Solve the Equation in the Real Number System

Introduction of solve the equation in the real number system:

Positive or negative, small or large, whole or decimal numbers are called real numbers.  Real number systems are represented by R.  Rational and irrational numbers are also called real numbers. Rational numbers are referred to as an integer or fraction. An irrational numbers cannot be expressed as a rational numbers. Example for irrational numbers are pi(3.14) and sqrt2. A real number system is a set of numbers that contains more operations such as addition, subtraction,  multiplication and so on. I like to share this Partial Differential Equation with you all through my article.

Example Problems – for Solve the Equation in the Real Number System

Example problem 1 - for solve the equation in the real number system

Solve the equation and find the value of x and y.

2x + y = 2

3x + y = 4

Solution:

2x + y = 2     -----------   1

3x + y = 4     -----------    2

2x + y = 2

3x + y = 4

_________

-x = -2

x = 2

substitute x = 2 in 1st equation

2(2) + y = 2

4 + y = 2

y  = -2

Answer: x=2, y= -2.

Example problem2 - for solve the equation in real number system

`sqrt(2)` is an irrational number. consider the equation of` x^2` +6x+1=0

` x^2` +6x+1=0

Solution:

-b`+-(sqrt(b^2 - 4ac))/(2a)`

-6 `+-` `(sqrt(6^2-4(1)(1)))/(2(1))`

-6 `+-` `(sqrt(36-4))/(2)`

-6 `+-` `(sqrt(32))/(2)`

-6 `+-` `(4sqrt(2))/(2)`

-6 /2 `+-` `(4sqrt(2))/(2)`

-3 `+-` `2sqrt(2)`

-3+`2sqrt(2)` , -3-`2sqrt(2)`

Both solutions are irrational numbers.

Example Problem3– for Solve the Equation in the Real Number System

Solve the equation and find the value of x, y and z.

2x + 4y + 4z = 0.

x + y + z = 6

x + 2y + 3z = 2

Solution:

2x + 4y + 4z = 0    -------------   1

x + y + z = 6          -------------    2

x + 2y + 3z = 2      -------------    3

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Take 2nd and 3rd equation and solve the problem

x + y +  z  = 6

x + 2y + 3z = 2

_______________

-y -2z = 4   -------------   4

_______________

Take 1st and 2nd equation

x + y + z =  6    ----------  Multiply 2 in 2nd equation

x = 12 Then, we get 2x + 2y + 2z  = 12.

2x + 4y + 4z = 0

2x + 2y + 2z = 12

________________

2Y + 2z = -12    ------------  5

________________

-y -2z = 4

2y + 2z = -12

_____________

Y = -8

_____________

Substitute y= -8 in 5th equation.

2(-8) + 2z = -12

-16 + 2z = -12

2z = -12 + 16

2z = 4

Z = 2

Substitute y=-8 and z=2 in 2nd equation

x + y +z = 6

x + (-8) +2 = 6

x -8 + 2 = 6

x – 6 = 6

x = 12

Practicing Problem - for Solve the Equation in the Real Number System

Practicing problem – for solve the equation in the real number system

Solve the equation and find the value of a, b, c.

2a + b = 8

2a + 2b =  4

Answer:  a = 6 , b = -4.