How do you write $y - \frac{2}{3} = - 2 (x - \frac{1}{4})$ $y-2/3 = -2( x-1/4 )$ in standard form?

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How do you write $y - \frac{2}{3} = - 2 (x - \frac{1}{4})$ $y-2/3 = -2( x-1/4 )$ in standard form?

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Answer 1 · 2015-09-18T21:17:21

$y = - 2 x + \frac{7}{6}$ $y = -2x + frac(7)(6)$

Explanation:

$y - \frac{2}{3} = - 2 (x - \frac{1}{4})$ $y-frac(2)(3) = -2(x-frac(1)(4))$
so, after opening the parentheses, we get
$y - \frac{2}{3} = - 2 x + \frac{2}{4}$ $y-frac(2)(3) = -2x + frac(2)(4)$
which simplifies to
$y - \frac{2}{3} = - 2 x + \frac{1}{2}$ $y-frac(2)(3) = -2x + frac(1)(2)$
and moving $- \frac{2}{3}$ $-frac(2)(3)$ on the other side of the equality
$y = - 2 x + \frac{1}{2} + \frac{2}{3}$ $y = -2x + frac(1)(2)+frac(2)(3)$
then performing the fraction calculation
$y = - 2 x + \frac{3}{6} + \frac{4}{6}$ $y = -2x + frac(3)(6)+frac(4)(6)$
which gives
$y = - 2 x + \frac{7}{6}$ $y = -2x + frac(7)(6)$

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How do you write $y - \frac{2}{3} = - 2 (x - \frac{1}{4})$ $y-2/3 = -2( x-1/4 )$ in standard form?

1 Answer

Explanation:

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How do you write $y - \frac{2}{3} = - 2 (x - \frac{1}{4})$ $y-2/3 = -2( x-1/4 )$ in standard form?