Solving Systems Of Equations

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
\( \Rightarrow \text{x}=\frac{5}{3+\sqrt{10}}=\left( \frac{5}{\sqrt{10}+3} \right)\times \left( \frac{\sqrt{10}-3}{\sqrt{10}-3} \right)\)
\(=5\left( \sqrt{10}-3 \right)\)
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
⇒ \(y=\frac{5\sqrt{30}}{\sqrt{2}}-\frac{16\sqrt{3}}{\sqrt{2}}\)
⇒ 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    ⇒    \(x\left( \frac{a}{b}-1 \right)\) = 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  ⇒ \(b\left( 1-\frac{y}{a} \right)\) = 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
⇒  \(\text{x}=\frac{-2ab}{-2b}-\frac{2{{b}^{2}}}{-2b}=a+b\)
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 = \(\frac { -2ab }{ a+b }\)