Solve the 3x equation. Solving equations in two variables
In the 7th grade mathematics course, they first meet with equations in two variables, but they are studied only in the context of systems of equations with two unknowns. That is why it falls out of sight whole line problems in which some conditions are introduced on the coefficients of the equation, limiting them. In addition, methods for solving problems such as "Solve an equation in natural or whole numbers" are also left without attention, although problems of this kind are found more and more often in the USE materials and on entrance exams.
Which equation will be called a two-variable equation?
So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are equations in two variables.
Consider the equation 2x - y = 1. It turns into true equality for x = 2 and y = 3, so this pair of values of the variables is a solution to the equation under consideration.
Thus, the solution to any equation with two variables is the set of ordered pairs (x; y), the values of the variables that this equation turns into a true numerical equality.
An equation with two unknowns can:
a) have one solution. For example, the equation x 2 + 5y 2 = 0 has only decision (0; 0);
b) have multiple solutions. For example, (5 - | x |) 2 + (| y | - 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);
v) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;
G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers, the sum of which is 3. The set of solutions to this equation can be written in the form (k; 3 - k), where k is any real number.
The main methods for solving equations with two variables are methods based on factoring expressions into factors, allocating a complete square, using the properties of a quadratic equation, limited expressions, and evaluative methods. The equation, as a rule, is transformed into a form from which a system for finding unknowns can be obtained.
Factorization
Example 1.
Solve the equation: xy - 2 = 2x - y.
Solution.
We group the terms for the purpose of factoring:
(xy + y) - (2x + 2) = 0. Move out the common factor from each parenthesis:
y (x + 1) - 2 (x + 1) = 0;
(x + 1) (y - 2) = 0. We have:
y = 2, x is any real number or x = -1, y is any real number.
Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.
Equality to zero is not negative numbers
Example 2.
Solve the equation: 9x 2 + 4y 2 + 13 = 12 (x + y).
Solution.
We group:
(9x 2 - 12x + 4) + (4y 2 - 12y + 9) = 0. Now each parenthesis can be folded using the squared difference formula.
(3x - 2) 2 + (2y - 3) 2 = 0.
The sum of two non-negative expressions is zero only if 3x - 2 = 0 and 2y - 3 = 0.
This means that x = 2/3 and y = 3/2.
Answer: (2/3; 3/2).
Evaluation method
Example 3.
Solve the equation: (x 2 + 2x + 2) (y 2 - 4y + 6) = 2.
Solution.
In each bracket, select a full square:
((x + 1) 2 + 1) ((y - 2) 2 + 2) = 2. Estimate 
the meaning of the expressions in parentheses.
(x + 1) 2 + 1 ≥ 1 and (y - 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:
(x + 1) 2 + 1 = 1 and (y - 2) 2 + 2 = 2, which means x = -1, y = 2.
Answer: (-1; 2).
Let's get acquainted with another method for solving equations with two variables of the second degree. This method is that the equation is considered as square with respect to any variable.
Example 4.
Solve the equation: x 2 - 6x + y - 4√y + 13 = 0.
Solution.
Solve the equation as square with respect to x. Let's find the discriminant:
D = 36 - 4 (y - 4√y + 13) = -4y + 16√y - 16 = -4 (√y - 2) 2. The equation will have a solution only for D = 0, that is, if y = 4. Substitute the value of y into the original equation and find that x = 3.
Answer: (3; 4).
Often in equations with two unknowns they indicate constraints on variables.
Example 5.
Solve the whole equation: x 2 + 5y 2 = 20x + 2.
Solution.
Rewrite the equation as x 2 = -5y 2 + 20x + 2. Right part the resulting equation when divided by 5 gives the remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number that is not divisible by 5 gives the remainder 1 or 4. Thus, equality is impossible and there are no solutions.
Answer: no roots.
Example 6.
Solve the equation: (x 2 - 4 | x | + 5) (y 2 + 6y + 12) = 3.
Solution.
Select the complete squares in each bracket:
((| x | - 2) 2 + 1) ((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided | x | - 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.
Answer: (2; -3) and (-2; -3).
Example 7.
For each pair of negative integers (x; y) satisfying the equation
x 2 - 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). In the answer, indicate the smallest of the amounts.
Solution.
Let's select complete squares:
(x 2 - 2xy + y 2) + (y 2 + 4y + 4) = 37;
(x - y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. The sum of the squares of two integers, equal to 37, is obtained if we add 1 + 36. Therefore:
(x - y) 2 = 36 and (y + 2) 2 = 1 
(x - y) 2 = 1 and (y + 2) 2 = 36.
Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).
Answer: -17.
Do not be discouraged if you have difficulty solving equations with two unknowns. With a little practice, you can tackle any equation.
