Solving Systems Of Equations By Elimination Method
Step I: Let the two 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 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 = (frac { 2 }{ 11 })
Substituting y = 2/11 in equation (1), we get
⇒ 4x – 3 × (frac { 2 }{ 11 }) = 4
⇒ 4x – (frac { 6 }{ 11 }) = 4
⇒ 4x = 4 + (frac { 6 }{ 11 })
⇒ 4x = (frac { 50 }{ 11 })
⇒ x = (frac { 50 }{ 44 }) = (frac { 25 }{ 22 })
Hence, solution of the given system of equation is :
x = (frac { 25 }{ 22 }), y = (frac { 2 }{ 11 })
(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 = (frac { 13 }{ 7 })
Putting x = (frac { 13 }{ 7 }) in equation (1), we get
5 × (frac { 13 }{ 7 }) – 6y = 8 ⇒ (frac { 65 }{ 7 }) – 6y = 8
⇒ 6y = (frac { 65 }{ 7 }) – 8 = (frac { 65-56 }{ 7 }) = (frac { 9 }{ 7 })
⇒ y = (frac { 9 }{ 42 }) = (frac { 3 }{ 14 })
Hence, the solution of the system of equations is x = (frac { 13 }{ 7 }) , y = (frac { 3 }{ 14 })
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.
(frac { x }{ y }) + (frac { 2y }{ 5 }) + 2 = 10; (frac { 2x }{ 7 }) – (frac { 5 }{ 2 }) + 1 = 9
Sol. The given system of equation is
(frac { x }{ y }) + (frac { 2y }{ 5 }) + 2 = 10 ⇒ (frac { x }{ y }) + (frac { 2y }{ 5 }) = 8 …(1)
(frac { 2x }{ 7 }) – (frac { 5 }{ 2 }) + 1 = 9 ⇒ (frac { 2x }{ 7 }) – (frac { 5 }{ 2 }) = 8 ….(2)
The equation (1) can be expressed as :
(frac { 5x+4y }{ 10 }) = 8 ⇒ 5x + 4y = 80 ….(3)
Similarly, the equation (2) can be expressed as :
(frac { 4x-7y }{ 14 }) = 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 = (-frac { 240 }{ 51 })
Putting y = (-frac { 240 }{ 51 }) in equation (5), we get ;
5x + 4 × (frac { -240 }{ 51 }) = 80 ⇒ 5x – (frac { 960 }{ 51 }) = 80
⇒ 5x = 80 + (frac { 960 }{ 51 }) = (frac { 4080+960 }{ 51 }) = (frac { 5040 }{ 51 })
⇒ x = (frac { 5040 }{ 255 }) = (frac { 1008 }{ 51 })= (frac { 336 }{ 17 }) ⇒ x = (frac { 336 }{ 17 })
Hence, the solution of the system of equations is, x = (frac { 336 }{ 17 }), y = (-frac { 80 }{ 17 }) .
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
( Rightarrow text{x}=frac{5}{3+sqrt{10}}=left( frac{5}{sqrt{10}+3} right)times left( frac{sqrt{10}-3}{sqrt{10}-3} right))
(=5left( sqrt{10}-3 right))
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=frac{5sqrt{30}}{sqrt{2}}-frac{16sqrt{3}}{sqrt{2}})
⇒ 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 :
(frac { ax }{ b }) – (frac { by }{ a }) = a + b ; ax – by = 2ab
Sol. The given system of equations is
(frac { ax }{ b }) – (frac { by }{ a }) = a + b ….(1)
ax – by = 2ab ….(2)
Dividing (2) by a, we get
x – (frac { by }{ a }) = 2b ….(3)
On subtracting (3) from (1), we get
(frac { ax }{ b }) – x = a – b ⇒ (xleft( frac{a}{b}-1 right)) = a – b
⇒ x = (frac{(a-b)b}{a-b}) = b ⇒ x = b
On substituting the value of x in (3), we get
b – (frac { by }{ a }) = 2b ⇒ (bleft( 1-frac{y}{a} right)) = 2b
⇒ 1 – (frac { y }{ a }) = 2 ⇒ (frac { y }{ a }) = 1 – 2
⇒ (frac { y }{ a }) = –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(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 = – 1/2
Hence x = –1/2, and y = 2
Example 8: 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
⇒ (text{x}=frac{-2ab}{-2b}-frac{2}{-2b}=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 = (frac { -2ab }{ a+b })
Hence, the solution is x = a + b,
y = (frac { -2ab }{ a+b })
