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Solve the following simultaneous linear equation 22 ÷ x +y + 15÷x-y=5,55÷x+y +40÷x-y = 13 |
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Answer» x=8, y=3Step-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 120Therefore, 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. 2we get,22/11 + 15/x-y =5---------> 2 + 15/x-y =5--------> 15/x-y =3--------> 15=3x-3yOn dividing the whole eq. of 15=3x-3y by 3, we get 5=x-y[6]now ADDING [5] and [6]11=x+y5=x-y______ [no CHANGE in signs, as we're adding]16=2x------> x=8substituting the value of 'x' in eq. 6, we get5=8-y-----> 5-8 = -y-----> -3 = -y ------> 3=yHope this answers your query. |
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