How to solve systems of the equations

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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Instruction

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.