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

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

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Answer 1 · 2018-04-22T04:27:35

$3 x + 4 y = - 11$ $3x+4y=-11$

Explanation:

Given:

$y + 2 = - \frac{3}{4} (x + 1)$ $y+2=-3/4(x+1)$

The standard form for a linear equation is:

$A x + B y = C$ $Ax+By=C$

To convert $y + 2 = - \frac{3}{4} (x + 1)$ $y+2=-3/4(x+1)$ into standard form, first distribute the slope $- \frac{3 x}{4}$ $-(3x)/4$ .

$y + 2 = - \frac{3 x}{4} - \frac{3}{4}$ $y+2=-(3x)/4-3/4$

Subtract $2$ $2$ from both sides.

$y = - \frac{3 x}{4} - \frac{3}{4} - 2$ $y=-(3x)/4-3/4-2$

Multiply both sides by $4$ $4$ .

$color(red)4xxy=color(black)cancel(color(red)(4))^1xx-(3x)/color(red)cancel(color(black)(4))^1-color(black)cancel(color(red)(4))^1xx3/color(red)cancel(color(black)(4))^1-4xx2$

Simplify.

$4y=-3x-3-8$

$4y=-3x-11$

Add $3x$ to both sides.

$3x+4y=-11$

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

1 Answer

Explanation:

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