## Tittle:- How to solve linear equations.

## Introduction:-

### Today in this article I will try to explain how to solve linear equations, hope this will help class(viii) to class (x) on going school students.

## Explanation:-

### To start this discussion let us first know,

## What is a linear equation?

### Answer:- An equation which has one or two variables and highest power of the variables are one and when plotted on the graph forms a straight line is called a linear equation. e.g; 3x +2y = 0 is a linear equation of two variables, where x, y are two variables.

again,ax + 5 = 0 is a linear equation of one / single variable of x.

The standard form of a linear equation is

### Now we will learn how to solve linear equations.

### Let us,

take a linear equation, 2x + 5 = x + 10

and we will solve it.

## Solve the following equations:-

## Question(1):-

### 2x + 5 = x + 10

Solution:- 2x + 5 = x + 10

=>2x - x = 10 -5

=> x = 5 solved.

## Rule(1):-

### (a) The variables are seperated in either side (here we have taken the variable 'x' in left hand side) along with numbers are seperated in right hand side and changing the signs of all applicable terms.

(b) Subtraction is made in both sides and we get x = 5

## Question (2):-

### ( 3x + 2) / 5 = (x + 1) / 2

=> 2(3x + 2) = 5( x + 1)[ using cross multiplication]

=> 6x + 4 = 5x + 5

=> 6x - 5x = 5 - 4

=> x = 1 solved

## Rule(2):-

### (a) Cross multiplication is applied in both sides.

(b) All terms in the bracket are multiplied by the specified number in both sides.

(c) Terms with variable are seperated in left hand side and numbers are seperated in right hand side.

(d) Subtraction is done in both sides and we get x= 1

## Question(3):-

### x/5 + ½ = x/2 - ⅙

=> x/5 - x/2 = - ⅙ - ½

=> [(10 ÷ 5)x - (10÷2)x] /10 = [(- 6 ÷ 6).1 - (6÷2).1]/6 [∵ L.C.M of 5,2 is 10 and that of 2,6 is 6]

=> (2x - 5x)/10 = ( - 1-3)/6

=> - 3x /10 = - 4/6

=> -3x/10 = - ⅔

=> 3x = 10.⅔ [!Removing '-' from both sides]

=>3x = 20/3

=> x = 20/(3.3)

=> x = 20/9 solved.

Rule(3):-

### (a) variable contain terms are seperated in left hand side.

(b) L.C.M of the denominators in both sides have been taken and then simplified

(c)Denominator of L.H.S is cross multiplied with R.H.S

(d) R.H.S number is divided by coefficient of L.H.S abd hence we get x = 20/9

# Important point:-

### When two linear equations (having two variables)combine together in a system, then they are called simultaneous linear equations.

e.g; 2x + y = 1 and x - y = 3 is a pair of simultaneous linear equations.

Now we will learn how to solve a pair simultaneous linear equations:-

A pair of simultaneous linear equations can be solved in the following methods:-

## Elimination Method:-

### In this method the variables x or y are removed first either by adding or subtracting two equations (if the coefficient of variable of equations are equal) or if variable variables are not equal then,we need to make them equal by taking L.C.M of variables and then multiplying them with suitable number.

## Solve the pair of linear equations by elimination method:-

### (a) x + y = 3; 2x - y = 0

Solution:-

x + y = 3…..(1)

2x - y = 0……(2)

—------------------------------------

(+) 3x = 3

=> x = 3/3

=> x = 1

Now putting x = 1 in equation (1)

we get,

1 + y = 3

=> y = 3 -1

=> y = 2

∴ x = 1 }

and y = 2 } Solved

## Rule(4):-

### (a) Two equations are written one after one vertically.

(b) Since, we have coefficient of y in both the equations are equal,i.e; 1,so we have added the equations.

(c)After addition we will find variable y will be cut.

