ELIMINATION METHOD OF SIMULTANEOUS LINEAR EQUATIONS –
In this method, we eliminate one of the unknowns.
Step.1) Multiply the given equations by suitable numbers so as to make the coefficients of one of the unknowns, numerically equal.
Step.2) Add the new equations, if the numerically equal coefficients are opposite in sign, otherwise, subtract them.
Step.3) The resulting equations is linear in one unknown. Solve it to obtain the value of one of the unknowns.
Step.4) Substitute the value of this unknown in any of the given equations. Solve it to get the value of the other unknowns.
Example.1) 5x + 3y = 12, 2x + y = 4
Ans.) The given equations are –
5x + 3y = 12 ……………………(i)
2x + y = 4 ………………………(ii)
[ If you observe above two equations very carefully then you will find numeric of two equations are different in both x & y. Now we have to equal the numeric in any one in x or y, then we have to multiply 2 in (i) and multiply 5 in (ii) and we will get 10x in both the equations or multiply by 1 in equations (i) and multiply 3 in (ii) and we will get 3y in both the equations. Then we would add or subtract depend on their signs. ]
Now we have to Multiplying (i) by 2 and (ii) by 5, and we get the following equations –
10x + 6y = 24 ……………………(iii)
10x + 5y = 20 ………………….. (iv)
Subtracting (iv) from (iii), and we get –
10x + 6y = 24
10x + 5y = 20
---------------------
Y = 4
Substituting y = 4 in (ii) –
=> 2x + y = 4
=> 2x + 4 = 4
=> 2x = 0
=> x = 0
Hence, x = 0, and y = 4 is the solution of the given equations. (Ans.)
5 3
Example.2) Solve ------- + 6y = 13, ------- + 4y = 7
x x
Ans.) The equations are –
5
------- + 6y = 13 ………………….. (i)
x
3
------- + 4y = 7 …………….…….. (ii)
x
[ If you observe above two equations very carefully then you will find numeric of two equations are different in both x & y. Now we have to equal the numeric in any one in x or y, then we have to multiply 3 in (i) and multiply 5 in (ii) and we will get 15/x in both the equations or multiply by 2 in equations (i) and multiply 3 in (ii) and we will get 12y in both the equations. Then we would add or subtract depend on their signs. ]
Multiplying (i) by 3 and (ii) by 5, and we get -
15
------- + 18y = 39 ………………….. (iii)
x
15
------- + 20y = 35 ………………….. (iv)
x
subtracting (iv) from (iii), and we get –
=> - 2y = 4
=> y = - 2
Substituting y = - 2 in (i), and we get
5
=> ------- + 6y = 13
x
5
=> ------- + 6 ( -2) = 13
x
5
=> ------- - 12 = 13
x
5
=> ------- = 13 + 12 = 25
x
5 1
=> x = -------- = --------
25 5
Hence, x = 1/5, and y = - 2 is the solution of the given equations. (Ans.)
1 1 1 1
Example.3) Solve ------- + ------- = 5, ------- - ------- = 7
7x 6y 2x 3y
Ans.) The given equations are –
1 1
------- + ------- = 5 …………………….. (i)
7x 6y
1 1
------- - ------- = 7 …………………….. (ii)
2x 3y
1 1
Putting -------- = u, and -------- = v, and these equation becomes
x y
1 1
------- u + ------- v = 5
7 6
=> 6u + 7v = 210 ………………….. (iii)
1 1
------- u - ------- v = 7
2 3
=> 3u + 2v = 42 ………………..(iv)
[ If you observe above two equations very carefully then you will find numeric of two equations are different in both x & y. Now we have to equal the numeric in any one in x or y, then we have to multiply 1 in (iii) and multiply 2 in (iv) and we will get 6u in both the equations or multiply by 2 in equations (iii) and multiply 7 in (iv) and we will get 14v in both the equations. Then we would add or subtract depend on their signs. ]
Multiplying (iv) by 2 and (iii) by 1, we get –
6u + 7v = 210 …………………..(v)
6u + 4v = 84 ………………..(vi)
Subtracting (vi) from (v), and we get –
6u + 7v = 210
6u + 4v = 84
-------------------
3v = 126
=> v = 42
Now substituting v = 42 in (v), we get –
6u + (7 X 42) = 210
=> 6u = 210 – 294 = - 84
=> u = - 14
1
Hence, u = ------- = - 14
x
1
=> x = - ------
14
1
Hence, v = ------ = 42
y
1
=> y = -------
42
Hence, x = - (1/14) and y = (1/42) are the solutions of the given equations. (Ans.)
Example.4) Solve, 3x + 2y = 2xy, 6x + 2y = 3xy, where x ≠ 0, and y ≠ 0
Ans.) The given equations are –
3x + 2y = 2xy ……………………. (i)
6x + 2y = 3xy ……………………. (ii)
On dividing each equations by xy, we get –
3 2
------- + ------- = 2 ……………………(iii)
y x
6 2
------- + ------- = 3 ……………………(iv)
y x
1 1
Putting, u = ------, and v = -------, and we get -
x y
2u + 3v = 2 ………………... (v)
2u + 6v = 3 ………………….. (vi)
Subtract (v) from (vi), and we get –
2u + 6v = 3
2u + 3v = 2
----------------
3v = 1
=> v = 1/3
Substituting, v = 1/3 in (vi), we get –
2u + 6v = 3
1
=> 2u + 6 x ------ = 3
3
=> 2u = 3 – 2
=> u = 1/2
1 1
=> ------ = ------
x 2
=> x = 2
And, as we obtained here v = 1/3
1 1
=> ------- = ------
y 3
so, y = 3
hence, x =2, and y = 3 is the solution of the given equations. (Ans.)