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