How to solve systems of the equations

Simply to solve system of the equations if to use the main ways of the decision of systems of the linear equations: by method of substitution and by method of addition.

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1

Let's consider methods of the decision of system of the equations on the example of system from two linear equations having two unknown values. In a general view such system registers as follows (at the left the equations unite a brace):

akh+bu=c

dx+ey=f, where

and, b, c, d, e, f - coefficients (concrete numbers), and x and at, as usual - unknown. Numbers and, b, with, d are called as coefficients at unknown, and with and f - free members. The solution of such system of the equations is found by two main methods.

Decision of system of the equations by a substitution method.

1

We take the first equation and we express one of unknown (x) through coefficients and another unknown (at): x = (with-by)/a

2

We substitute received for x expression in the second equation: d (with-by)/a+ey=f

3

Solving the received equation, we find expression for at: at = (af-CD) / (ae-bd)

4

We substitute the received expression for at in expression for x: x = (se-bf) / (ae-bd)

Example: it is required to solve system of the equations:

3kh-2u=4

h+3u=5

We find value x from the first equation:

x = (2u+4)/3

We substitute the received expression in the second equation and we receive the equation from one variable (at):

(2u+4)/3+3u=5, from where we receive:

u=1

Now we substitute the found value at in expressions for a variable x:

x = (2*1+4)/3=2

Answer: h=2, u=1.

2

Decision of system of the equations by method of addition (subtraction).

This method is reduced to multiplication of both parts of the equations by such numbers (parameters) that as a result coefficients at one of variables coincided (perhaps with an opposite sign).

Generally, both parts of the first equation need to be increased on (-d), and both parts of the second equation on and. As a result we receive:

- to an adx-bd =-sd

adx+aey=af

Having put the received equations, we will receive:

- bdu+ayeu =-sd+af,

from where we receive expression for a variable at:

at = (af-CD) / (ae-bd),

substituting expression for at in any equation of system, we receive:

ax+b (af-CD) / (ae-bd) =c?

from this equation we find the second unknown:

x = (se-bf) / (ae-bd)

Example. To solve by method of addition or subtraction system of the equations:

3kh-2u=4

h+3u=5

Let's increase the first equation on (-1), and the second on 3:

- 3kh+2u =-4

3kh+9u=15

Having put (term by term) both equations, we receive:

11u=11

From where we receive:

u=1

We substitute the received value for at in any of the equations, for example, in the second, we receive:

3kh+9=15, from where

h=2

Answer: h=2, u=1.