Solve equation 3x. Solving equations with two variables
In the 7th grade mathematics course, they first meet with equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why it is out of sight whole line problems in which certain conditions are introduced on the coefficients of the equation that restrict them. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are encountered more and more often in the materials of the Unified State Examination and at entrance exams.
Which equation will be called an equation with two variables?
So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are two-variable equations.
Consider the equation 2x - y = 1. It turns into a true equality at x = 2 and y = 3, so this pair of variable values is a solution to the equation under consideration.
Thus, the solution of 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 whose sum is 3. The set of solutions to this equation can be written as (k; 3 - k), where k is any real number.
The main methods for solving equations with two variables are methods based on the decomposition of expressions into factors, the selection of a full square, the use of the properties of a quadratic equation, the boundedness of expressions, and evaluation 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. Take out the common factor from each bracket:
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.
In this way, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.
Zero is not negative numbers
Example 2
Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).
Solution.
Grouping:
(9x 2 - 12x + 4) + (4y 2 - 12y + 9) = 0. Now each parenthesis can be collapsed using the square 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.
So 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 the full square:
((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Estimate 
the meaning of the expressions in brackets.
(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, so 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 some variable.
Example 4
Solve the equation: x 2 - 6x + y - 4√y + 13 = 0.
Solution.
Let's solve the equation as a quadratic one 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 only have a solution if D = 0, that is, if y = 4. We substitute the value of y into the original equation and find that x = 3.
Answer: (3; 4).
Often in equations with two unknowns indicate restrictions on variables.
Example 5
Solve the equation in integers: x 2 + 5y 2 = 20x + 2.
Solution.
Let's rewrite the equation as x 2 = -5y 2 + 20x + 2. Right part the resulting equation, when divided by 5, gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number that is not divisible by 5 gives a remainder of 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.
Let's select the full 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 if |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). Answer the smallest amount.
Solution.
Select full 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, we get 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.
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Service assignment. Matrix Calculator designed to solve systems linear equations in a matrix way (see an example of solving similar problems).Instruction. For online solutions it is necessary to choose the type of equation and set the dimension of the corresponding matrices.
where A, B, C are given 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 #1. Exercise. Find a solution to a matrix equation
Solution. Denote:
Then matrix equation will be written in the form: A X B = C.
The determinant of matrix A is detA=-1
Since A is a nonsingular 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 are looking for the matrix X by the formula: X = A -1 C B -1 
Answer:
Example #2. Exercise. Solve matrix equation 
Solution. 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 degenerate matrix (the determinant is 0), therefore, the equation has no solution.
Example #3. Exercise. Find a solution to a matrix equation
Solution. 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 nonsingular matrix, there is an inverse matrix A -1 . Multiply on the right both sides of the equation by A -1: X A A -1 = B A -1 , from which we find that X = B A -1
Find the inverse matrix A -1 .
Transposed matrix A T: 
Inverse matrix A -1: 
We are looking for the matrix X by the formula: X = B A -1 
Answer: > 
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. Power or exponential equations are called equations in which the variables are in powers, and the base is a number. For instance:
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 bases 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.
Suppose we are given an exponential equation of the following form:
It is worth starting the solution of this equation with an analysis of the base. The bases are different - 2 and 4, and for the solution we need them to be the same, so we transform 4 according to the following formula - \ [ (a ^ n) ^ m = a ^ (nm): \]
Add to the original equation:
Let's take out the brackets \
Express \
Since the degrees are the same, we discard them:
Answer: \
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I. ax 2 \u003d 0 – incomplete quadratic equation (b=0, c=0 ). Solution: x=0. Answer: 0.
Solve equations.
2x·(x+3)=6x-x 2 .
Solution. Expand the brackets by multiplying 2x for each term in brackets:
2x2 +6x=6x-x2 ; moving the terms from the right side to the left side:
2x2 +6x-6x+x2=0; Here are similar terms:
3x 2 =0, hence x=0.
Answer: 0.
II. ax2+bx=0 –incomplete quadratic equation (s=0 ). Solution: x (ax+b)=0 → x 1 =0 or ax+b=0 → x 2 =-b/a. Answer: 0; -b/a.
5x2 -26x=0.
Solution. Take out the common factor X for brackets:
x(5x-26)=0; each factor can be zero:
x=0 or 5x-26=0→ 5x=26, divide both sides of the equality by 5 and we get: x \u003d 5.2.
Answer: 0; 5,2.
Example 3 64x+4x2=0.
Solution. Take out the common factor 4x for 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. Applying the formula for the square of the difference of two expressions, open the brackets:
x 2 -6x+9+5x=9; transform to the form: x 2 -6x+9+5x-9=0; Here are similar terms:
x2-x=0; endure X outside the brackets, we get: x (x-1)=0. From here or x=0 or x-1=0→ x=1.
Answer: 0; 1.
III. ax2+c=0 –incomplete quadratic equation (b=0 ); Solution: ax 2 \u003d -c → x 2 \u003d -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 \u003d 49, from here x=±7. Answer:-7; 7.
Example 6 9x2-4=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 reciprocals 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 \u003d-p; x 1 ∙ x 2 \u003d q.
Express across p and q:
1) the sum of the squares of the roots of the equation x2+px+q=0;
2) the sum of the cubes of the roots of the equation x2+px+q=0.
Solution.