(d) R.H.S number 3 has been divided by the coefficient of x and we have got x = 1

(d) Now putting the value of x in equation(1)(you can put in any equation either(1)or (2) as your choice) we have got y=2

(b) x + 2y =4; 2x -y = 3

Solution:-

x + 2y =4……(1)

2x -y = 3…….(2)

(1)×2 =>2x + 4y = 8

(2)×1 =>2x - y = 3

(-) ( +) (-)

—---------------------------------

(-) 5y = 5

=> y = 5/5

=> y = 1

Now,

putting the value of y in equation(1) weget,

x + 2.1 = 4

=> x + 2 = 4

=> x = 4 -2

=> x = 2

∴ x = 2}

y = 1} Solved

## Rule(5):-

### (a) Two equations are written one after one vertically

(b)∵ the coefficients of(neither x nor y) are equal, so will make coefficient of x or y equal.

(c) ∵ the L.C.M of coefficients of x (1,2)is 2 (here the L.C.M of coefficients of y (1,2) is also 2 )

(d)Equation(1)is multiplied by 2 and equation(2)is multiplied by 1 to make x's coefficient equal.(you may make y's coefficient equal also)

(e) Now subtracting we get y =1

(f) Now putting y = 1 in equation (1)( you may put in equation(2) also) we get x = 2

## Substitution Method:-

### In this method the value of one variable of one equation (which may be either x or y) is put in place of x or y in the other equation.

## Solve by substitution method:-

### x + y = 5 ; 2x - y =1

∵ x + y = 5 ; 2x - y =1….(2)

=> x = 5 - y…..(1)

Now putting the value of x in equation (2) we get,

2(5-y) - y = 1

=> 10 -2y -y =1

=> 10 - 3y = 1

=> - 3y = 1 - 10

=> - 3y = -9

=> y = - 9 / - 3

=> y = 3

Now,

from (1) => x = 5 - 3 [ ∵ putting y = 3]

### => x = 2

∴ x = 2]

y = 3] Solved

## Rule (6):-

### (a)We will find the value of variable x or y keeping x or y in L.H.S and all other terms will be transferred in R.H.S in any equation (1) or (2) [ Here, we have taken equation (1)]

(b) We hat put the value of x in equation (2)

(c) After simplifying and doing all necessary steps we have found y = 3

(d) Now putting y = 3 in equation (1) we have got x = 2

## Cross multiplication Method :-

### In this process the coefficients of x and y and constant term are placed in such a way that we can multiply them in a perticular manner to get the values of x and y.

## Sove by Cross multiplication Method:-

### 3x -y = 2; x + 2y = 3

solution:-

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

=> 3x - y - 2 = 0; =>x + 2y - 3 = 0

………..(1) ………….(2)

### x - y 1

=>---------- = —-------- = —------------

### (-1)(-3) - 2(-2) 3(-3)-1(-2) 3×2 -1(-1)

x - y 1

=>—------ = —------- = —-----------

3 +4 - 9 + 2 6+1

x - y 1

=> ----- = —------- = —---------

7 -7 7

x y 1

=> —-- = —------- = —---------

7 7 7

x 1 y 1

∴ —---- = —------- and —-- = —---

7 7 7 7

=> x = 7/7 ∴ y = 7/7

=> x =1 =>y = 1

∴ x = 1]

y = 1] Solved

## Rule(7) :-

### (a) Equations are made in standard form

(b) Putting x below it a table is made with the coefficient of y and constant terms and this process is done for y and constant term or all are arranged in the form of a matrix.

(c)Product of coefficients are made from upper to lower direction first, then lower to upper for x,y and constant term respectively.

(d) After doing these we have found, x/7 = y/7 = 1/7

(e) Then equating x/7 = 1/7, we have found x=1 similarly we also have got y =1

## Graphical Method :-

### In this process different values of x and y which form points are put in a cartesian plan or graph paper for both the equations, (generally three different sets of x and y's value for both equations).Then the points are joined, they form a straight line, for two equations there is formed two straight lines and they intersect each-other at a perticular point and this point is the solution of the given equations.