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Service purpose. Matrix calculator designed to solve systems linear equations matrix method (see an example of solving similar problems).Instruction. For online solutions it is necessary 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 required 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 you must 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 that you familiarize yourself with the basic operations on matrices.Example # 1. Exercise... Find the solution to the matrix equation
Solution... Let's denote:
Then matrix equation will be written as: A X B = C.
The determinant of the matrix A is equal to detA = -1
Since A is a non-degenerate 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: 
Find the inverse matrix B -1.
Transpose matrix B T:
Inverse matrix B -1: 
We look for the matrix X by 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 as: A X = B.
The determinant of the matrix A is equal to detA = 0
Since A is a degenerate 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 as: X A = B.
The determinant of the matrix A is equal to detA = -60
Since A is a non-degenerate matrix, there is an inverse matrix A -1. We multiply both sides of the equation on the right by A -1: X A A -1 = B A -1, whence we find that X = B A -1
Find the inverse matrix A -1.
Transpose matrix A T: 
Inverse matrix A -1: 
We look for the matrix X by the formula: X = B A -1 
Answer:> 
The use of equations is widespread in our life. They are used in many calculations, building construction, and even sports. Man used equations in ancient times and since then their application has only increased. Power or exponential equations are equations in which the variables are in powers and the base is a number. For example:
Solving the exponential equation comes down to 2 fairly simple steps:
1. It is necessary to check whether the bases of the equation on the right and on the left are the same. If the grounds are not the same, we are looking for options to solve this example.
2. After the bases become the same, we equate the degrees and solve the resulting new equation.
Let's say an exponential equation of the following form is given:
It is worth starting the solution of this equation with the analysis of the base. The bases are different - 2 and 4, and for the solution we need to be the same, so we transform 4 according to the following formula - \ [(a ^ n) ^ m = a ^ (nm): \]
Add to the original equation:
Take out the brackets \
We express \
Since the degrees are the same, we discard them:
Answer: \
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I. ax 2 = 0 – incomplete quadratic equation (b = 0, c = 0 ). Solution: x = 0. Answer: 0.
Solve equations.
2x (x + 3) = 6x-x 2.
Solution. Let's expand the brackets by multiplying 2x for each term in brackets:
2x 2 + 6x = 6x-x 2; we transfer the terms from the right to the left:
2x 2 + 6x-6x + x 2 = 0; we give similar terms:
3x 2 = 0, hence x = 0.
Answer: 0.
II. ax 2 + bx = 0 –incomplete quadratic equation (c = 0 ). Solution: x (ax + b) = 0 → x 1 = 0 or ax + b = 0 → x 2 = -b / a. Answer: 0; -b / a.
5x 2 -26x = 0.
Solution. Take out the common factor NS outside the brackets:
x (5x-26) = 0; each factor can be zero:
x = 0 or 5x-26 = 0→ 5x = 26, we divide both sides of the equality by 5 and we get: x = 5.2.
Answer: 0; 5,2.
Example 3. 64x + 4x 2 = 0.
Solution. Take out the common factor 4x outside the brackets:
4x (16 + x) = 0. We have three factors, 4 ≠ 0, therefore, or x = 0 or 16 + x= 0. From the last equality we get x = -16.
Answer: -16; 0.
Example 4.(x-3) 2 + 5x = 9.
Solution. Using the formula for the square of the difference of two expressions, we open the brackets:
x 2 -6x + 9 + 5x = 9; transform to the form: x 2 -6x + 9 + 5x-9 = 0; we give similar terms:
x 2 -x = 0; take out NS brackets, we get: x (x-1) = 0. Hence or x = 0 or x-1 = 0→ x = 1.
Answer: 0; 1.
III. ax 2 + c = 0 –incomplete quadratic equation (b = 0 ); Solution: ax 2 = -c → x 2 = -c / a.
If (-c / a)<0 , then there are no real roots. If (-s / a)> 0
Example 5. x 2 -49 = 0.
Solution.
x 2 = 49, hence x = ± 7. Answer:-7; 7.
Example 6. 9x 2 -4 = 0.
Solution.
It is often required to find the sum of squares (x 1 2 + x 2 2) or the sum of cubes (x 1 3 + x 2 3) of the roots of a quadratic equation, less often - the sum of the inverse values of the squares of the roots or the sum of arithmetic square roots from the roots of the quadratic equation:
Vieta's theorem can help with this:
x 2 + px + q = 0
x 1 + x 2 = -p; x 1 ∙ x 2 = q.
Let us express across p and q:
1) the sum of the squares of the roots of the equation x 2 + px + q = 0;
2) the sum of the cubes of the roots of the equation x 2 + px + q = 0.
Solution.
