Instruction

1

Let's say you have

2x+y=10

x-y=2

It can be solved in several ways.

**the equation**:2x+y=10

x-y=2

It can be solved in several ways.

2

Way to postuniversity one variable and substitute it into another equation. It is possible to Express any variable of your choice. For example, Express "from the second equation:

x-y=2 => y=x-2Затем all the substitute in the first equation:

2+(x-2)=10Перенесите all the numbers without "x to the right side and calculate:

2x+x=10+2

3=12 Then, to find x , divide both sides of equation by 3:

x=4.So, you have found "x . Find ". To do this, substitute the "x in the equation from where you have expressed ":

y=x-2=4-2=2

y=2.

x-y=2 => y=x-2Затем all the substitute in the first equation:

2+(x-2)=10Перенесите all the numbers without "x to the right side and calculate:

2x+x=10+2

3=12 Then, to find x , divide both sides of equation by 3:

x=4.So, you have found "x . Find ". To do this, substitute the "x in the equation from where you have expressed ":

y=x-2=4-2=2

y=2.

3

Make the check. To do this, substitute the resulting values into the equations:

2*4+2=10

4-2=2

Unknown found right!

2*4+2=10

4-2=2

Unknown found right!

4

Method of addition or subtraction uraninites from some variable. In our case it is easier to do with ".

As in the first equation "the sign", and the second " - then you can perform the operation of addition, i.e. the left part of the fold on the left and the right on the right:

2x+y+(x-y)=10+2Преобразуйте:

2x+y+x-y=10+2

3=12

x=4Подставьте "x in any equation and find ":

2*4+y=10

8+y=10

at=10-8

y=2 on the 1st way you can check that the roots are found true.

As in the first equation "the sign", and the second " - then you can perform the operation of addition, i.e. the left part of the fold on the left and the right on the right:

2x+y+(x-y)=10+2Преобразуйте:

2x+y+x-y=10+2

3=12

x=4Подставьте "x in any equation and find ":

2*4+y=10

8+y=10

at=10-8

y=2 on the 1st way you can check that the roots are found true.

5

If there is no clearly defined variables, it is necessary to convert the equation.

In the first equation have a "2 , and the second just "x . To the addition or subtraction of "x was reduced, multiply the second equation by 2:

x-y=2

2-2U=4Затем subtract the first equation from the second:

2x+y-(2x-2Y)=10-4Заметим, if before the parenthesis is minus, then after the disclosure adjust the marks to the opposite:

2x+y-2x+2Y=6

3y=6

y=2"x, find the expressing of any equation, i.e.,

x=4

In the first equation have a "2 , and the second just "x . To the addition or subtraction of "x was reduced, multiply the second equation by 2:

x-y=2

2-2U=4Затем subtract the first equation from the second:

2x+y-(2x-2Y)=10-4Заметим, if before the parenthesis is minus, then after the disclosure adjust the marks to the opposite:

2x+y-2x+2Y=6

3y=6

y=2"x, find the expressing of any equation, i.e.,

x=4

# Advice 2: How to solve linear equation with two variables

**The equation**, in General written ax+by+C=0, is called a linear equation with two

**variables**. This equation in itself contains an infinite number of solutions, so the task it is always complemented by either another equation or bounding condition. Depending on conditions, will be given a task, solve a linear equation with two

**variables**follows different ways.

You will need

- - linear equation with two variables;
- - the second equation or additional terms.

Instruction

1

If given a system of two linear equations, solve it in the following way. Select one of the equations in which the coefficients in front of

**variables**is smaller and Express one of the variables, e.g., x. Then substitute this value contains, in the second equation. In the resulting equation will have only one variable y, move all the parts from the left side, and the free members to the right. Find and replace any of the original equations to find X.2

To solve a system of two equations in another way. Multiply one of the equations on a number below the coefficient in front of one of the variables, for example, before x was the same in both equations. Then subtract one equation from the other (if the right side is not equal to 0, don't forget to deduct the same and right parts). You will see that the variable x disappeared, leaving only one variable. Solve the resulting equation, and substitute the value found in any of the original equations. Find H.

3

The third method of solving systems of two linear equations – graphical. Draw a coordinate system and draw the graphs of two straight lines whose equations are provided in your system. For this, substitute any two values of x in the equation and find the corresponding y – it will be the coordinates of the points belonging to a straight line. It is most convenient to find the intersection with the coordinate axes is sufficient to substitute the values x=0 and y=0. The coordinates of the point of intersection of these two lines will be the solution of the problem.

4

If the conditions of the problem, only one linear equation, so you are given additional conditions that can help you find a solution. Carefully read the task to find these terms. If

**variables**x and y denoted the distance, speed, age, and weight – feel free to put the constraint x≥0 and y≥0. It is possible, under x or hides the number of children, apples, trees, etc. – then values can only be integers. If x is the age of the son, it is clear that he can't be older than my father, so specify it in terms of the problem.5

Plot the straight line, the corresponding linear equation. Look at the chart, perhaps it will be only a few solutions satisfying all conditions – for example, integers and positive integers. They will be solutions of your equation.

# Advice 3: How to solve an equation with three unknowns

In itself

**an equation**with three**unknowns**has many solutions, so often it is complemented by two more equations or conditions. Depending on what the source data will depend largely on the progress of the solution.You will need

- - a system of three equations with three unknowns.

Instruction

1

If two of the three equations of the system have only two unknown of the three, try to Express some variables through others and substitute them in

**the equation**with three**unknown**. Your goal is to turn it into a normal**equation**with one unknown. If this is successful, then the solution is quite simple – substitute the value found into the other equation and find all the other unknown.2

Some systems of equations can be solved by subtracting one equation from the other. Let's see if there are opportunities to multiply one expression by the number or variable so that when subtraction was reduced from two unknown. If so, use it, most likely, a subsequent decision is not difficult. Don't forget that when multiplying by a number, multiply both the left and right. Similarly, when you subtract the equations you need to remember that the right side must also be deducted.

3

If the previous methods did not help, use the General method of solutions of any equations with three

**unknowns**. To do this, rewrite the equation in the form а11х1+а12х2+а13х3=b1, а21х1+а22х2+а23х3=b2, а31х1+а32х2+а33х3=b3. Now make a matrix of the coefficients x (A), the unknown matrix (X) matrix and free members (In). Please note, multiplying the coefficient matrix by a matrix is unknown, you will get a matrix equal to the matrix free members, that is, A*X=B.4

Find the matrix a of degree (-1) after finding the determinant of a matrix, note that it should not be zero. Then, multiply the resulting matrix on the matrix, as a result you get the desired matrix X, showing all the values.

5

Find the solution of the system of three equations using the method of Kramer. To do this, find the determinant of the third order ∆, the corresponding matrix system. Then find three more of the determinant ∆1, ∆2 and ∆3, by substituting values of the corresponding columns of the values of the free members. Now find x: x1=∆1/∆, x2=∆2/∆, X3=∆3/∆.