Mathematics

Step I : Let the two linear equations obtained be
a₁x + b₁y + c₁ = 0 ....(1)
a₂x + b₂y + c₂ = 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 simultaneous linear equations
• Equation of the form ax + by + c and bx + ay = d where a and b are not equal
  
• Equation of the form a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂ and a₃x + b₃y + c₃z = d₃ 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(b² + a²) y = 4 (a² + b²) ⇒ y = 2
Multiplying (1) by a and (2) by b and adding, we get
2(a² + b²) x + a² + b² = 0
⇒ 2(a² + b²) x = – (a² + b²) ⇒ 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 =