## Solve by graphical method:-

### 2x - y = 3; 3x + 2y - 1 = 0

Solution:-

∵ 2x - y = 3

=> y = 2x - 3……(1)

Let,

x = 1

from(1)=> y = 2×1 - 3 [ putting x = 1]

=> y = 2 - 3

=> y = - 1

So, when x = 1

y = - 1

Let,

x = 0

from (1) => y = 2×0 - 3 [ putting x = 0]

=> y = 0 -3

=> y = - 3

So, when x = 0

y = - 3

Let,

x = 4

from (1) => y = 2×4 -3 [ putting x = 4]

=> y = 8 - 3

=> y = 5

So, when x = 4

y = 5

∴ Three solutions of equation(1) are:-

x = 1 ] x = 0] x = 4]

y = - 1]; y = - 3]; y = 5]

Again,

3x + 2y - 1 = 0

=> 2y = - 3x + 1

=> y = (- 3x + 1)/ 2…….(2)

Let,

x = 1

from(2)=> y = (- 3×1 + 1)/ 2[ putting x = 1]

=> y = (- 3 + 1) /2

=> y = - 2 /2

=> y = -1

So, when x = 1

y = - 1

Let,

x = - 1

from(2)=> y = [- 3×(-1) + 1] / 2[ putting x = -1]

=> y = (3 +1) /2

=> y = 4 / 2

=> y = 2

So, when x = - 1

y = 2

Let,

x = 3

from(2)=> y = (- 3×3 + 1) / 2 [ putting x = 3]

=> y = (- 9 + 1) /2

=> y = - 8 /2

=> y = - 4

So, when x = 3

y = - 4

∴ Three solutions of equation(2) are:-

x = 1 ] x = -1] x = 3]

y = - 1]; y = 2]; y = - 4]

Now putting all solutions of equation (1) and (2) in the graph paper…

From the graph we get, point of intersection of the straight lines is P and the co -ordinate of point P = ( 1, - 1)

∴ x = 1 ]

y = -1] which is the solution of the given equations.

Description of the Graph :-

XOX⁻ —----> X –axis

YOY⁻ —-----> Y —axis

O(0,0) is their origin. AB straight line is the graph of equation (1) and CD straight line is the graph of equation (2). These two straight lines intersect each other at point 'P' So, 'P' is their point of intersection and co-ordinate of P = ( 1, -1)

∴ x = 1

and y = -1 solved

## Conclusion:-

### Two simultaneous linear equations of two variables can be solved using the above mentioned methods.......

## FREQUENTLY ASKED QUESTIONS:-
## What is the easiest way to solve linear equations?

Answer:- Elimination method.
In this method if co-efficient of one variable of both the equations are given equal,then simply we add or subtract the equations according to the sum demands and we get the value of other variable.

## What are the 4 methods of solving linear equations?

Answer:- The 4 methods of solving linear equations are :-
(a) Elimination method.
(b) Substitution method.
(c)Cross multiplication method.
(d) Graphical method.

## How to solve linear equations with two variables?

Answer:- The linear equations with two variables can be solved in any of the following given methods.The methods are:-
(a) Elimination method.
(b) Substitution method.
(c)Cross multiplication method.
(d) Graphical method.

## What is a standard linear equation?

Answer:- A standard linear equation is ,
ax + by + c = 0

##
What is the degree of a linear equation?

Answer:- The degree of a linear equation is one (1)

## What is the easiest way to solve linear equations?

Answer:- Elimination method. In this method if co-efficient of one variable of both the equations are given equal,then simply we add or subtract the equations according to the sum demands and we get the value of other variable.

## What are the 4 methods of solving linear equations?

Answer:- The 4 methods of solving linear equations are :- (a) Elimination method. (b) Substitution method. (c)Cross multiplication method. (d) Graphical method.

## How to solve linear equations with two variables?

Answer:- The linear equations with two variables can be solved in any of the following given methods.The methods are:- (a) Elimination method. (b) Substitution method. (c)Cross multiplication method. (d) Graphical method.

## What is a standard linear equation?

Answer:- A standard linear equation is , ax + by + c = 0

## What is the degree of a linear equation?

Answer:- The degree of a linear equation is one (1)

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