Online equation solver with detailed solutions. How to solve an equation with variables (unknowns) on both sides of the equation. Roots of a quadratic equation

In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems in which certain conditions are introduced on the coefficients of the equation that limit them fall out of sight. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are found more and more often in the Unified State Examination materials and in 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 equations in two variables.

Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values ​​is a solution to the equation in question.

Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values ​​of the variables that turn this equation 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 a unique solution (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 equal to 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, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.

Factorization

Example 1.

Solve the equation: xy – 2 = 2x – y.

Solution.

We group the terms for the purpose of factorization:

(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y – 2) = 0. We have:

y = 2, x – any real number or x = -1, y – any real number.

Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality of non-negative numbers to zero

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 bracket 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 x = 2/3 and y = 3/2.

Answer: (2/3; 3/2).

Estimation method

Example 3.

Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.

Solution.

In each bracket we select a complete square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s 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 consists of treating the equation 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 equation for 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 when 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 they indicate restrictions on variables.

Example 5.

Solve the equation in whole numbers: x 2 + 5y 2 = 20x + 2.

Solution.

Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of 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 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 highlight 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 every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.

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. We get the sum of the squares of two integers equal to 37 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.

Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle any equation.

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2x 4 + 5x 3 - 11x 2 - 20x + 12 = 0

First you need to find one root using the selection method. Usually it is a divisor of the free term. In this case, the divisors of the number 12 are ±1, ±2, ±3, ±4, ±6, ±12. Let's start substituting them one by one:

1: 2 + 5 - 11 - 20 + 12 = -12 ⇒ number 1

-1: 2 - 5 - 11 + 20 + 12 = 18 ⇒ number -1 is not a root of a polynomial

2: 2 ∙ 16 + 5 ∙ 8 - 11 ∙ 4 - 20 ∙ 2 + 12 = 0 ⇒ number 2 is the root of the polynomial

We have found 1 of the roots of the polynomial. The root of the polynomial is 2, which means the original polynomial must be divisible by x - 2. In order to perform the division of polynomials, we use Horner’s scheme:

	2	5	-11	-20	12
2

The coefficients of the original polynomial are displayed in the top line. The root we found is placed in the first cell of the second row 2. The second line contains the coefficients of the polynomial that results from division. They are counted like this:

2 5 -11 -20 12 2 2	In the second cell of the second row we write the number 2, simply by moving it from the corresponding cell of the first row.
2 5 -11 -20 12 2 2 9	2 ∙ 2 + 5 = 9
2 5 -11 -20 12 2 2 9 7	2 ∙ 9 - 11 = 7
2 5 -11 -20 12 2 2 9 7 -6	2 ∙ 7 - 20 = -6
2 5 -11 -20 12 2 2 9 7 -6 0	2 ∙ (-6) + 12 = 0

The last number is the remainder of the division. If it is equal to 0, then we have calculated everything correctly.

2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2)(2x 3 + 9x 2 + 7x - 6)

But this is not the end. You can try to expand the polynomial in the same way 2x 3 + 9x 2 + 7x - 6.

Again we are looking for a root among the divisors of the free term. Number divisors -6 are ±1, ±2, ±3, ±6.

1: 2 + 9 + 7 - 6 = 12 ⇒ number 1 is not a root of a polynomial

-1: -2 + 9 - 7 - 6 = -6 ⇒ number -1 is not a root of a polynomial

2: 2 ∙ 8 + 9 ∙ 4 + 7 ∙ 2 - 6 = 60 ⇒ number 2 is not a root of a polynomial

-2: 2 ∙ (-8) + 9 ∙ 4 + 7 ∙ (-2) - 6 = 0 ⇒ number -2 is the root of the polynomial

Let's write the found root into our Horner scheme and start filling in the empty cells:

