An equation with one unknown, which, after opening the brackets and bringing similar terms, takes the form
ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we’ll figure out how to solve these linear equations.
For example, all equations:
2x + 3= 7 – 0.5x; 0.3x = 0; x/2 + 3 = 1/2 (x – 2) - linear.
The value of the unknown that turns the equation into a true equality is called decision or root of the equation .
For example, if in the equation 3x + 7 = 13 instead of the unknown x we substitute the number 2, we obtain the correct equality 3 2 +7 = 13. This means that the value x = 2 is the solution or root of the equation.
And the value x = 3 does not turn the equation 3x + 7 = 13 into a true equality, since 3 2 +7 ≠ 13. This means that the value x = 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.
Let's move the free term from the left side of the equation to the right, 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.
Let's move 2 from the left side of the equation to the right, changing the sign in front of 2 to the opposite, we get
3x = 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.
This means that the value x = 3 is the solution or root of the equation.
Answer: x = 3.
If a = 0 and b = 0, then we get the equation 0x = 0. This equation has infinitely many solutions, since when we multiply any number by 0 we get 0, but b is also equal to 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 = 3x – 12 + 2x ‒ 1.
5x – 3x ‒ 2x = – 12 ‒ 1 + 15 ‒ 2.
Here are some similar terms:
0x = 0.
Answer: x - any number.
If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when we multiply any number by 0 we get 0, but b ≠ 0.
Example 3. Solve the equation x + 8 = x + 5.
Let’s group terms containing unknowns on the left side, and free terms on the right side:
x – x = 5 – 8.
Here are some similar terms:
0х = ‒ 3.
Answer: no solutions.
On Figure 1 shows a diagram for solving a linear equation
Let's compose general scheme solving equations with one variable. Let's consider the solution to Example 4.
Example 4. Suppose we need to 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 terms containing unknown and free terms, open the brackets:
4x – 16 + 6x + 6 – 12 = 30x – 90 + 24x – 22x – 86.
4) Let us group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x – 30x – 24x + 22x = ‒ 90 – 86 + 16 – 6 + 12.
5) Let us present similar terms:
- 22х = - 154.
6) Divide by – 22, We get
x = 7.
As you can see, the root of the equation is seven.
Generally such equations can be solved using the following scheme:
a) bring the equation to its integer form;
b) open the 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 similar terms.
However, this scheme is not necessary for every equation. When solving many simpler equations, you have to start not from the first, but from the second ( Example. 2), third ( Example. 1, 3) and even from the fifth stage, as in example 5.
Example 5. Solve the equation 2x = 1/4.
Find the unknown x = 1/4: 2,
x = 1/8 .
Let's look at solving some linear equations found in the main state exam.
Example 6. Solve the 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) = 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 = 6 – 2, x = 4.
If x = 4 then
f(6) = 3 7-4 = 3 3 = 27
Answer: 27.
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Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.
Before studying specific solution methods, note that all quadratic equations can be divided into three classes:
- Have no roots;
- Have exactly one root;
- They have two different roots.
This is an important difference between quadratic equations and linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.
Discriminant
Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.
You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:
- If D< 0, корней нет;
- If D = 0, there is exactly one root;
- If D > 0, there will be two roots.
Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:
Task. How many roots do quadratic equations have:
- x 2 − 8x + 12 = 0;
- 5x 2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16
So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.
The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.
Discriminant equal to zero- there will be one root.
Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.
By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.
Roots of a quadratic equation
Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:
Basic formula for the roots of a quadratic equation
When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.
- x 2 − 2x − 3 = 0;
- 15 − 2x − x 2 = 0;
- x 2 + 12x + 36 = 0.
First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.
D > 0 ⇒ the equation has two roots. Let's find them:
Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.
D > 0 ⇒ the equation again has two roots. Let's find them
\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]
Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.
D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:
As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of errors.
Incomplete quadratic equations
It happens that a quadratic equation is slightly different from what is given in the definition. For example:
- x 2 + 9x = 0;
- x 2 − 16 = 0.
It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:
The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.
Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.
Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:
Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:
- If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
- If (−c /a)< 0, корней нет.
As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.
Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:
Taking the common factor out of bracketsThe product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:
Task. Solve quadratic equations:
- x 2 − 7x = 0;
- 5x 2 + 30 = 0;
- 4x 2 − 9 = 0.
x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.
