LINEAR EQUATIONS math capsule

Do you want to learn something about linear equations and how to play with them?
Here is your guide to it . 

Tips : grab a cup of coffee and enjoy this long article .

so here we begin ,

EQUATION

An equation is a statement of equality of two algebraic polynomial(monomials) involving one or more variable.
or
An equation is a statement in which two algebraic expressions are equal.
e.g. 7x+5 = 0 is an equation but only 7x+5 is an expression.


LINEAR EQUATIONS IN ONE VARIABLE

The expression of the form Ax+B , where A and B are real numbers and A≄0, is a linear polynomial and equation involving only linear polynomial are called as linear equations.
e.g. 
(1) 5x+8 = 9 - x
(2) 2y/3 + 7 = y/2
(3) t + 3t = 9 - t
All the above linear equations in one variable.

*Graph of linear equations in one variable.
*Linear equations involving only one variable is called a linear equation in that variable.
*x=0 is the equation of Y-axis and y=0 is the equation of X-axis.


I think you are a good learner , even i could not learn this much at one go! Nice buddy , keep learning.


RULES FOR SOLVING A LINEAR EQUATION

1. If same number is added to both the sides of an equation, the equality remains same.

2. If same number is subtracted to both the sides of an equation, the equality remains the same.

3. If same number is multiplied both the sides of the equation the equality remains the same.

4. If both sides are divided by a some non zero number the equality remains the same.

5. Any term of an equation can be taken to the other side with its sign changed, without affecting the equality.

ELIMINATION METHOD-

In this method, one variable is eliminated from both equations to have an equations in one variable, so it is called elimination method.

STEPS USED IN THIS METHOD ARE GIVEN BELOW
STEP 1
Firstly, make the coefficient of one variable(x or y) numerically equal by multiplying both equations by some suitable non zero constant.
STEP 2
Now, add or subtract both equations, so that one variable is eliminated and remained equation has one variable.
STEP 3
Solve the equation in one variable to get the value of this variable(x or y).
STEP 4
Substitute the value(x or y) in any one of the given equations to get the value of other variable.


Q. The solutions of the given system of equations 2x + 5y = 11 and 3x + 4y = 13
A. given equation are ,
2x + 5y = 11 .......(i)         and 3x + 4y = 13.........(ii)
on multiplying eq. (i) by 3 and eq. (ii) by 2 we get
6x + 15y = 33 .......(iii)  and  6x + 8y = 26 ........ (iv)

on subtracting eq (iv) from eq (iii) we get 
(6x +15y)-(6x + 8y) = 33-26
7y = 7
So, y=1
 putting y = 1 in eq (i)
2x +5 * 1 = 11
on solving this we have x=3
 solution of system of equations is (3 , 1).


Hussshhh ! my coffee cup is empty now , you study I'm going to fill it up .

METHOD OF COMPARISON-

* In this method , using both the equations , find the value of one of the variables(same in both case) in terms of the other.
* Equate the results and solve the resulting equation.
* Substitute the values in either of the results obtained in first step an find the value of other variable.

Q. The system of equations 3x - 5y = 1 and 4x + 3y = 11 has the solution
A. As, 3x - 5y =1 => 3x= 1+5y => x =( 1+5y )/ 3
also 4x + 3y =11 => 4x = 11 - 3y
=> x = (11 - 3y) / 4

equating both values of x, we get 
(1 + 5y) / 3  = (11 - 3y) / 4
on solving this we have y = 1 
also
x = (1 + 5y) / 3 => putting y =1, we have , x = 2.
So, (2 , 1) is the required solution.