2 5 -11 -20 12 2 2 9 7 -6 0 -2 2	In the second cell of the third row we write the number 2, simply by moving it from the corresponding cell of the second row.
2 5 -11 -20 12 2 2 9 7 -6 0 -2 2 5	-2 ∙ 2 + 9 = 5
2 5 -11 -20 12 2 2 9 7 -6 0 -2 2 5 -3	-2 ∙ 5 + 7 = -3
2 5 -11 -20 12 2 2 9 7 -6 0 -2 2 5 -3 0	-2 ∙ (-3) - 6 = 0

Thus, we factored the original polynomial:

2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2)(x + 2)(2x 2 + 5x - 3)

Polynomial 2x 2 + 5x - 3 can also be factorized. To do this, you can solve the quadratic equation through the discriminant, or you can look for the root among the divisors of the number -3. One way or another, we will come to the conclusion that the root of this polynomial is the number -3

2 5 -11 -20 12 2 2 9 7 -6 0 -2 2 5 -3 0 -3 2	In the second cell of the fourth row we write the number 2, simply by moving it from the corresponding cell of the third row.
2 5 -11 -20 12 2 2 9 7 -6 0 -2 2 5 -3 0 -3 2 -1	-3 ∙ 2 + 5 = -1
2 5 -11 -20 12 2 2 9 7 -6 0 -2 2 5 -3 0 -3 2 -1 0	-3 ∙ (-1) - 3 = 0

Thus, we decomposed the original polynomial into linear factors:

2x 4 + 5x 3 - 11x 2 - 20x + 12 = (x - 2)(x + 2)(x + 3)(2x - 1)

And the roots of the equation are.

Goals:

1. Systematize and generalize knowledge and skills on the topic: Solutions of equations of the third and fourth degree.
2. Deepen your knowledge by completing a number of tasks, some of which are unfamiliar either in type or method of solution.
3. Forming an interest in mathematics through the study of new chapters of mathematics, nurturing a graphic culture through the construction of graphs of equations.

Lesson type: combined.

Equipment: graphic projector.

Visibility: table "Viete's Theorem".

During the classes

1. Oral counting

a) What is the remainder when dividing the polynomial p n (x) = a n x n + a n-1 x n-1 + ... + a 1 x 1 + a 0 by the binomial x-a?

b) How many roots can a cubic equation have?

c) How do we solve equations of the third and fourth degrees?

d) If b is an even number in a quadratic equation, then what is the value of D and x 1; x 2

2. Independent work (in groups)

Write an equation if the roots are known (answers to tasks are coded) “Vieta’s Theorem” is used

1 group

Roots: x 1 = 1; x 2 = -2; x 3 = -3; x 4 = 6

Make up an equation:

B=1 -2-3+6=2; b=-2

c=-2-3+6+6-12-18= -23; c= -23

d=6-12+36-18=12; d= -12

e=1(-2)(-3)6=36

x 4 -2 x 3 - 23x 2 - 12 x + 36 = 0(this equation is then solved by group 2 on the board)

Solution . We look for whole roots among the divisors of the number 36.

р = ±1;±2;±3;±4;±6…

p 4 (1)=1-2-23-12+36=0 The number 1 satisfies the equation, therefore =1 is the root of the equation. According to Horner's scheme

p 3 (x) = x 3 - x 2 -24x -36

p 3 (-2) = -8 -4 +48 -36 = 0, x 2 = -2

p 2 (x) = x 2 -3x -18=0

x 3 =-3, x 4 =6

Answer: 1;-2;-3;6 sum of roots 2 (P)

2nd group

Roots: x 1 = -1; x 2 = x 3 =2; x 4 =5

Make up an equation:

B=-1+2+2+5-8; b= -8

c=2(-1)+4+10-2-5+10=15; c=15

D=-4-10+20-10= -4; d=4

e=2(-1)2*5=-20;e=-20

8+15+4x-20=0 (group 3 solves this equation on the board)

р = ±1;±2;±4;±5;±10;±20.

p 4 (1)=1-8+15+4-20=-8

р 4 (-1)=1+8+15-4-20=0

p 3 (x) = x 3 -9x 2 +24x -20

p 3 (2) = 8 -36+48 -20=0

p 2 (x) = x 2 -7x +10 = 0 x 1 = 2; x 2 =5

Answer: -1;2;2;5 sum of roots 8(P)