1) Expression x 1 2 + x 2 2 is obtained by squaring both sides of the equality x 1 + x 2 = -p;
(x 1 + x 2) 2 = (- p) 2; expand the brackets: x 1 2 + 2x 1 x 2 + x 2 2 = p 2; express the required sum: x 1 2 + x 2 2 = p 2 -2x 1 x 2 = p 2 -2q. We got a useful equality: x 1 2 + x 2 2 = p 2 -2q.
2) Expression x 1 3 + x 2 3 we represent by the formula the sum of cubes in the form:
(x 1 3 + x 2 3) = (x 1 + x 2) (x 1 2 -x 1 x 2 + x 2 2) = - p (p 2 -2q-q) = - p (p 2 -3q).
Another useful equality: x 1 3 + x 2 3 = -p · (p 2 -3q).
Examples.
3) x 2 -3x-4 = 0. Without solving the equation, calculate the value of the expression x 1 2 + x 2 2.
Solution.
x 1 + x 2 = -p = 3, and the work x 1 ∙ x 2 = q =in example 1) equality:
x 1 2 + x 2 2 = p 2 -2q. We have -p= x 1 + x 2 = 3 → p 2 = 3 2 = 9; q = x 1 x 2 = -4. Then x 1 2 + x 2 2 = 9 - 2 (-4) = 9 + 8 = 17.
Answer: x 1 2 + x 2 2 = 17.
4) x 2 -2x-4 = 0. Calculate: x 1 3 + x 2 3.
Solution.
By Vieta's theorem, the sum of the roots of this reduced quadratic equation x 1 + x 2 = -p = 2, and the work x 1 ∙ x 2 = q =-4. Let's apply the received by us ( in example 2) equality: x 1 3 + x 2 3 = -p (p 2 -3q) = 2 (2 2 -3 (-4)) = 2 (4 + 12) = 2 16 = 32.
Answer: x 1 3 + x 2 3 = 32.
Question: what if we are given a non-reduced quadratic equation? Answer: it can always be "reduced" by dividing by the first coefficient.
5) 2x 2 -5x-7 = 0. Without deciding, calculate: x 1 2 + x 2 2.
Solution. We are given a complete quadratic equation. Divide both sides of the equality by 2 (the first coefficient) and get the reduced quadratic equation: x 2 -2.5x-3.5 = 0.
By Vieta's theorem, the sum of the roots is 2,5 ; the product of the roots is -3,5 .
We solve in the same way as an example 3) using equality: x 1 2 + x 2 2 = p 2 -2q.
x 1 2 + x 2 2 = p 2 -2q = 2,5 2 -2∙(-3,5)=6,25+7=13,25.
Answer: x 1 2 + x 2 2 = 13,25.
6) x 2 -5x-2 = 0. Find:
We transform this equality and, by Vieta's theorem, replacing the sum of roots by -p, and the product of roots through q, we get another useful formula. When deriving the formula, equality 1 was used): x 1 2 + x 2 2 = p 2 -2q.
In our example x 1 + x 2 = -p = 5; x 1 ∙ x 2 = q =-2. We substitute these values into the resulting formula:
7) x 2 -13x + 36 = 0. Find:
We transform this sum and get a formula by which it will be possible to find the sum of arithmetic square roots from the roots of a quadratic equation.
We have x 1 + x 2 = -p = 13; x 1 ∙ x 2 = q = 36. Substitute these values into the derived formula:
Advice : always check the possibility of finding the roots of a quadratic equation by suitable way, after all 4 reviewed useful formulas allow you to quickly complete the task, especially in cases where the discriminant is an "inconvenient" number. In all simple cases, find roots and operate on them. For example, in the last example, we select the roots according to Vieta's theorem: the sum of the roots must be equal to 13 , and the product of roots 36 ... What are these numbers? Of course, 4 and 9. Now calculate the sum of the square roots of these numbers: 2+3=5. That's it!
I. Vieta's theorem for the reduced quadratic equation.
The sum of the roots of the reduced quadratic equation x 2 + px + q = 0 is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term:
x 1 + x 2 = -p; x 1 ∙ x 2 = q.
Find the roots of the reduced quadratic equation using Vieta's theorem.
Example 1) x 2 -x-30 = 0. This is the reduced quadratic equation ( x 2 + px + q = 0), the second coefficient p = -1 and the free term q = -30. First, let's make sure that given equation has roots, and that the roots (if any) will be expressed in integers. For this, it is sufficient that the discriminant is the perfect square of an integer.
Find the discriminant D= b 2 - 4ac = (- 1) 2 -4 ∙ 1 ∙ (-30) = 1 + 120 = 121 = 11 2 .
Now, according to Vieta's theorem, the sum of the roots should be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:
x 1 + x 2 = 1; x 1 ∙ x 2 = -30. We need to choose two numbers so that their product is equal -30 , and the sum is unit... These are numbers -5 and 6 . Answer: -5; 6.
