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Question Number 74371 by liki last updated on 23/Nov/19

Commented by liki last updated on 23/Nov/19

$. . . . p l z h e l p m e 16 a,, e m e r g e r c e$
	Answered by MJS last updated on 23/Nov/19

	$k y = k x + x + 7$ $y = x + \frac{x}{k} + \frac{7}{k}$ $y = x (1 + \frac{1}{k}) + \frac{7}{k}$ $1 + \frac{1}{k} = 2 \Rightarrow k = 1$ $\Rightarrow y = 2 x + 7$
	Commented by liki last updated on 23/Nov/19

	$. . . T h a n k s s i r$
	Answered by peter frank last updated on 24/Nov/19

	$k y = k x + x + 7$ ${k y}^{^{'}} = k + 1$ $y^{^{'}} = 2$ $2 k = k + 1$ $k = 1$ $y - i n t e r c e p t x = 0$ $k y = 7$ $y = 7$
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