1) Expression x 1 2 + x 2 2 obtained by squaring both sides of the equation x 1 + x 2 \u003d-p;
(x 1 +x 2) 2 \u003d (-p) 2; open the brackets: x 1 2 +2x 1 x 2 + x 2 2 =p 2; we express the desired amount: x 1 2 +x 2 2 \u003d p 2 -2x 1 x 2 \u003d p 2 -2q. We have a useful equation: x 1 2 +x 2 2 \u003d p 2 -2q.
2) Expression x 1 3 + x 2 3 represent by the formula of 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 equation: x 1 3 + x 2 3 \u003d-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 \u003d-p \u003d 3, and the work x 1 ∙x 2 \u003d q \u003din example 1) equality:
x 1 2 +x 2 2 \u003d 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 \u003d-p \u003d 2, and the work x 1 ∙x 2 \u003d q \u003d-4. Let us apply what we have obtained ( in example 2) equality: x 1 3 +x 2 3 \u003d-p (p 2 -3q) \u003d 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 term by term by the first coefficient.
5) 2x2 -5x-7=0. Without solving, calculate: x 1 2 + x 2 2.
Solution. We are given a complete quadratic equation. Divide both sides of the equation by 2 (the first coefficient) and get the following quadratic equation: x 2 -2.5x-3.5 \u003d 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 the equality: x 1 2 +x 2 2 \u003d 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) x2 -5x-2=0. Find:
Let us transform this equality and, by replacing the sum of the roots in terms of the Vieta theorem, -p, and the product of the roots through q, we get another useful formula. When deriving the formula, we used equality 1): x 1 2 +x 2 2 \u003d p 2 -2q.
In our example x 1 + x 2 \u003d -p \u003d 5; x 1 ∙x 2 \u003d q \u003d-2. Substitute these values into the resulting formula:
7) x 2 -13x+36=0. Find:
Let's 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 \u003d -p \u003d 13; x 1 ∙x 2 \u003d q \u003d 36. Substitute these values into the derived formula:
Advice : always check the possibility of finding the roots of the quadratic equation by suitable way, after all 4 reviewed useful formulas allow you to quickly complete the task, first of all, in cases where the discriminant is an “inconvenient” number. In all simple cases, find the roots and operate on them. For example, in the last example, we select the roots using the Vieta theorem: the sum of the roots should be equal to 13 , and the product of the roots 36 . What are these numbers? Certainly, 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 \u003d-p; x 1 ∙ x 2 \u003d q.
Find the roots of the given 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 as integers. For this, it is sufficient that the discriminant be the full square of an integer.
Finding the discriminant D=b 2 - 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .
Now, according to the Vieta theorem, the sum of the roots must 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 \u003d -30. We need to choose such two numbers so that their product is equal to -30 , and the sum is unit. These are the 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 free member q=8. Make sure that there are integer roots. Let's find the discriminant D1 D1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , so the roots of this equation are integers. We choose the roots according to the Vieta theorem: the sum of the roots is equal to –p=-6, and the product of the roots is q=8. These are the numbers -4 and -2 .
Actually: -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. Let's find the discriminant D1, since the second coefficient is even number. D1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of a number, so we do conclusion: the roots of this equation are not integers and cannot be found using Vieta's theorem. So, we solve this equation, as usual, according to the formulas (in this case, according to the formulas). We get:
Example 4). Write a quadratic equation using its roots if x 1 \u003d -7, x 2 \u003d 4.
Solution. The desired equation will be written in the form: x 2 +px+q=0, moreover, based on the Vieta theorem –p=x1 +x2=-7+4=-3 →p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x2 +3x-28=0.
Example 5). Write a quadratic equation using its roots if:
II. Vieta's theorem for the complete quadratic equation ax2+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 \u003d -b / a; x 1 ∙ x 2 \u003d c / a.
Example 6). Find the sum of the roots of a quadratic equation 2x2 -7x-11=0.
Solution.
We are convinced that this equation will have roots. To do this, it is enough to write 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 . And 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 a quadratic equation 3x2 +8x-21=0.
Solution.
Let's find the discriminant D1, since the second coefficient ( 8 ) is an even number. D1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to the Vieta theorem, the product of the roots x 1 ∙ x 2 \u003d c: a=-21:3=-7.
I. ax 2 +bx+c=0 is a 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) 2x2 +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.
4x2 +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. ax2+bx+c=0 – special quadratic equation for an even second
coefficient b
Example 3) 3x2 -10x+3=0.
Solution. a=3; b\u003d -10 (even number); c=3.
Example 4) 5x2-14x-3=0.
Solution. a=5; b= -14 (even number); c=-3.
Example 5) 71x2 +144x+4=0.
Solution. a=71; b=144 (even number); c=4.
Example 6) 9x 2 -30x+25=0.
Solution. a=9; b\u003d -30 (even number); c=25.
III. ax2+bx+c=0 – quadratic equation private type, provided: a-b+c=0.
The first root is always minus one, and the second root is minus With divided by a:
x 1 \u003d -1, x 2 \u003d - c / a.
Example 7) 2x2+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 \u003d -1, x 2 \u003d -c / a \u003d -7 / 2 \u003d -3.5. Answer: -1; -3,5.
IV. ax2+bx+c=0 – quadratic equation of a particular form under the condition : a+b+c=0.
The first root is always equal to one, and the second root is equal to With divided by a:
x 1 \u003d 1, x 2 \u003d c / a.
Example 8) 2x2 -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 \u003d 1, x 2 \u003d c / a \u003d 7/2 \u003d 3.5. Answer: 1; 3,5.
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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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