Step I : Let the **two linear equations** obtained be

a_{1}x + b_{1}y + c_{1} = 0 ....(1)

a_{2}x + b_{2}y + c_{2} = 0 ....(2)

Step II : Multiplying the given equation so as to make the co-efficients of the variable to be eliminated equal.

Step III : Add or subtract the equations so obtained in Step II, as the terms having the same co-efficients may be either of opposite or the same sign.

Step IV : Solve the equations in one varibale so obtained in Step III.

Step V : Substitute the value found in Step IV in any one of the given equations and then copmpute the value of the other variable.

**Solving Simultaneous Linear Equations in one variables****Solving a system of equations which is reducibe to a system of s****imultaneous linear equations****Equation of the form ax + by + c and bx + ay = d where a and b are not equal****Equation of the form a**_{1}x + b_{1}y + c_{1}z = d_{1}, a_{2}x + b_{2}y + c_{2}z = d_{2 }and a_{3}x + b_{3}y + c_{3}z = d_{3 }here a and b are not equal

**Solving Simultaneous Linear Equations in one variables**

Exampe: Solve the following system of linear equations by applying the method of elimination by equating the co-efficients :

(i) 4x – 3y = 4 and 2x + 4y (ii) 5x – 6y = 8 and 3x + 2y = 6

**Solution:** (i) We have the linear equations are

4x – 3y = 4 ....(1)

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

Let us decide to **eliminate** x from the given equation. Here, the co-efficients of x are 4 and 2 respectively. We find the L.C.M. of 4 and 2 is 4. Then, make the co-efficients of x equal to 4 in the two equations.

Multiplying equation (1) with 1 and equation (2) with 2, we get ;

4x – 3y = 4 ....(3)

4x + 8y = 6 ....(4)

Subtracting equation (4) from (3), we get ;

–11y = –2

⇒ y =

Substituting y = 2/11 in equation (1), we get;

4x – 3 × = 4

⇒ 4x – = 4

⇒ 4x = 4 +

⇒ 4x =

⇒ x =

Hence, solution of the given system of equation is :

x = , y =

(ii) We have;

5x – 6y = 8 ....(1)

3x + 2y = 6 ....(2)

Let us eliminate y from the given system of equations. The co-efficients of y in the given equations are 6 and 2 respectively. The L.C.M. of 6 and 2 is 6. We have to make the both coefficients equal to 6. So, multiplying both sides of equation (1) with 1 and equation (2) with 3, we get ;

5x – 6y = 8 ....(3)

9x + 6y = 18 ....(4)

Adding equation (3) and (4), we get ;

14x = 26 ⇒ x =

Putting x = 13/7 in equation (1), we get ;

5× – 6y = 8 ⇒ – 6y = 8

⇒ 6y = – 8 = =

⇒ y =

Hence, the solution of the system of equations is x = , y =.

Example: Solve the following system of equations by using the method of elimination by equating the co-efficients. + + 2 = 10; – + 1 = 9

**Solution:** The given system of equation is

+ + 2 = 10 ⇒ + = 8 ...(1) and

**– **+ 1- 9 ⇒ **– **= 8 ....(2)

The equation (1) can be expressed as : = 8 ⇒ 5x + 4y = 80 ....(3)

Similarly, the equation (2) can be expressed as : = 8 ⇒ 4x – 7y = 112 ....(4)

Now the new system of equations is

5x + 4y = 80 ....(5)

4x – 7y = 112 ....(6)

Now multiplying equation (5) by 4 and equation (6) by 5, we get ;

20x – 16y = 320 ....(7)

20x + 35y = 560 ....(8)

Subtracting equation (7) from (8), we get ;

y =

Putting y = in equation (5), we get ;

5x + 4 × = 80 ⇒ 5x – = 80

⇒ 5x = 80 + =

⇒ x =⇒ x =

Hence, the solution of the system of equations is, x = , y = .

**Solve the following system of linear equations by usnig the method of elimination by equating the coefficients : 3x + 4y = 25 ; 5x – 6y = – 9**

Sol. The given system of equations is

3x + 4y = 25 ....(1)

5x – 6y = – 9 ....(2)

Let us eliminate y. The coefficients of y are 4 and – 6. The LCM of 4 and 6 is 12. So, we make the coefficients of y as 12 and – 12.

Multiplying equation (1) by 3 and equation (2) by 2, we get

9x + 12y = 75 ....(3)

10x – 12y = – 18 ....(4)

Adding equation (3) and equation (4), we get

19x = 57 ⇒ x = 3.

Putting x = 3 in (1), we get,

3 × 3 + 4y = 25

⇒ 4y = 25 – 9 = 16 ⇒ y = 4

Hence, the solution is x = 3, y = 4.

**Verification :** Both the equations are satisfied by x = 3 and y = 4, which shows that the solution is correct.

**Ex.23 Solve the following system of equations : 15x + 4y = 61; 4x + 15y = 72 **

Sol. The given system of equation is

15x + 4y = 61 ....(1)

4x + 15y = 72 ....(2)

Let us eliminate y. The coefficients of y are 4 and 15. The L.C.M. of 4 and 15 is 60. So, we make the coefficients of y as 60. Multiplying (1) by 15 and (2) by 4, we get

225x + 60y = 915 ....(3)

16x + 60y = 288 ....(4)

Substracting (4) from (3), we get

209x = 627 ⇒ x = = 3

Putting x = 3 in (1), we get

15 × 3 + 4y = 61 ⇒ 45 + 4y = 61

⇒ 4y = 61 – 45 = 16 ⇒ y = = 4

Hence, the solution is x = 3, y = 4.

**Verification :** On putting x = 3 and y = 4 in the given equations, they are satisfied. Hence, the solution is correct.

**Ex.26 Solve the following system of linear equations : 2(ax – by) + (a + 4b) = 0 and 2(bx + ay) + (b – 4a) = 0**

Sol. 2ax – 2by + a + 4b = 0 .... (1)

2bx + 2ay + b – 4a = 0 .... (2)

Multiplyng (1) by b and (2) by a and subtracting, we get

2(b^{2} + a^{2}) y = 4 (a^{2} + b^{2}) ⇒ y = 2

Multiplying (1) by a and (2) by b and adding, we get

2(a^{2} + b^{2}) x + a^{2} + b^{2} = 0

⇒ 2(a^{2} + b^{2}) x = – (a^{2} + b^{2}) ⇒ x = –

Hence x = –, and y = 2

Ex.27 Solve (a – b) x + (a + b) y = a2 – 2ab – b2

(a + b) (x + y) = a2 + b2

Sol. The given system of equation is

(a – b) x + (a + b) y = a2 – 2ab – b2 ....(1)

(a + b) (x + y) = a2 + b2 ....(2)

(a + b) x + (a + b) y = a2 + b2 ....(3)

Subtracting equation (3) from equation (1), we get

(a – b) x – (a + b) x = (a2 – 2ab– b2) – (a2 + b2)

–2bx = – 2ab – 2b2

x = – = a + b

Putting the value of x in (1), we get

(a – b) (a + b) + (a + b) y = a2 – 2ab – b2

(a + b) y = a2 – 2ab – b2 – (a2 – b2)

(a + b) y = – 2ab

y =

Hence, the solution is x = a + b,

y =

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