CROSS MULTIPLICATION METHOD-

Let a general system of two simultaneous linear equations be
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
where , a₁ ≉ b₁ has a unique solution given by,
             a₂    b₂ 

x = (b₁ * c₂ - b₂ *c₁ ) / (a₁ * b₂ - a₂ * b₁)         and    y = (c₁* a₁ - c₂ * a₁) / (a₁ * b₂ - a₂ * b₁ )

REMEMBERING TECHNIQUE (CROSS-MULTIPLICATION)

The following diagram helps in remembering the above solution.
remembering technique (cross multiplication)

In the above pattern the arrow (→) between the two number b₁ and c₂ indicates that the two numbers are to be multiplied are arrow (→) between the two numbers b₁ and c₁ indicates that the two numbers are to be multiplied, but this product is to be subtracted from the first product.



Okay! How can you learn this fast ? Might be i can give some pat to myself for being a good instructor !

EQUATION REDUCIBLE TO A PAIR OF LINEAR EQUATIONS-

Some of the situations are given below ,
(1) If equation are in the form of 1/x , 1/y , then put 1/x = p, 1/y = q to convert it linear form. After solving , put the value of p and q in above substitutions, to get the values of x and y .

(2) If equations are in the form 1/ x±a = p, 1/ y±b = q , then put 1 / x±a = p , 1 / y±b = q to convert it into linear form . After solving put the values of p and q as in above substitutions to get the values of x and y.


oh! Its high time to see some applications on linear equations .......

APPLICATION OF LINEAR EQUATION-

PROBLEMS BASED ON AGES----------------
If the problem involves finding out the ages of two persons, take the present age of one person as x and of the other as y .
Then, 'a' year ago , age of Ist person was (x - a) year and that of 2nd person , was (y - a) year.
After b year, age of Ist person will be (x + b) year and the of 2nd person will be (y + b) year.
Formulate the equations and then solve them.

Q. Age is X is six times that of Y. After 4 yr, X is 4 times elder to to Y .What is the present age of Y?
A. let the age of X and Y yr respectively.
condition 1

x=6y...........(i)
after 4 yr,  age of X=(x+4) yr
after 4 yr, age of Y =(y+4) yr
condition 2

(x + 4)=(y + 4)*4
x+4= 4y +16...........(ii)
put the value of x = 6yin eq (ii) we get
=> 6y+4 = 4y+16
=> y = 6
so the present age of Y is 6 yr.

PROBLEMS BASED ON NUMBERS AND DIGITS--------------

Let the digit in unit's place be x and that in ten's place be y. Then the two digits number is given by 10y+x . On interchanging the positions of the digits, the digit in unit is place becomes y and in ten's place becomes x, it becomes 10x+y. Formulate the equations and then solve them.

PROBLEMS BASED ON FRACTIONS -----------------

Let the numerator of the fraction be x and denominator be y , then the fraction is x / y.
Formulate the linear equations on the basis of conditions given and solve for x and y to get the value o the fraction.

PROBLEM BASED ON DISTANCE, SPEED AND TIME--------------

Useful formulae to be used for problem is 

Speed = Distance / Time

To solve problem related to speed of motor boat going downstream and upstream , let the speed of boat in still water be x km/h and speed of stream be y km/h.

 Then the speed of boat downstream = (x+y) km/h
and speed of boat downstream = (x - y) km/h

PROBLEMS BASED ON COMMERCIAL MATHEMATICS----------

The fare of 1 full ticket may be taken as rs x and the reservation charge may be taken as rs y , so that 1 full fare = x + y and 1 half fare = x/2 + y
Formulate the equations and then solve them.


Informative article , Isn't it buddy ?
So after this great blog post , i think you should do some exercises related to this blog post!
Hard ?
NO
These are very simple for you , believe me !
I have attached some images of solutions too , but don't cheat , Okay ?  

 
linear equations 1
linear equations 1

linear equations 2
linear equations 2

linear equations 3
linear equations 3

linear equations 4
linear equations 4

linear equations 5
linear equations 5

linear equations 6
linear equations 6 
linear equations 7
linear equations 7 
linear equations 8
linear equations 8

linear equations 9
linear equations 9

linear equations 10
linear equations 10

linear equations 11
linear equations 11


THANK YOU FOR READING THIS AT Math Capsule

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