3 group

Roots: x 1 = -1; x 2 =1; x 3 = -2; x 4 =3

Make up an equation:

В=-1+1-2+3=1;В=-1

с=-1+2-3-2+3-6=-7;с=-7

D=2+6-3-6=-1; d=1

e=-1*1*(-2)*3=6

x 4 - x 3- 7x 2 + x + 6 = 0(group 4 solves this equation later on the board)

Solution. We look for whole roots among the divisors of the number 6.

р = ±1;±2;±3;±6

p 4 (1)=1-1-7+1+6=0

p 3 (x) = x 3 - 7x -6

р 3 (-1) = -1+7-6=0

p 2 (x) = x 2 - x -6 = 0; x 1 = -2; x 2 =3

Answer: -1;1;-2;3 Sum of roots 1(O)

4 group

Roots: x 1 = -2; x 2 = -2; x 3 = -3; x 4 = -3

Make up an equation:

B=-2-2-3+3=-4; b=4

c=4+6-6+6-6-9=-5; с=-5

D=-12+12+18+18=36; d=-36

e=-2*(-2)*(-3)*3=-36;e=-36

x 4 +4x 3 – 5x 2 – 36x -36 = 0(this equation is then solved by group 5 on the board)

Solution. We look for whole roots among the divisors of the number -36

р = ±1;±2;±3…

p(1)= 1 + 4-5-36-36 = -72

p 4 (-2) = 16 -32 -20 + 72 -36 = 0

p 3 (x) = x 3 +2x 2 -9x-18 = 0

p 3 (-2) = -8 + 8 + 18-18 = 0

p 2 (x) = x 2 -9 = 0; x=±3

Answer: -2; -2; -3; 3 Sum of roots-4 (F)

5 group

Roots: x 1 = -1; x 2 = -2; x 3 = -3; x 4 = -4

Write an equation

x 4+ 10x 3 + 35x 2 + 50x + 24 = 0(this equation is then solved by group 6 on the board)

Solution . We look for whole roots among the divisors of the number 24.

р = ±1;±2;±3

p 4 (-1) = 1 -10 + 35 -50 + 24 = 0

p 3 (x) = x- 3 + 9x 2 + 26x+ 24 = 0

p 3 (-2) = -8 + 36-52 + 24 = O

p 2 (x) = x 2 + 7x+ 12 = 0

Answer: -1;-2;-3;-4 sum-10 (I)

6 group

Roots: x 1 = 1; x 2 = 1; x 3 = -3; x 4 = 8

Write an equation

B=1+1-3+8=7;b=-7

c=1 -3+8-3+8-24= -13

D=-3-24+8-24= -43; d=43

x 4 - 7x 3- 13x 2 + 43x - 24 = 0 (this equation is then solved by group 1 on the board)

Solution . We look for whole roots among the divisors of the number -24.

p 4 (1)=1-7-13+43-24=0

p 3 (1)=1-6-19+24=0

p 2 (x)= x 2 -5x - 24 = 0

x 3 =-3, x 4 =8

Answer: 1;1;-3;8 sum 7 (L)

3. Solving equations with a parameter

1. Solve the equation x 3 + 3x 2 + mx - 15 = 0; if one of the roots is equal to (-1)

Write the answer in ascending order

R=P 3 (-1)=-1+3-m-15=0

x 3 + 3x 2 -13x - 15 = 0; -1+3+13-15=0

By condition x 1 = - 1; D=1+15=16

P 2 (x) = x 2 +2x-15 = 0

x 2 = -1-4 = -5;

x 3 = -1 + 4 = 3;

Answer: - 1; -5; 3

In ascending order: -5;-1;3. (b N S)

2. Find all roots of the polynomial x 3 - 3x 2 + ax - 2a + 6, if the remainders from its division into binomials x-1 and x +2 are equal.

