Solve the following simultaneous linear equation 22 ÷ x +y
+ 15÷x-y=5,55÷x+y +40÷x-y = 13
Answers
Answer:
x=8, y=3
Step-by-step explanation:
22/x+y + 15/x-y = 5 [1]
55/x+y + 40/x-y = 13 [2]
LCM of 40 and 15 is 120
Therefore, multiplying [1] by 8 and [2] by 3, we get
[1] 8 x (22/x+y + 15/x-y =5]-------> 176/x+y + 120/x-y = 140 [3]
[2] 3 x (55/x+y + 40/x-y =13]------>165/x+y + 120/x+y = 139 [4]
Now, Subtracting [4] from [3], we get.
176/x+y + 120/x-y = 140
165/x+y + 120/x-y = 139
- - - [Changing the signs, as we're subtracting]
____________________
11/x+y = 1 ----------------> 11=x + y [5]
now, substituting the value of (x+y) in eq. 2
we get,
22/11 + 15/x-y =5---------> 2 + 15/x-y =5--------> 15/x-y =3--------> 15=3x-3y
On dividing the whole eq. of 15=3x-3y by 3, we get 5=x-y[6]
now adding [5] and [6]
11=x+y
5=x-y
______ [no change in signs, as we're adding]
16=2x------> x=8
substituting the value of 'x' in eq. 6, we get
5=8-y-----> 5-8 = -y-----> -3 = -y ------> 3=y
Hope this answers your query.