SOLVED DIFFERENT PROBLEMS USING ELIMINATION METHOD
Solved Different Problems Using Elimination Method :
In this section, we will learn, how to solve linear equations using elimination method.
Question 1 :
Solve the pair of linear equations using elimination method.
10/(x + y) + 2/(x –y) = 4 and 15/(x + y) – 5/(x – y) = -2
Solution :
1/(x + y) = a 1/(x – y) = b
10 a + 2 b = 4 --------(1)
15 a – 5 b = -2 --------(2)
(1) ⋅ 5 => 50 a + 10 b = 20
(2) ⋅ 2 => 30 a – 10 b = -4
-------------------------
80 a = 16
a = 16/80
a = 1/5
Substitute a = 1/5 in the first equation
10(1/5) + 2b = 4
2 + 2b = 4
2b = 2
b = 1
|
1/(x + y) = 1/5 5 = x + y x + y = 5 -------(3) |
1/(x – y) = 1 x – y = 1 -------(4) |
(3) + (4)
x + y = 5
x – y = 1
------------
2x = 6
x = 3
By applying the value of x in (3), we get
3 + y = 5
y = 2
Question 2 :
Solve the pair of linear equations using elimination method.
1/(3x + y) + 1/(3x – y) = 3/4 and 1/2(3x + y) - 1/2(3x – y) = -1/8
Solution :
1/(3x + y) = a and 1/(3x – y) = b
a + b = 3/4
4(a + b) = 3
4 a + 4 b = 3 ------(1)
(a/2) – (b/2) = -1/8
4a – 4b = -1------(2)
4a + 4b = 3
4a – 4b = 1
------------------
8 a = 4 ==> a = 1/2
By applying the value of a in (1), we get
4(1/2) + 4b = 3
2 + 4b = 3
4 b = 1
b = 1/4
|
1/(3 x + y) = 1/2 2 = 3 x + y 3 x + y = 2-------(3) |
1/(3 x – y) = 3/4 4 = 3 x – y 3 x – y = 4 --------(4) |
3 x + y = 2
3 x – y = 4
--------------
6 x = 6
x = 1
By applying x = 1 in the (3) equation, we get
3(1) + y = 2
y = 2 – 3
y = -1
After having gone through the stuff and examples, we hope that the students would have understood, how to solve linear equations using elimination method.
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