Solution: R=P 3 (1) = P 3 (-2)

P 3 (1) = 1-3 + a- 2a + 6 = 4-a

P 3 (-2) = -8-12-2a-2a + 6 = -14-4a

x 3 -Zx 2 -6x + 12 + 6 = x 3 -Zx 2 -6x + 18

x 2 (x-3)-6(x-3) = 0

(x-3)(x 2 -6) = 0

The product of two factors is equal to zero if and only if at least one of these factors is equal to zero, and the other makes sense.

2nd group. Roots: -3; -2; 1; 2;

3 group. Roots: -1; 2; 6; 10;

4 group. Roots: -3; 2; 2; 5;

5 group. Roots: -5; -2; 2; 4;

6 group. Roots: -8; -2; 6; 7.

In simple algebraic equations, a variable is on only one side of the equation, but in more complex equations, variables can be on both sides of the equation. When solving such equations, always remember that any operation that is performed on one side of the equation must also be performed on the other side. Using this rule, variables can be moved from one side of an equation to the other to isolate them and calculate their values.

Steps

Solving equations with one variable on both sides of the equation

1. Apply the distributive law (if necessary). This law states that a (b + c) = a b + a c (\displaystyle a(b+c)=ab+ac). The distributive law allows you to open the brackets by multiplying the term outside the brackets by each term in the brackets.

   • For example, given an equation, use the distributive law to multiply the term outside the parentheses by each term in the parentheses:
     2 (10 − 2 x) = 4 (2 x + 2) (\displaystyle 2(10-2x)=4(2x+2))
     

2. Get rid of a variable on one side of the equation. To do this, subtract or add the same term with the variable. For example, if a variable term is subtracted, add the same term to get rid of it; if a term with a variable is added, subtract the same term to get rid of it. It is usually easier to get rid of the variable with the smaller coefficient.

   • For example, in Eq. 20 − 4 x = 8 x + 8 (\displaystyle 20-4x=8x+8) get rid of your penis − 4 x (\displaystyle -4x); for this add 4 x (\displaystyle 4x):
     20 − 4 x + 4 x = 8 x + 8 (\displaystyle 20-4x+4x=8x+8).

3. Make sure that equality is not violated. Any mathematical operation performed on one side of the equation must also be performed on the other side. So if you add or subtract a term to get rid of a variable on one side of the equation, add or subtract the same term on the other side of the equation.

   • For example, if you add to one side of the equation 4 x (\displaystyle 4x) to get rid of a variable, you need to add 4 x (\displaystyle 4x) and to the other side of the equation:
     

4. Simplify the equation by adding or subtracting like terms. At this point, the variable should be on one side of the equation.

   • For example:
     20 − 4 x + 4 x = 8 x + 8 + 4 x (\displaystyle 20-4x+4x=8x+8+4x)
     

5. Move the free terms to one side of the equation (if necessary). It is necessary to make sure that the term with the variable is on one side, and the free term is on the other. To move the dummy term (and get rid of it on one side of the equation), add or subtract it from both sides of the equation.

   • For example, to get rid of a free member + 8 (\displaystyle +8) on the variable side, subtract 8 from both sides of the equation:
     20 = 12 x + 8 (\displaystyle 20=12x+8)
     20 − 8 = 12 x + 8 − 8 (\displaystyle 20-8=12x+8-8)
     

6. Get rid of the coefficient on the variable. To do this, perform the opposite operation of the operation between the coefficient and the variable. In most cases, simply divide both sides of the equation by the coefficient of the variable. Remember that any mathematical operation performed on one side of the equation must also be performed on the other side.

   • For example, to get rid of the factor 12, divide both sides of the equation by 12:
     12 = 12 x (\displaystyle 12=12x)
     12 12 = 12 x 12 (\displaystyle (\frac (12)(12))=(\frac (12x)(12)))
     1 = x (\displaystyle 1=x)

7. Check the answer. To do this, substitute the found value into the original equation. If equality is satisfied, the answer is correct.

