An equation with one unknown, which, after opening the brackets and reducing like terms, takes the form
ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.
For example, all equations:
2x + 3 \u003d 7 - 0.5x; 0.3x = 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.
The value of the unknown that turns the equation into a true equality is called decision or the root of the equation .
For example, if in the equation 3x + 7 \u003d 13 we substitute the number 2 instead of the unknown x, then we get the correct equality 3 2 + 7 \u003d 13. This means that the value x \u003d 2 is the solution or the root of the equation.
And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 2 + 7 ≠ 13. Therefore, the value x \u003d 3 is not a solution or a root of the equation.
Solution of any linear equations reduces to solving equations of the form
ax + b = 0.
We transfer the free term from the left side of the equation to the right side, while changing the sign in front of b to the opposite, we get
If a ≠ 0, then x = – b/a .
Example 1 Solve the equation 3x + 2 =11.
We transfer 2 from the left side of the equation to the right side, while changing the sign in front of 2 to the opposite, we get
3x \u003d 11 - 2.
Let's do the subtraction, then
3x = 9.
To find x, you need to divide the product by a known factor, that is,
x = 9:3.
So the value x = 3 is the solution or the root of the equation.
Answer: x = 3.
If a = 0 and b = 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0, we get 0, but b is also 0. The solution to this equation is any number.
Example 2 Solve the equation 5(x - 3) + 2 = 3 (x - 4) + 2x - 1.
Let's expand the brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.
5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.
Here are similar members:
0x = 0.
Answer: x is any number.
If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when multiplying any number by 0, we get 0, but b ≠ 0.
Example 3 Solve the equation x + 8 = x + 5.
Let us group the terms containing unknowns on the left side, and the free terms on the right side:
x - x \u003d 5 - 8.
Here are similar members:
0x = - 3.
Answer: no solutions.
On the figure 1 the scheme for solving the linear equation is shown
Let us compose a general scheme for solving equations with one variable. Consider the solution of example 4.
Example 4 Let's solve the equation
1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.
2) After reduction we get
4 (x - 4) + 3 2 (x + 1) - 12 = 6 5 (x - 3) + 24x - 2 (11x + 43)
3) To separate members containing unknown and free members, open the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) We group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) Here are similar members:
- 22x = - 154.
6) Divide by - 22 , We get
x = 7.
As you can see, the root of the equation is seven.
In general, such equations can be solved as follows:
a) bring the equation to an integer form;
b) open brackets;
c) group the terms containing the unknown in one part of the equation, and the free terms in the other;
d) bring similar members;
e) solve an equation of the form aх = b, which was obtained after bringing like terms.
However, this scheme is not required for every equation. When solving many more simple equations you have to start not from the first, but from the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.
Example 5 Solve the equation 2x = 1/4.
We find the unknown x \u003d 1/4: 2,
x = 1/8 .
Consider the solution of some linear equations encountered in the main state exam.
Example 6 Solve equation 2 (x + 3) = 5 - 6x.
2x + 6 = 5 - 6x
2x + 6x = 5 - 6
Answer: - 0.125
Example 7 Solve the equation - 6 (5 - 3x) \u003d 8x - 7.
– 30 + 18x = 8x – 7
18x - 8x = - 7 +30
Answer: 2.3
Example 8 Solve the Equation
3(3x - 4) = 4 7x + 24
9x - 12 = 28x + 24
9x - 28x = 24 + 12
Example 9 Find f(6) if f (x + 2) = 3 7's
Solution
Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.
We solve the linear equation x + 2 = 6,
we get x \u003d 6 - 2, x \u003d 4.
If x = 4 then
f(6) = 3 7-4 = 3 3 = 27
Answer: 27.
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We will analyze two types of solving systems of equations:
1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.
In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. From any equation, we express one variable.
