Q. Solve the following pair of linear equations by the elimination method
(and the substitution method)
Soln.
Elimination method
Multiplying eqn (i) by 2, we get
–
y =
+
–
… (iii)
Subtracting eqn (ii) from eqn (iii)
(i) x + y - 5 = 0 and 2x – 3y -4 = 0
x
+
y
=
5
–
=
4
2x
+
2y
=
10
2x
–
3y
=
4
5y
=
6
6
5
Substituting y = in (i)
6
5
... (i)
... (ii)
x
+
6
5
=
5
Solution is
x
=
19
5
y =
6
5
1
1
2
2x
3
3y
... (ii)
–
=
4
2x
3y
+
=
10
2x
2y
+
+
Multiplying throughout by 5, we get
To remove 5 from denominator multiplying throughout by 5
5x
+
6
=
25
5x
=
25
-
6
5x
=
19
x
=
19
5
Constant → R.H.S
6 → -6 to R.H.S
… (iii)
... (ii)
In eliminition method, keep variable on L.H.S and constant on R.H.S
Check the coefficient of the variables
Are the coefficient same ?
No
To make the coefficient of x same we will have to multiplying equation (i) by 2
To make the coefficient of y same we will have to multiplying equation (i) by 3
OR
Now check the coefficient of the variables
Coefficient of x is same
and
Coefficient of y is different
We can eliminate x by subtracting
As Signs are same