥ 3

and equal to − 2 x + 3 if x< 3 .

x< −1

the equation

is tantamount to

equation

- x -1 -

(− 2 x + 3 ) = 1, which implies that

x = 5. But the number 5 is not

satisfies the condition x< − 1, следовательно,

at x< − 1 данное

the equation has no solutions.

−1 ≤ x<

the equation

is tantamount to

equation

x + 1− (2x + 3) = 1, which implies that x = 1;

number 1 satisfies-

no condition − 1 ≤ x<

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations

x ≥

the equation

is tantamount to

equation

x + 1 − (− 2 x − 3 ) = 1, which has a solution x = 3. And since the number 3

satisfies the condition x ≥

then it is a solution to the equation.

x+2

c) If the numerator and denominator of a fraction

have the same

x − 1

signs, then the fraction is positive, and if different, then it is negative, i.e.

x+2

x+2

If x ≤ − 2, if x > 1,

x − 1

x − 1

x+2

If − 2< x < 1.

−1

For x ≤ − 2

ypre x > 1

the original equation is equivalent to the equation

x+2

X=1, x+2

X (x -1) = x -1, x 2 - x +3 =0.

x − 1

The last equation has no solutions.

At − 2< x < 1 данное уравнение равносильно уравнению

x+2

X \u003d 1, - x -2 + x 2 - x \u003d x -1, x 2 -3 x -1 \u003d 0.

x − 1

Let's find the roots of this equation:

x = 3 ± 9 + 4 = 3 ± 13 .

inequalities

− 2 < x < 1 удовлетворяет число 3 − 13

Follow-

Therefore, this number is a solution to the equation.

x ≥ 0 given

the equation

is tantamount to

equation

x2 - x -6 = 0,

whose roots are the numbers 3 and - 2. The number 3

satisfies the condition x > 0,

and the number - 2 does not satisfy this

law, therefore, only the number 3 is a solution to the original

x< 0 удовлетворяет число − 3 и не удовлетворяет число 2.

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations
		x ≥ − 1 given	the equation	is tantamount to				equation
6 x 2 − x − 1 = 0, find its roots: x = 1 ±				25 , x = 1 , x				= −1 .
Both roots satisfy the condition x ≥ − 1,					therefore, they are
are solutions of this equation. At				x< − 1 данное уравнение
is equivalent to the equation 6 x 2 + x + 1 = 0, which has no solutions.
Let the expressions f (x , a ) and g (x , a ) be given,					dependent on change
x	and a.	Then the equation	f (x, a) = g(x, a)		regarding change-

noah x is called equation with parameter a. To solve an equation with a parameter means, for any admissible value of the parameter, to find all solutions of this equation.

Example 2. Solve the equation for all valid values ​​of the parameter a :

a) ax 2 - 3 \u003d 4 a 2 - 2 x 2; b) (a - 3) x 2 = a 2 - 9;

c) (a − 1 ) x2 + 2 (a + 1 ) x + (a − 2 ) = 0.

x 2 =	4a 2 + 3	Expression 4 a 2		3 > 0 for any a ; for a > − 2 we have
	a + 2
we have two solutions: x =			4a 2 + 3	and x = −	4a 2		If	a + 2< 0, то
			a + 2		a + 2

expression 4 a 2 + 3< 0, тогда уравнение не имеет решений. a + 2

	Answer: x = ±	4a 2 + 3	For a > − 2;	for a ≤ − 2 there are no solutions.
		a + 2
				then x 2 = a + 3. If a + 3 = 0,
	b) If a = 3, then x. If a ≠ 3,
	those. if a = − 3,	then the equation has a unique solution x = 0.

whether a< − 3, то уравнение не имеет решений. Если a >− 3 and a ≠ 3, then the equation has two solutions: x 1 = a + 3 and x 2 = − a + 3.

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations

a = 1 this equation takes the form

4x − 1 = 0,

x=1

is his solution. At

a ≠ 1 this equation is

square, its discriminant D 1 is

(a + 1 ) 2 − (a − 1 )(a − 2 ) = 5 a − 1.

If 5 a − 1< 0, т.е. a < 1 ,

then this equation has no solutions.

