Question

1/6x + 1/16y = 6; 1/12y - 1/9x = 2 find the value of x and y​

Answer 1

Solution :

• $\sf{\dfrac{1}{6x} + \dfrac{1}{16y} = 6}$

• $\sf{\dfrac{1}{12y} - \dfrac{1}{9x} = 2}$

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,

• The value of 1/x = a
• The value of 1/y = b

Now by substituting the value 1/x and 1/y in the equation, we get :

• $\sf{\dfrac{a}{6} + \dfrac{b}{16} = 6}$⠀⠀⠀⠀⠀⠀Eq.(i)

• $\sf{\dfrac{b}{12} - \dfrac{a}{9} = 2}$⠀⠀⠀⠀⠀⠀Eq.(ii)

By solving Eq.(i) , we get :

$:\implies \sf{\dfrac{a}{6} + \dfrac{b}{16} = 6}$

$:\implies \sf{\dfrac{8a + 3b}{48} = 6}$

$:\implies \sf{8a + 3b = 6 \times 48}$

$:\implies \sf{8a + 3b = 288}$

$\boxed{\therefore \sf{8a + 3b = 288}}$⠀⠀⠀⠀⠀⠀Eq.(iii)

By solving Eq.(ii) , we get :

$:\implies \sf{\dfrac{b}{12} - \dfrac{a}{9} = 2}$

$:\implies \sf{\dfrac{3a - 4b}{36} = 2}$

$:\implies \sf{3b - 4a = 2 \times 36}$

$:\implies \sf{3b - 4a = 72}$

$\boxed{\therefore \sf{-4a + 3b = 72}}$⠀⠀⠀⠀⠀⠀Eq.(iv)

By subtracting Eq.(iv) from Eq.(iii), we get :

$:\implies \sf{(8a + 3b) - (-4a + 3b) = 288 - 72}$

$:\implies \sf{8a + 3b + 4a - 3b = 288 - 72}$

$:\implies \sf{8a + 4a = 288 - 72}$

$:\implies \sf{12a = 216}$

$:\implies \sf{a = \dfrac{216}{12}}$

$:\implies \sf{a = 18}$

$\boxed{\therefore \sf{a = 18}}$

Hence the value of a is 18.

By substituting the value of a in Eq.(iii), we get :

$:\implies \sf{8a + 3b = 288}$

$:\implies \sf{8(18) + 3b = 288}$

$:\implies \sf{144 + 3b = 288}$

$:\implies \sf{3b = 288 - 144}$

$:\implies \sf{3b = 144}$

$:\implies \sf{b = \dfrac{144}{3}}$

$:\implies \sf{b = 48}$

$\boxed{\therefore \sf{b = 48}}$

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,

• 1/x = a = 18

By solving the above equation, we get :

$:\implies \bf{\dfrac{1}{x} = 18}$

$:\implies \bf{\dfrac{1}{18} = x}$

$\boxed{\therefore \sf{x = \dfrac{1}{18}}}$

Hence the value of x is 1/18.

• 1/y = b = 48

By solving the above equation, we get :

$:\implies \bf{\dfrac{1}{y} = 48}$

$:\implies \bf{\dfrac{1}{48} = y}$

$\boxed{\therefore \sf{y = \dfrac{1}{48}}}$

Hence the value of y is 1/48.