How do you write $y - 7 = 4 (x + 4)$ $y - 7 = 4(x + 4)$ in standard form?

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How do you write $y - 7 = 4 (x + 4)$ $y - 7 = 4(x + 4)$ in standard form?

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Answer 1 · 2018-05-29T11:52:24

$4 x - y = - 23$ $4x-y=-23$

Explanation:

$y - 7 = 4 (x + 4)$ $y-7=4(x+4)$

This is in the format of $y - y_{1} = m (x - x_{1})$ $y-y_1= m(x-x_1)$ which is known as the point-slope form. To change this to standard form ( $A x + B y = C$ $Ax+By=C$ ), you need to find the values of the points $(x_{1}, y_{1})$ $(x_1,y_1)$ and the slope ( $m$ $m$ ):

$y - 7 = 4 (x + 4)$ $y-7=4(x+4)$
$y - y_{1} = m (x - x_{1})$ $y-y_1= m(x-x_1)$

Therefore,

$(x_{1}, y_{1}) = (- 4, 7)$ $(x_1,y_1) = (-4, 7)$ and $m = 4$ $m=4$

$y - 7 = 4 (x + 4)$ $y-7=4(x+4)$

Now simplify and rearrange to get $x$ $x$ and $y$ $y$ terms on one side and constants on the other side of equal sign:

$y - 7 = 4 (x + 4)$ $y-7=4(x+4)$
$y - 7 = 4 x + 16$ $y-7=4x+16$
$y = 4 x + 16 + 7$ $y=4x+16+7$
$y = 4 x + 23$ $y=4x+23$
$- 4 x + y = 23$ $-4x+y=23$

This is almost in the format of $A x + B y = C$ $Ax+By=C$ but $A x$ $Ax$ cannot have a negative sign in front of it. To get rid of this negative, divide both sides of the equation by $- 1$ $-1$ :

$\frac{- 4 x + y}{- 1} = \frac{23}{- 1}$ $(-4x+y)/-1=23/-1$
$4 x - y = - 23$ $4x-y=-23$

Now the equation is in standard form because $A = 4$ $A=4$ , $B = - 1$ $B=-1$ and $C = - 23$ $C=-23$ .

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How do you write $y - 7 = 4 (x + 4)$ $y - 7 = 4(x + 4)$ in standard form?

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How do you write y - 7 = 4(x + 4)y−7=4(x+4)y - 7 = 4(x + 4) in standard form?

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How do you write $y - 7 = 4 (x + 4)$ $y - 7 = 4(x + 4)$ in standard form?