1/6x + 1/16y = 6; 1/12y - 1/9x = 2 find the value of x and y
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1/6x + 1/16y = 6; 1/12y - 1/9x = 2 find the value of x and y
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Solution :
From the above equation, we get that 1/x and 1/y are common in both the equations , thus we can represent them by a single variable i.e,
Now by substituting the value 1/x and 1/y in the equation, we get :
By solving Eq.(i) , we get :
[tex]:\implies \sf{\dfrac{a}{6} + \dfrac{b}{16} = 6} \\ \\ \\ [/tex]
[tex]:\implies \sf{\dfrac{8a + 3b}{48} = 6} \\ \\ \\ [/tex]
[tex]:\implies \sf{8a + 3b = 6 \times 48} \\ \\ \\ [/tex]
[tex]:\implies \sf{8a + 3b = 288} \\ \\ \\ [/tex]
[tex]\boxed{\therefore \sf{8a + 3b = 288}}[/tex]⠀⠀⠀⠀⠀⠀Eq.(iii)
By solving Eq.(ii) , we get :
[tex]:\implies \sf{\dfrac{b}{12} - \dfrac{a}{9} = 2} \\ \\ \\ [/tex]
[tex]:\implies \sf{\dfrac{3a - 4b}{36} = 2} \\ \\ \\ [/tex]
[tex]:\implies \sf{3b - 4a = 2 \times 36} \\ \\ \\ [/tex]
[tex]:\implies \sf{3b - 4a = 72} \\ \\ \\ [/tex]
[tex]\boxed{\therefore \sf{-4a + 3b = 72}}[/tex]⠀⠀⠀⠀⠀⠀Eq.(iv)
By subtracting Eq.(iv) from Eq.(iii), we get :
[tex]:\implies \sf{(8a + 3b) - (-4a + 3b) = 288 - 72} \\ \\ \\ [/tex]
[tex]:\implies \sf{8a + 3b + 4a - 3b = 288 - 72} \\ \\ \\ [/tex]
[tex]:\implies \sf{8a + 4a = 288 - 72} \\ \\ \\ [/tex]
[tex]:\implies \sf{12a = 216} \\ \\ \\ [/tex]
[tex]:\implies \sf{a = \dfrac{216}{12}} \\ \\ \\ [/tex]
[tex]:\implies \sf{a = 18} \\ \\ \\ [/tex]
[tex]\boxed{\therefore \sf{a = 18}} \\ \\ \\ [/tex]
Hence the value of a is 18.
By substituting the value of a in Eq.(iii), we get :
[tex]:\implies \sf{8a + 3b = 288} \\ \\ \\ [/tex]
[tex]:\implies \sf{8(18) + 3b = 288} \\ \\ \\ [/tex]
[tex]:\implies \sf{144 + 3b = 288} \\ \\ \\ [/tex]
[tex]:\implies \sf{3b = 288 - 144} \\ \\ \\ [/tex]
[tex]:\implies \sf{3b = 144} \\ \\ \\ [/tex]
[tex]:\implies \sf{b = \dfrac{144}{3}} \\ \\ \\ [/tex]
[tex]:\implies \sf{b = 48} \\ \\ \\ [/tex]
[tex]\boxed{\therefore \sf{b = 48}} \\ \\ \\ [/tex]
Hence the value of b is 48.
But we have taken the value of 1/x as a and the value of 1/y as b.
So,
By solving the above equation, we get :
[tex]:\implies \bf{\dfrac{1}{x} = 18} \\ \\ [/tex]
[tex]:\implies \bf{\dfrac{1}{18} = x} \\ \\ [/tex]
[tex]\boxed{\therefore \sf{x = \dfrac{1}{18}}} \\ \\ \\ [/tex]
Hence the value of x is 1/18.
By solving the above equation, we get :
[tex]:\implies \bf{\dfrac{1}{y} = 48} \\ \\ [/tex]
[tex]:\implies \bf{\dfrac{1}{48} = y} \\ \\ [/tex]
[tex]\boxed{\therefore \sf{y = \dfrac{1}{48}}} \\ \\ \\ [/tex]
Hence the value of y is 1/48.