2. Substitute. We substitute in another equation instead of the expressed variable, the resulting value.
3. We solve the resulting equation with one variable. We find a solution to the system.
To solve system by term-by-term addition (subtraction) necessary:
1. Select a variable for which we will make the same coefficients.
2. We add or subtract the equations, as a result we get an equation with one variable.
3. We solve the resulting linear equation. We find a solution to the system.
The solution of the system is the intersection points of the graphs of the function.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by the substitution method
Solving the system of equations by the substitution method2x+5y=1 (1 equation)
x-10y=3 (2nd equation)
1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y
2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1
3. We solve the resulting equation with one variable.
2(3+10y)+5y=1 (open brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution of the system of equations is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first paragraph where we expressed we substitute y there.
x=3+10y
x=3+10*(-0.2)=1
It is customary to write points in the first place, we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)
Example #2:
Let's solve by term-by-term addition (subtraction).
Solving a system of equations by the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. Select a variable, let's say we select x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.
3x-2y=1 |*2
6x-4y=2
2x-3y=-10 |*3
6x-9y=-30
2. From the first equation, subtract the second to get rid of the variable x. Solve the linear equation.
__6x-4y=2
5y=32 | :five
y=6.4
3. Find x. We substitute the found y in any of the equations, let's say in the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6
The point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)
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What are irrational equations and how to solve them
Equations in which the variable is contained under the sign of the radical or under the sign of raising to a fractional power are called irrational. When we deal with a fractional power, we deprive ourselves of many mathematical operations for solving the equation, so irrational equations are solved in a special way.
Irrational equations are usually solved by raising both sides of the equation to the same power. At the same time, raising both sides of the equation to the same even degree is an equivalent transformation of the equation, and to an even one is a non-equivalent one. Such a difference is obtained due to such features of exponentiation, such as if raised to an even power, then negative values"lost".
The point of raising both sides of an irrational equation to a power is to get rid of "irrationality". Thus, we need to raise both sides of the irrational equation to such a degree that all fractional powers both parts of the equation turned into integers. Then you can look for a solution given equation, which will coincide with the solutions of the irrational equation, with the difference that in the case of raising to an even power, the sign is lost and the final solutions will require verification and not all will work.
Thus, the main difficulty is associated with raising both parts of the equation to the same even power - due to the non-equivalence of the transformation, extraneous roots may appear. Therefore, it is obligatory to check all found roots. Checking the found roots is most often forgotten by those who solve an irrational equation. It is also not always clear to what degree it is necessary to raise an irrational equation in order to get rid of irrationality and solve it. Our smart calculator is just designed to solve an irrational equation and automatically check all the roots, which will eliminate forgetfulness.
Free online calculator of irrational equations
Our free solver will allow you to solve an irrational equation online of any complexity in a matter of seconds. All you need to do is just enter your data into the calculator. You can also learn how to solve the equation on our website. And if you have any questions, you can ask them in our VKontakte group.
Instruction
Note:π is written as pi; square root as sqrt().
Step 1. Enter the given example consisting of fractions.
Step 2 Click the "Solve" button.
Step 3 Get detailed results.
In order for the calculator to calculate fractions correctly, enter the fraction through the sign: “/”. For example: . The calculator will calculate the equation and even show on the graph why such a result was obtained.
What is a fractional equation
A fractional equation is an equation in which the coefficients are fractional numbers. Linear equations with fractions are solved according to the standard scheme: the unknowns are transferred in one direction, and the known ones in the other.
Let's look at an example:
Fractions with unknowns are moved to the left, and the rest of the fractions are moved to the right. When numbers are transferred beyond the equal sign, then the sign of the numbers changes to the opposite:
Now you need to perform only the actions of both parts of the equality:
The result is an ordinary linear equation. Now you need to divide the left and right parts by the coefficient of the variable.
Solve the equation with fractions online updated: October 7, 2018 by: Scientific Articles.Ru
At the stage of preparation for the final testing, high school students need to improve their knowledge on the topic "Exponential Equations". The experience of past years indicates that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of preparation, need to carefully master the theory, memorize the formulas and understand the principle of solving such equations. Having learned to cope with this type of tasks, graduates will be able to count on high scores when passing the exam in mathematics.
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The main definitions and formulas are presented in the "Theoretical Reference" section.
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