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Methods of Solving Simultaneous Linear Equations

There are different methods for solving simultaneous linear Equations:

I. Elimination of a variable

II. Substitution

III. Cross-multiplication

IV. Evaluation of proportional value of variables 


This topic is purely based upon numerical examples. So, let us solve some examples based upon solving linear equations in two variables.

I: Solved example on simultaneous linear Equations using elimination method:

Solve for β€˜x’ and β€˜y’:

3x + 2y = 18.  

4x + 5y = 25.  

Solution: 

3x + 2y = 18 ............. (i)

4x + 5y = 25 ............. (ii)

Let us multiply equation (i) by 4 on both sides and equation (ii) by 3 on both sides, so as to make coefficients of β€˜x’ equal.

On multiplying, we get;

4(3x + 2y) = 4 βˆ™ 18

or, 12x + 8y = 72 ............. (iii)

and

3(4x + 5y) = 3 βˆ™ 25

or, 12x + 15y = 75 ............. (iv)

Subtracting (iii) from (iv), we get;

12x + 15y - (12x + 8y) = 75 - 72

or, 12x + 15y - 12x - 8y = 3

or, 7y = 3

or, y = 37.

Substituting value of β€˜y’ in equation (i), we get;

3x + 2(37) = 18.

or, 3x + 67 = 18

or, 3x = 18 – 67

or, 3x = 1207.

or, x = 12021.

or, x = 407.

Hence, x = 407 and y = 37.


II: Solved examples on simultaneous linear Equations using substitution method:

1.  Solve for β€˜x’ and β€˜y’:

            x + 3y = 9

           3x + 4y = 20

Solution:

x + 3y = 9 ............. (i)

3x + 4y = 20 ............. (ii)

Taking first equation in reference, i.e,

x + 3y = 9

x = 9 - 3y ............. (iii)

Substituting this value of β€˜x’ from previous equation in 2nd equation, we get;

3x + 4y = 20.

or, 3(9 - 3y) + 4y = 20

or, 27 - 9y + 4y = 20

or, -5y = 20 - 27

or, -5y = -7

or, 5y = 7

or, y = 75

Substituting this value of y into equation of x in (iii), we get;

x = 9 - 3y

or, x = 9 - 375

or, x = 9 – 215

or, x = 2475.

Hence, x = 2475 and y = 75.


2. Solve for β€˜x’ and β€˜y’; 

x + y = 5

4x + y = 10

Solution:

x + y = 5 ............. (i)

4x + y = 10 ............. (ii)

From equation (i),  we get value of y as:

y = 5 - x

substituting this value of y in equation (ii), weget

4x + (5 - x) = 10

or, 4x + 5 - x = 10

or, 3x = 10 - 5

or, 3x = 5

x = 53.

Substituting this value of x as 53 in equation y = 5 - x , we get;

y = 5 - 53

or, y = 103.

Hence, x = 53 and y = 103.


III. Solved example on simultaneous linear Equations using Cross-multiplication method:

Solve for β€˜x’ and β€˜y’;

3x + 5y - 25 = 0.

5x + 3y – 35 = 0.

Solution:

Assume two linear equations be

Ax + B1y + C= 0, and

A2x + B2y + C= 0.

The coefficients of x are: Aand  A2.

The coefficients of y are: B1 and B2.

The constant terms are: C1 and  C2.

To solve the equations in a simplified way, we use following table:

Simultaneous Linear Equations

xB1C2βˆ’B2C1=yC1A2βˆ’C2A1=1A1B2βˆ’A2B1

In the given equations,

The coefficients of x are 3 and 5.

The coefficients of y are 5 and 3.

The constant terms are -25 and -35.

On substituting the respective values, we get

x5Γ—(βˆ’35)βˆ’3Γ—(βˆ’25)=y(βˆ’25)Γ—5βˆ’(βˆ’35)Γ—3=13Γ—3βˆ’5Γ—5.

or, xβˆ’175+75=yβˆ’125+105=19βˆ’25.

or, xβˆ’100=yβˆ’20=1βˆ’16.

On equating x term with constant term, we get;

x = 254.

On equating y term with constant term, we get;

y = 54.


IV: Solved example on simultaneous linear Equations using evaluation method:

Solve for x and y:

2x + 3y = 4; 3x - 5y = -2

Solution:

The given equations are 

2x + 3y = 4 .......... (1)

3x - 5y = -2.......... (2)

Multiplying equation (2) by 2, we get

6x - 10y = -4.......... (3)

Now we add equations (1) and (3) we get

8x - 7y = 0

or, 8x = 7y

or, x7 = y8 (= k)

Substituting x = 7k and y = 8k in equation (1) we get

2 βˆ™ 7k + 3 βˆ™ 8k = 4

or, 14k + 24k = 4

or, 38k = 4

or, K = 438

or, k = 219

Therefore, x = 7 βˆ™  219 and y = 8 βˆ™ 219

Thus, x = 1419 and y = 1619





9th Grade Math

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