If a =

then the equation has a unique solution

a+1

x = −

a − 1

−1

If a >

and a ≠ 1,

then this equation has two solutions:

x = − (a + 1 ) ± 5 a − 1 .

a − 1

−(a +1 ) ±

1 at

a = 1; x=3

for a

; x=

5a − 1

a − 1

for a > 1

and a ≠ 1; for a< 1

the equation has no solutions.

§7. Solution of systems of equations. Solving problems that reduce to quadratic equations

In this section, we consider systems that contain equations of the second degree.

Example 1. Solve a system of equations

2x + 3y = 8

xy = 2.

In this system, the equation 2 x + 3 y = 8 is the equation of the first degree, and the equation xy = 2 is the second. We solve this system by the method

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations

substitutions. From the first equation of the system, we express x in terms of y and substitute this expression for x into the second equation of the system:

	8 − 3y	4 −
				y 4			y y = 2.

The last equation reduces to the quadratic equation

8y − 3y 2 = 4, 3y 2 − 8y + 4 = 0.

Finding its roots:

4 ± 4

4 ± 2

Y=2, y

From the condition x = 4 −

we get x = 1, x

Answer: (1;2) and

Example 2. Solve the system of equations:

x 2 + y 2 \u003d 41,

xy = 20.

Multiply both sides of the second equation by 2 and add to the first

system equation:	x 2 + y 2 + 2xy \u003d 41 + 20 2,				(x + y) 2 = 81, whence
it follows that x + y = 9 or x + y = − 9.
If x + y = 9 then	x = 9 − y . Substitute this expression for x in
second equation of the system:
(9 − y ) y = 20, y 2 − 9 y + 20 = 0,
y = 9 ± 81 − 80 = 9 ± 1 , y = 5, y				4, x=4, x=5.
From the condition x + y = − 9 we obtain the solutions (− 4; − 5) and (− 5; − 4 ) .
Answer: (± 4; ± 5) , (± 5; ± 4) .
Example 3. Solve the system of equations:
		y=1
	x -
	x − y

We write the second equation of the system in the form

( x − y )( x + y ) = 5.

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations

Using the equation x − y = 1, we obtain: x + y = 5. Thus, we obtain a system of equations equivalent to the given

	x -	y=1
		y=5.

We add these equations, we get: 2 x \u003d 6,		x=3, x=9.
Substituting the value x = 9 into the first equation			systems, receiving
we have 3 − y = 1, which implies that y = 4.
Answer: (9;4) .	(x + y)(x		Y −4 ) = −4,
Example 4. Solve the system of equations: (x 2 + y 2 ) xy \u003d - 160.
				xy=v;
Let's introduce new variables	x + y = u
x2 + y2 = x2 + y2 + 2 xy − 2 xy = (x + y) 2 − 2 xy = u2 − 2 v,
u (u −4 ) = −4,
the system is reduced to the form (u 2 − 2 v ) v = − 160.
We solve the equation:
u (u − 4) = − 4, u 2 − 4u + 4 = 0, (u − 2) 2 = 0, u = 2.
We substitute this value for u into the equation:
(u 2 - 2v ) v = - 160, (4 - 2v ) v = - 160, 2v 2 - 4v - 160 = 0,
v 2 − 2v − 80 = 0, v = 1± 1 + 80 = 1 ± 9, v= 10, v						= −8.
We solve two systems of equations:
	x + y = 2,
x + y = 2,
	And
xy = 10	xy = − 8.
We solve both systems by the substitution method. For the first system we have:
x= 2 − y, ( 2 − y) y= 10, y2 − 2 y+ 10 = 0.

The resulting quadratic equation has no solutions. For the second system we have: x= 2 − y, (2 − y) y= − 8, y2 − 2 y− 8 = 0.

y= 1 ± 1 + 8 = 1 ± 3, y1 = 4, y2 = − 2. Thenx1 = − 2 Andx2 = 4. Answer: (− 2;4 ) And(4; − 2 ) .

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

multiplied by 3, we get:

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations

Example 5 Solve the system of equations:

x 2 + 4 xy = 3,

y 2 + 3 xy = 2.

