Solving an equation of the first class in one unknown
, One of the easiest mathematical equations
5 x -8 = 2
And resolved to take the 8, the other side opposite signal
then divisible
be 5
5 x = 2+8
5 x = 10
And dividing the 5 parties be
Q = 2
Solving equations of the first class in the unknown
5 x+ 4 y = 21
X+y = 5
In this case you should solve the equations together and there are two ways
First
Multiplying the second equation in 5 and ask them first equation
Become equations
5 x+5 y = 25
-
5 x+4 y = 21
=
P = 4
The compensation value of p in the second equation
Q 4 = 5
X = 4-5
X = 1
If you solve the equation x = 1 and y = 4
To make sure Awad in these values equations
The second way is the way of compensation
From the second equation
X+y = 5
x = 5 -y
And compensation in the first equation
5 (5-y) 4y = 21
25-5y 4y = 21
25-21 = 5y-4y
4 = y
Compensation in the second equation
X+y = 5
x+ 4 = 5
X = 1
Another example
5x+ 4y = 15
4x+ 3y = 2
Find the value of X & Y
Multiplying the first equation in 4 and hit the second equation in 5 put equations
And to ensure consistency with values of X.
20X+ 16y = 60
-
20x+ 15y = 10
=
y = 50
And compensation in the second equation
4x 3 * 50 = 2
4x + 150 = 2
4x = 2-150
4x = -148
Dividing 4
x = -37
And it can be solved other way but it will be difficult
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