**Solving Systems Of Equations By Elimination Method**

**Step I:** Let the two 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 coefficients 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.

**Elimination Method Examples**

**Example 1:** Solve the following system of linear equations by applying the method of elimination by equating the coefficients :

(i) 4x – 3y = 4 (ii) 5x – 6y = 8

2x + 4y = 3 3x + 2y = 6

**Sol.** **(i)** We have,

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 = 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 2:** 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 equation 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.

**Example 3:** 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.

**Example 4:** Solve the following system of equations by using the method of elimination by equating the co-efficients.

+ + 2 = 10; – + 1 = 9

**Sol.** The given system of equation is

+ + 2 = 10 ⇒ + = 8 …(1)

– + 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 = .

**Example 5:** Solve the following system of linear equatoins by using the method of elimination by equating the coefficients

√3x – √2y = √3 = ; √5x – √3y = √2

**Sol.** The given equations are

√3x – √2y = √3 ….(1)

√5x – √3y = √2 ….(2)

Let us eliminate y. To make the coefficients of equal, we multiply the equation (1) by √3 and equation (2) by √2 to get

3x – √6y = 3 ….(3)

√10x + √6y = 2 ….(4)

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

3x + √10x = 5 ⇒ (3 + √10) x = 5

Putting x = 5( √10– 3) in (1) we get

√3 × 5(√10 – 3) –√2 y = √3

⇒ 5√30 – 15√3 – √2y = √3

⇒ √2y = 5√30 – 15√3 – √3

⇒ √2y = 5√30 – 16√3

⇒

⇒ y = 5√15 – 8√6

Hence, the solution is x = 5( √10– 3) and y = 5√15 – 8√6

**Example 6: ** Solve for x and y :

– = a + b ; ax – by = 2ab

**Sol. ** The given system of equations is

– = a + b ….(1)

ax – by = 2ab ….(2)

Dividing (2) by a, we get

x – = 2b ….(3)

On subtracting (3) from (1), we get

– x = a – b ⇒ = a – b

⇒ x = = b ⇒ x = b

On substituting the value of x in (3), we get

b – = 2b ⇒ = 2b

⇒ 1 – = 2 ⇒ = 1 – 2

⇒ = –1 ⇒ y = –a

Hence, the solution of the equations is

x = b, y = – a

**Example 7: **Solve the following system of linear equations :

2(ax – by) + (a + 4b) = 0

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 = – 1/2

Hence x = –1/2, and y = 2

**Example 8: **Solve (a – b) x + (a + b) y = a^{2} – 2ab – b^{2}

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

**Sol.** The given system of equation is

(a – b) x + (a + b) y = a^{2} – 2ab – b^{2} ….(1)

(a + b) (x + y) = a^{2} + b^{2} ….(2)

⇒ (a + b) x + (a + b) y = a^{2} + b^{2} ….(3)

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

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

⇒ –2bx = – 2ab – 2b^{2}

⇒

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

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

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

⇒ (a + b) y = – 2ab

⇒ y =

Hence, the solution is x = a + b,

y =