From the first equation multiplied by 2, subtract the second equation,

2 x 2 − xy − 3 y 2 = 0.

If y= 0, then and x= 0, but a couple of numbers (0;0 ) is not a solution to the original system. We divide both parts of the equation in the resulting equation

leadership on y2 ,

1 ± 5 , x = 2 y And x = − y .

−3

= 0,

y

Substitute

meaning

x =

3y

first equation

9 y2 + 6 y2 = 3, 11y2 = 4, y=

, x=

, x= −

We substitute the value x= − y into the first equation of the system: y2 − 4 y2 = 3, − 3 y2 = 3.

There are no solutions.

Example 9 Find all parameter values a, for which the system of equations

x 2 + ( y − 2 ) 2 = 1,

y = ax 2 .

has at least one solution.

This system is called a system with a parameter. They can be solved analytically, i.e. using formulas, or you can use the so-called graphical method.

Note that the first equation defines a circle centered at the point (0;2 ) with radius 1. The second equation for a≠ 0 defines a parabola with vertex at the origin.

If a 2

In case a), the parabola touches the circle. From the second equation of the system,

em what x2 = y/ a,

substitute these values ​​for

x 2

into the first equation:

1

+(y−2 )

= 1,

+ y

− 4 y+ 4 = 1, y

4 − ay+ 3

= 0.

In the case of tangency, due to symmetry, there is a unique value y, so the discriminant of the resulting equation should be

is 0. Since the ordinate y the touch point is positive, and because

y = 2

− a

we get

> 0; D

1 2

4 − a

4 − a

− 12 = 0,

4 − a

> 0

we get: 4

= 2

= 4 −2

a =

4 + 2 3

4 + 2 3

2 +

( 4 − 2 3)( 4 + 2 3) =

16 − 12 =

4 − 2 3

If a> 2 + 2 3 , then the parabola will intersect the circle at 4 points

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 5, 8 cells. Mathematics. Quadratic equations

Therefore, the system has at least one solution if

a≥ 2 + 2 3 .

Example 10 The sum of the squares of the digits of some natural two-digit number is 9 more than twice the product of these digits. After dividing this two-digit number by the sum of its digits, the quotient is 4 and the remainder is 3. Find this two-digit number.

Let the two digit number be 10 a+ b, Where a And b are the digits of this number. Then from the first condition of the problem we get: a2 + b2 = 9 + 2 ab, and from the second condition we get: 10 a+ b= 4 (a+ b) + 3.

a 2 + b 2 = 9 + 2 ab ,

We solve the system of equations: 6 a− 3 b= 3.

From the second equation of the system we obtain

6a− 3b= 3, 2a− b= 1, b= 2a− 1.

We substitute this value for b into the first equation of the system:

a2 + ( 2a− 1) 2 = 9 + 2a( 2a− 1) , 5a2 − 4a+ 1 = 9 + 4a2 − 2a,

a2 − 2a− 8 = 0, D1 = 1 + 8 = 9, a= 1 ± 3, a1 = 4, a2 = − 2 < 0, b1 = 7.

Answer: 47.

Example 11. After mixing two solutions, one of which contained 48 g and the other 20 g, of anhydrous potassium iodide, 200 g of a new solution were obtained. Find the concentration of each of the initial solutions if the concentration of the first solution was 15% greater than the concentration of the second.

Denote by x% is the concentration of the second solution, and through (x+ 15 ) % is the concentration of the first solution.

(x+ 15 )%			x %
I solution			II solution

In the first solution, 48 g is (x+ 15 ) % by weight of the whole solution,

so the weight of the solution is x48 + 15 100. In the second solution, 20 g of co-

© 2011, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

10x - 5y - 3z = - 9,

6 x + 4 y - 5 z = - 1.3 x - 4 y - 6 z = - 23.

We equalize the coefficients at x in the first and second equations, for this we multiply both parts of the first equation by 6, and the second equation by 10, we get:

60x - 30 y - 18z = - 54.60x + 40 y - 50z = - 10.

We subtract from the second equation of the resulting system the first equation

we get: 70 y - 32 z = 44, 35 y - 16 z = 22.

