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Solve : 2x+23y=16\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6} and 3x+2y=0\dfrac{3}{x} + \dfrac{2}{y} = 0.

Hence, find 'm' for which y = mx - 4.

Linear Equations

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Answer

Lets take 1x\dfrac{1}{x} = u and 1y\dfrac{1}{y} = v. The given equation becomes,

2u + 23\dfrac{2}{3} v = 16\dfrac{1}{6} ⇒ 12u + 4v = 1

And, 3u + 2v = 0

Multiply second equation by 4 and subtract from first equation, we get

(3u + 2v = 0) x 4

12u+4v=112u+8v=04v=104v=1\begin{matrix} & 12u & + & 4v & = & 1 \ & 12u & + & 8v & = & 0 \ & - & - & & & - \ \hline & & - & 4v & = & 1 - 0 \ \Rightarrow & & - & 4v & = & 1 \end{matrix}

⇒ v = -14\dfrac{1}{4}

Substituting the value of v in equation (2), we get:

⇒ 3u + 2 ×(14)\times (-\dfrac{1}{4}) = 0

⇒ 3u - 12\dfrac{1}{2} = 0

⇒ 3u = 12\dfrac{1}{2}

⇒ u = 16\dfrac{1}{6}

So, x = 1u\dfrac{1}{u} = 6 and y = 1v\dfrac{1}{v} = -4

Substituting the value of x and y in y = mx - 4, we get:

⇒ -4 = m ×\times 6 - 4

⇒ -4 + 4 = 6m

⇒ 0 = 6m

⇒ m = 06\dfrac{0}{6}

⇒ m = 0

Hence, the value of x = 6 , y = -4 and m = 0.

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