5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.
4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.
What are irrational equations and how to solve them
Equations in which the variable is contained under the radical sign or under the sign of raising to a fractional power are called irrational. When we deal with fractional powers, we deprive ourselves of many mathematical operations to solve 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. In this case, raising both sides of the equation to the same odd power is an equivalent transformation of the equation, and raising it to an even power is a unequal transformation. This difference is obtained due to such features of raising to a power, such as if raised to an even power, then negative values are “lost”.
The point of raising both sides of an irrational equation to a power is the desire to get rid of “irrationality.” Thus, we need to raise both sides of the irrational equation to such a degree that all fractional powers of both sides of the equation turn 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 be suitable.
Thus, the main difficulty is associated with raising both sides of the equation to the same even power - due to the inequality of the transformation, extraneous roots may appear. Therefore, it is necessary to check all found roots. Those who solve an irrational equation most often forget to check the found roots. It is also not always clear to what power an irrational equation must be raised in order to get rid of irrationality and solve it. Our smart calculator was created specifically to solve irrational equations and automatically check all the roots, which will eliminate forgetfulness.
Free online irrational equations calculator
Our free solver will solve an irrational equation online any complexity in seconds. All you need to do is simply enter your data into the calculator. You can also find out how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group.
Let us analyze two types of solutions to systems of equations:
1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.
In order to solve the system of equations by substitution method you need to follow a simple algorithm:
1. Express. From any equation we express one variable.
2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
3. Solve the resulting equation with one variable. We find a solution to the system.
To decide system by term-by-term addition (subtraction) method need to:
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.
The solution to the system is the intersection points of the function graphs.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by substitution method
Solving a system of equations using 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, which means that it is easiest to express the variable x from the second equation.
x=3+10y
2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1
3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution to the equation system 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 point where we expressed it, 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 using the term-by-term addition (subtraction) method.
Solving a system of equations using the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. We choose a variable, let’s say we choose 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. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y=2
5y=32 | :5
y=6.4
3. Find x. We substitute the found y into any of the equations, let’s say into 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 intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)
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Purpose of the service. The matrix calculator is designed to solve systems of linear equations using a matrix method (see example of solving similar problems).Instructions. To solve online, you need to select the type of equation and set the dimension of the corresponding matrices. where A, B, C are the specified matrices, X is the desired matrix. Matrix equations of the form (1), (2) and (3) are solved through the inverse matrix A -1. If the expression A·X - B = C is given, then it is necessary to first add the matrices C + B and find a solution for the expression A·X = D, where D = C + B. If the expression A*X = B 2 is given, then the matrix B must first be squared.
It is also recommended to familiarize yourself with the basic operations on matrices.Example No. 1. Exercise. Find the solution to the matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: A·X·B = C.
The determinant of matrix A is equal to detA=-1
Since A is a non-singular matrix, there is an inverse matrix A -1 . Multiply both sides of the equation on the left by A -1: Multiply both sides of this equation on the left by A -1 and on the right by B -1: A -1 ·A·X·B·B -1 = A -1 ·C·B -1 . Since A A -1 = B B -1 = E and E X = X E = X, then X = A -1 C B -1
Inverse matrix A -1: 
Let's find the inverse matrix B -1.
Transposed matrix B T:
Inverse matrix B -1: 
We look for matrix X using the formula: X = A -1 ·C·B -1 
Answer:
Example No. 2. Exercise. Solve matrix equation 
Solution. Let's denote:
Then the matrix equation will be written in the form: A·X = B.
The determinant of matrix A is detA=0
Since A is a singular matrix (the determinant is 0), therefore the equation has no solution.
Example No. 3. Exercise. Find the solution to the matrix equation
Solution. Let's denote:
Then the matrix equation will be written in the form: X A = B.
The determinant of matrix A is detA=-60
Since A is a non-singular matrix, there is an inverse matrix A -1 . Let's multiply both sides of the equation on the right by A -1: X A A -1 = B A -1, from where we find that X = B A -1
Let's find the inverse matrix A -1 .
Transposed matrix A T: 
Inverse matrix A -1: 
We look for matrix X using the formula: X = B A -1 
Answer: > 