Subtract the third equation multiplied by 2 from the second equation of the original system, we get: 4 y + 8 y − 5 z + 12 z = − 1 + 46,

12y + 7z = 45.

Now we solve a new system of equations:

35y − 16z = 22.12y + 7z = 45.

To the first equation of the new system, multiplied by 7, we add the second equation, multiplied by 16, we get:

35 7y + 12 16y = 22 7 + 45 16,

Now we substitute y = 2, z = 3 into the first equation of the original system

topics, we get: 10x - 5 2 - 3 3 = - 9, 10x - 10 - 9 = - 9, 10x = 10, x = 1.

Answer: (1; 2; 3) . ▲

§ 3. Solution of systems with a parameter and with modules

ax + 4y = 2a,

Consider the system of equations

x + ay = a.

2010-2011 academic year year., No. 3, 8 cells. Mathematics. Systems of equations.

In this system, in fact, there are three variables, namely: a , x , y . The unknowns are x and y , and a is called a parameter. It is required to find solutions (x , y ) of this system for each value of the parameter a .

Let us show how such systems are solved. Let's express the variable x from the second equation of the system: x = a − ay . We substitute this value for x into the first equation of the system, we get:

a (a − ay) + 4 y = 2 a,

(2 − a )(2 + a ) y = a (2 − a ) .

If a = 2, then we get the equation 0 y = 0. Any number y satisfies this equation, and then x = 2 − 2 y , i.e., for a = 2, the pair of numbers (2 − 2 y ; y ) is a solution to the system . Since y can be

any number, then the system for a = 2 has infinitely many solutions.

If a = − 2, then we get the equation 0 y = 8. This equation has no solution.

If now a ≠ ± 2,				then y =		a (2 − a)
						(2 − a )(2 + a )		2 + a
x = a − ay = a −
				2 + a

Answer: For a = 2, the system has infinitely many solutions of the form (2 − 2 y ; y ) , where y is any number;

for a = − 2 the system has no solutions;
for a ≠ ± 2, the system has a unique solution						. ▲
		2 + a		2 + a

We have solved this system and established for what values ​​of the parameter a the system has one solution, when it has infinitely many solutions, and for what values ​​of the parameter a it has no solutions.

Example 1. Solve the system of equations

© 2010, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 3, 8 cells. Mathematics. Systems of equations.

−3

y − 1

3x − 2y = 5.

From the second equation of the system, we express x in terms of y, we get

2y + 5

we substitute this value for x into the first equation of the sys-

topics, we get:

2y+5

−3

y − 1

−3

−1

5 = 0

Expression

y = −

y > −

; If

−5

= −y

The expression y − 1 = 0,

if y = 1. If

y > 1, then

y − 1

Y − 1, and

whether y< 1, то

y − 1

1 − y .

If y ≥ 1 then

y − 1

Y −1 and

we get the equation:

−3 (y

− 1) = 3,

−3 y

3, −

(2 2 +

5 ) = 3. The number 2 > 1, so the pair (3;2) is re-

system.

Let now

5 ≤ y<1,

y − 1

− y;

finding

we get

the equation

3y−3

4y + 10

3y=6

13y=8

© 2010, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 3, 8 cells. Mathematics. Systems of equations.

(2y + 5) =

But less than

so a couple of numbers

is the solution to the system.

y< −

then we get the equation:

3y−3

4y-

3y=6

5y=

28 , y = 28 .

meaning

so there are no solutions.

Thus, the system has two solutions (3;2) and 13 27 ; 13 8 . ▲

§ 4. Solution of problems with the help of systems of equations

Example 1. A car travels from a city to a village in 2.5 hours. If he increases his speed by 20 km/h, then in 2 hours he will cover a distance of 15 km more than the distance from the city to the village. Find this distance.

Denote by S the distance between the city and the village and by V the speed of the car. Then, to find S, we obtain a system of two equations

2.5V=S

(V + 20) 2 = S + 15.

© 2010, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

2010-2011 academic year year., No. 3, 8 cells. Mathematics. Systems of equations.

into the second equation:		S+202	S+15,		S=25	S = 125.