   • For example, if 1 = x (\displaystyle 1=x), substitute 1 (instead of the variable) into the original equation:
     2 (10 − 2 x) = 4 (2 x + 2) (\displaystyle 2(10-2x)=4(2x+2))
     2 (10 − 2 (1)) = 4 (2 (1) + 2) (\displaystyle 2(10-2(1))=4(2(1)+2))
     2 (10 − 2) = 4 (2 + 2) (\displaystyle 2(10-2)=4(2+2))
     20 − 4 = 8 + 8 (\displaystyle 20-4=8+8)
     16 = 16 (\displaystyle 16=16)

   Solving a system of equations with two variables

   1. Isolate the variable in one equation. Perhaps in one of the equations the variable will already be isolated; otherwise, use mathematical operations to isolate the variable on one side of the equation. Remember that any mathematical operation performed on one side of the equation must also be performed on the other side.

      • For example, the equation is given. To isolate a variable y (\displaystyle y), subtract 1 from both sides of the equation:
        y + 1 = x − 1 (\displaystyle y+1=x-1)
        y + 1 − 1 = x − 1 − 1 (\displaystyle y+1-1=x-1-1)
        

   2. Substitute the value (as an expression) of the isolated variable into the other equation. Be sure to substitute the entire expression. The result is an equation with one variable that is easy to solve.

      • For example, the first equation has the form , and the second equation is reduced to the form y = x − 2 (\displaystyle y=x-2). In this case, in the first equation instead y (\displaystyle y) substitute x − 2 (\displaystyle x-2):
        2 x = 20 − 2 y (\displaystyle 2x=20-2y)
        

   3. Find the value of the variable. To do this, move the variable to one side of the equation. Then move the free terms to the other side of the equation. Then isolate the variable using a multiplication or division operation.

      • For example:
        2 x = 20 − 2 (x − 2) (\displaystyle 2x=20-2(x-2))
        2 x = 20 − 2 x + 4 (\displaystyle 2x=20-2x+4)
        2 x = 24 − 2 x (\displaystyle 2x=24-2x)
        2 x + 2 x = 24 − 2 x + 2 x (\displaystyle 2x+2x=24-2x+2x)
        4 x = 24 (\displaystyle 4x=24)
        4 x 4 = 24 4 (\displaystyle (\frac (4x)(4))=(\frac (24)(4)))
        x = 6 (\displaystyle x=6)

   4. Find the value of another variable. To do this, substitute the found value of the variable into one of the equations. The result is an equation with one variable that is easy to solve. Keep in mind that the found value of a variable can be substituted into any equation.

      • For example, if x = 6 (\displaystyle x=6), substitute 6 (instead of x (\displaystyle x)) into the second equation:
        y = x − 2 (\displaystyle y=x-2)
        y = (6) − 2 (\displaystyle y=(6)-2)
        y = 4 (\displaystyle y=4)

   5. Check the answer. To do this, substitute the values ​​of both variables into one of the equations. If equality is satisfied, the answer is correct.

      • For example, if you find that x = 6 (\displaystyle x=6) And y = 4 (\displaystyle y=4), substitute these values ​​into one of the original equations:
        2 x = 20 − 2 y (\displaystyle 2x=20-2y)
        2 (6) = 20 − 2 (4) (\displaystyle 2(6)=20-2(4))
        12 = 20 − 8 (\displaystyle 12=20-8)
        12 = 12 (\displaystyle 12=12)

   Solving equations

   1. Solve the following one-variable equation using the distributive law: .

      • 
        5 (x + 4) = 6 x − 5 (\displaystyle 5(x+4)=6x-5)
        
      • Get rid of 5 x (\displaystyle 5x) on the left side of the equation; to do this, subtract 5 x (\displaystyle 5x) from both sides of the equation:
        5 x + 20 = 6 x − 5 (\displaystyle 5x+20=6x-5)
        5 x + 20 − 5 x = 6 x − 5 − 5 x (\displaystyle 5x+20-5x=6x-5-5x)
        