Example 2) x 2 + 6x + 8 = 0. We have the reduced quadratic equation with the second coefficient p = 6 and a free member q = 8... Let's make sure there are integer roots. Find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 ... The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us choose roots according to Vieta's theorem: the sum of the roots is equal to –P = -6, and the product of the roots is q = 8... These are numbers -4 and -2 .
In fact: -4-2 = -6 = -p; -4 ∙ (-2) = 8 = q. Answer: -4; -2.
Example 3) x 2 + 2x-4 = 0... In this reduced quadratic equation, the second coefficient p = 2 and the free term q = -4... Find the discriminant D 1, since the second coefficient is even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, therefore, we do output: the roots of this equation are not integers and cannot be found by Vieta's theorem. This means that we will solve this equation, as usual, using the formulas (in this case, using the formulas). We get:
Example 4). Make a quadratic equation for its roots if x 1 = -7, x 2 = 4.
Solution. The required equation will be written in the form: x 2 + px + q = 0, and, on the basis of Vieta's theorem –P = x 1 + x 2=-7+4=-3 → p = 3; q = x 1 ∙ x 2=-7∙4=-28 ... Then the equation will take the form: x 2 + 3x-28 = 0.
Example 5). Write a quadratic equation for its roots if:
II. Vieta's theorem for the complete quadratic equation ax 2 + bx + c = 0.
The sum of the roots is minus b divided by a, the product of the roots is with divided by a:
x 1 + x 2 = -b / a; x 1 ∙ x 2 = c / a.
Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11 = 0.
Solution.
We make sure that this equation will have roots. To do this, it is enough to compose an expression for the discriminant, and, without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 ... Now let's use theorem Vieta for complete quadratic equations.
x 1 + x 2 = -b: a=- (-7):2=3,5.
Example 7)... Find the product of the roots of the quadratic equation 3x 2 + 8x-21 = 0.
Solution.
Find the discriminant D 1, since the second coefficient ( 8 ) is an even number. D 1=4 2 -3∙(-21)=16+63=79>0 ... The quadratic equation has 2 root, according to Vieta's theorem the product of roots x 1 ∙ x 2 = c: a=-21:3=-7.
I. ax 2 + bx + c = 0- general quadratic equation
Discriminant D = b 2 - 4ac.
If D> 0, then we have two real roots:
If D = 0, then we have a single root (or two equal roots) x = -b / (2a).
If D<0, то действительных корней нет.
Example 1) 2x 2 + 5x-3 = 0.
Solution. a=2; b=5; c=-3.
D = b 2 - 4ac= 5 2 -4 ∙ 2 ∙ (-3) = 25 + 24 = 49 = 7 2> 0; 2 real roots.
4x 2 + 21x + 5 = 0.
Solution. a=4; b=21; c=5.
D = b 2 - 4ac= 21 2 - 4 ∙ 4 ∙ 5 = 441-80 = 361 = 19 2> 0; 2 real roots.
II. ax 2 + bx + c = 0 – partial quadratic equation with an even second
coefficient b
Example 3) 3x 2 -10x + 3 = 0.
Solution. a=3; b= -10 (even number); c=3.
Example 4) 5x 2 -14x-3 = 0.
Solution. a=5; b= -14 (even number); c=-3.
Example 5) 71x 2 + 144x + 4 = 0.
Solution. a=71; b= 144 (even number); c=4.
Example 6) 9x 2 -30x + 25 = 0.
Solution. a=9; b= -30 (even number); c=25.
III. ax 2 + bx + c = 0 – quadratic equation private view provided: a-b + c = 0.
The first root is always minus one and the second root is always minus with divided by a:
x 1 = -1, x 2 = -c / a.
Example 7) 2x 2 + 9x + 7 = 0.
Solution. a=2; b=9; c= 7. Let's check the equality: a-b + c = 0. We get: 2-9+7=0 .
Then x 1 = -1, x 2 = -c / a = -7 / 2 = -3.5. Answer: -1; -3,5.
IV. ax 2 + bx + c = 0 – quadratic equation of a particular form provided : a + b + c = 0.
The first root is always one and the second root is with divided by a:
x 1 = 1, x 2 = c / a.
Example 8) 2x 2 -9x + 7 = 0.
Solution. a=2; b=-9; c= 7. Let's check the equality: a + b + c = 0. We get: 2-9+7=0 .
Then x 1 = 1, x 2 = c / a = 7/2 = 3.5. Answer: 1; 3,5.
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At the stage of preparation for the final test, senior students need to improve their knowledge on the topic "Exponential Equations". The experience of past years shows that such tasks cause certain difficulties for schoolchildren. Therefore, high school students, regardless of their level of training, need to thoroughly master the theory, memorize formulas and understand the principle of solving such equations. Having learned how to cope with this type of problems, graduates will be able to count on high scores when passing the exam in mathematics.
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