Answer: 125 km. ▲

Example 2. The sum of the digits of a two-digit number is 15. If these digits are interchanged, you get a number that is 27 more than the original. Find these numbers.

Let the given number ab , i.e. the number of tens is a , and the number of units is b . From the first condition of the problem we have: a + b = 15. If we subtract the number ab from the number ba, then we get 27, from here we get the second equation: 10 b + a − (10 a + b) = 27. x

2010-2011 academic year year., No. 3, 8 cells. Mathematics. Systems of equations.

Multiply both sides of the equation by 20, we get: x + 8 y = 840. To find x and y, we got a system of equations

Answer: 40 tons, 100 tons ▲

Example 4. A computer operator, working with a student, processes a task in 2 hours and 24 minutes. If the operator will work for 2 hours, and the student for 1 hour, then

children completed 2 3 of all work. How long will it take for an operator

ru and the student separately to process the task?

Let's denote all work as 1, operator performance as x, and student performance as y . We take into account that

2 hours 24 minutes = 2 5 2 hours = 12 5 hours.

It follows from the first condition of the problem that (x+y ) 12 5 = 1. From the second condition of the problem it follows that 2 x + y = 2 3 . Got a system of equations

(x+y)

2 x + y =

We solve this system using the substitution method:

− 2 x ;

−2 x

−x

− 1;

; x=

; y=

© 2010, FZFTSH at MIPT. Compiled by: Yakovleva Tamara Kharitonovna

slide 2

.

Solving equations with parameters and modules, applying the properties of functions in unexpected situations and mastering geometric techniques for solving problems. Non-standard equations The purpose of the lesson.

slide 3

The absolute value or modulus of the number a is the number a if a>0, the number -a if a 0 ׀ a ׀=( 0 if a=0 -a if a 0) is equivalent to the double inequality -a 0. Inequality ׀ x ׀>a, (if a>0) is equivalent to two inequalities - Inequality׀ x׀>a, (if a

slide 4

To solve an equation with parameters means to indicate at what values ​​of the parameters solutions exist and what they are. a) determine the set of admissible values ​​of the unknown and parameters; b) for each admissible system of parameter values, find the corresponding sets of solutions to the equation. Repetition of the most important theoretical material on the topics "Solution of equations with parameters"

slide 5

1. Solve the equation ׀ x-2 ׀ =5; Answer 7;-3 ׀ x-2 ׀ =-5; The answer of the decision is no ׀ x-2 ׀ =x+5; ; The answer is no; 1.5 ׀ x-2 ׀ \u003d ׀ x + 5 ׀; The answer is no; -1.5; there is no solution; -1.5; oral exercises.

slide 6

2. Solve equations=1; Answer. If a=0, then there is no solution; if a=0, then x=1/ a 1.3. Solve the equation (a²-1) x \u003d a + 1. 1) a \u003d 1; then the equation takes the form Ox = 2 and has no solution 2) a = 1; we get Ox = O, and obviously x is any. 1 3) if a \u003d ± 1, then x \u003d - a-1 Answer. If a \u003d -1, then x is any; if a \u003d 1, then there is no solution 1 if a \u003d ± 1, then x \u003d - a-1

Slide 7

2. Solve the equation ׀ x + 3 ׀ + ׀ y -2 ׀ = 4; . 2 3. 4. 1

Slide 8

3 3 2 x y 0 1 Answer: (-3; 2).

Slide 9

2. Solve the equations ax=1;

Answer. If a=0, then there is no solution; if a=0, then x=1/ a 1.3. Solve the equation (a²-1) x \u003d a + 1. 1) a \u003d 1; then the equation takes the form Ox = 2 and has no solution 2) a = 1; we get Ox = O, and obviously x is any. 1 3) if a \u003d ± 1, then x \u003d - a-1 Answer. If a \u003d -1, then x is any; if a \u003d 1, then there is no solution 1 if a \u003d ± 1, then x \u003d - a-1

Slide 10

3 Construct a graph of the function

y x Y=IxI 1 2 -3 -4 -1 1 -2 2 3 0 -5 4 5 6 -1 -2 Y=Ix+3I-2 Y=Ix-2I Y=Ix+5I Y=Ix-2I + 3