      • Isolate the variable; To do this, add 5 to both sides of the equation:
        20 = x − 5 (\displaystyle 20=x-5)
        20 + 5 = x − 5 + 5 (\displaystyle 20+5=x-5+5)
        25 = x (\displaystyle 25=x)

   2. Solve the following fraction equation: .

      • Get rid of the fraction. To do this, multiply both sides of the equation by the expression (or number) in the denominator of the fraction:
        − 7 + 3 x = 7 − x 2 (\displaystyle -7+3x=(\frac (7-x)(2)))
        2 (− 7 + 3 x) = 2 (7 − x 2) (\displaystyle 2(-7+3x)=2((\frac (7-x)(2))))
        
      • Get rid of − x (\displaystyle -x) on the right side of the equation; for this add x (\displaystyle x) to both sides of the equation:
        − 14 + 6 x = 7 − x (\displaystyle -14+6x=7-x)
        − 14 + 6 x + x = 7 − x + x (\displaystyle -14+6x+x=7-x+x)
        
      • Move the free terms to one side of the equation; To do this, add 14 to both sides of the equation:
        − 14 + 7 x = 7 (\displaystyle -14+7x=7)
        − 14 + 7 x + 14 = 7 + 14 (\displaystyle -14+7x+14=7+14)
        
      • Get rid of the coefficient on the variable; To do this, divide both sides of the equation by 7:
        7 x = 21 (\displaystyle 7x=21)
        7 x 7 = 21 7 (\displaystyle (\frac (7x)(7))=(\frac (21)(7)))
        x = 3 (\displaystyle x=3)

   3. Solve the following system of equations: 9 x + 15 = 12 y; 9 y = 9 x + 27 (\displaystyle 9x+15=12y;9y=9x+27)

      • Isolate a variable y (\displaystyle y) in the second equation:
        
        9 y = 9 (x + 3) (\displaystyle 9y=9(x+3))
        9 y 9 = 9 (x + 3) 9 (\displaystyle (\frac (9y)(9))=(\frac (9(x+3))(9)))
        y = x + 3 (\displaystyle y=x+3)
      • In the first equation instead y (\displaystyle y) substitute x + 3 (\displaystyle x+3):
        9 x + 15 = 12 y (\displaystyle 9x+15=12y)
        9 x + 15 = 12 (x + 3) (\displaystyle 9x+15=12(x+3))
      • Use the distributive law to open the brackets:
        
      • Get rid of the variable on the left side of the equation; to do this, subtract 9 x (\displaystyle 9x) from both sides of the equation:
        9 x + 15 = 12 x + 36 (\displaystyle 9x+15=12x+36)
        9 x + 15 − 9 x = 12 x + 36 − 9 x (\displaystyle 9x+15-9x=12x+36-9x)
        
      • Move the free terms to one side of the equation; To do this, subtract 36 from both sides of the equation:
        15 = 3 x + 36 (\displaystyle 15=3x+36)
        15 − 36 = 3 x + 36 − 36 (\displaystyle 15-36=3x+36-36)
        
      • Get rid of the coefficient on the variable; To do this, divide both sides of the equation by 3:
        − 21 = 3 x (\displaystyle -21=3x)
        − 21 3 = 3 x 3 (\displaystyle (\frac (-21)(3))=(\frac (3x)(3)))
        − 7 = x (\displaystyle -7=x)
      • Find the value y (\displaystyle y); To do this, substitute the found value x (\displaystyle x) into one of the equations:
        9 y = 9 x + 27 (\displaystyle 9y=9x+27)
        9 y = 9 (− 7) + 27 (\displaystyle 9y=9(-7)+27)
        9 y = − 63 + 27 (\displaystyle 9y=-63+27)
        9 y = − 36 (\displaystyle 9y=-36)
        9 y 9 = − 36 9 (\displaystyle (\frac (9y)(9))=(\frac (-36)(9)))
        y = − 4 (\displaystyle y=-4)

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